Sponsor Area
direction cosines are 0, 
Can a directed line have direction angles 45°, 60°, 120°?
A line makes an angle of 
with each of y-axis and z-axis. Find the angle made by it with x-axis.
direction cosines of the line are 
The x-axis makes angles 0°, 90° and 90° with x, y and z-axis.
∴ direction cosines of x-axis are cos 0°, cos 90°, cos 90° i.e. 1, 0. 0
Again y-axis makes angles 90°, 0°, 90° with x, y and z-axis.
∴ direction cosines of y-axes are cos 90°, cos 0°, cos 90° i.e. 0, 1,0.
Also z-axis makes angles 90°, 90°, 0° with x, y and z-axis.
∴ direction cosines of z-axis and cos 90°, cos 90°, cos 0° i.e. 0, 0, 1.
required direction-cosines are 
If a line has direction ratios 2, -1, -2, determine its direction cosines.
the direction cosines of line are 
If a line has direction ratios 2, -1, -2, determine its direction cosines.
the direction-cosines of the line are 
the direction cosines of the line are 
we get the direction cosines of the line as 
we get, the direction cosines of the line as
The direction ratios of the line joining the points (–2, 4, –5) and (1. 2, 3) are
1 + 2, 2 – 4, 3 + 5 i.e. 3, – 2, 8.
Dividing each by 
we get the direction cosines of the line as 
Find the direction cosines of the line joining the points (2, 1, 2) and (4, 2, 0).
the direction-cosines of the line are 
A line makes angles of 45° and 60° with the positive axes of x and y respectively. What angle does it make with the positive axis of z?
direction - cosines of the line are 
∵ the line makes angles α, β, γ with the axes
∴ cos2 α + cos2 β + cos2 γ = 1
∴ 1 – sin2 α + 1 – sin2 β + 1 – sin2 γ = 1 or – sin2 α – sin2 β – sin2 γ = –2
∴ sin2 α + sin2 β + sin2 γ = 2.
Show that the points (2, 3, 4), (–1, –2, 1), (5, 8, 7) are collinear.
Sponsor Area
Let P(– 1, 2, – 3), Q(4, 5, 1) , R(9, 8, 5) be given points.
The direction ratios of PQ are 4 + 1, 5 – 2, 1 + 3 i.e. 5, 3, 4,
The direction ratios of PR are 9 + 1, 8 – 2, 5 + 3, i.e. 10, 6, 8.
Since 
∴ lines PQ and PR are parallel.
But P is a common point on both lines.
∴ points P, Q, R are collinear.
The given points are A(2, 3, – 4), B(1, –2, 3), C(3, 8, –11)
The direction ratios of the AB are 1 – 2, – 2 – 3, 3 + 4 i.e. – 1 – 5,1
The direction ratios of AC are 3 – 2, 8 – 3, –11 + 4 i.e. 1, 5, – 7
∴ direction ratios of AB and AC are proportional
∴ AB is parallel to AC.
But A is a common point on both the lines.
∴ A, B, C are collinear points.
Find the length of the projection of the line segment joining the points P (3, -1, 2) and Q(2, 4, -1) on the line with direction ratios -1, 2, -2.
projection of PQ on AB
Find the projection of the line segment joining the points (1, 2, 3), (4, 3, 1) on the line with direction ratios 3, –6, –2.
P, Q are the points (2, –3, 0), (0, 4, 5) and direction cosines of line AB are 
projection of PQ on AB = 
A directed line segment makes angles 45° and 60° with x-axis and y-axis and an acute angle with z-axis. If P (– 1, 2, – 3) and Q (4, 3, 1) are two points in space, find the projection of PQ on the given line.
direction cosines of line AB are 
projection of PQ on AB = (4+1). 
direction cosines of PQ are 
projection of AB on PQ = 
its direction cosines are 
.
projection of AB on PQ = (4 - 2). 
+ (2+1). 
If P, Q, R, S are the points (– 2, 3, 4), (– 4, 4, 6), (4, 3, 5), (0, 1, 2), prove by projection that PQ is perpendicular to RS.
The given points are P (– 2, 3, 4), Q (– 4, 4, 6), R (4, 3, 5) and S (0, 1, 2).
Direction ratios of RS are. 0 – 4, 1 – 3, 2 – 5 i.e., – 4, – 2, – 3
∴ direction-cosines of RS are
projection of PQ on RS
∴ PQ is perpendicular to RS.
[∵ projection of a line perpendicular to it is zero]
Let P (1, –1, 2), Q (3, 4, –2), R (0, 3, 2) and S (3, 5, 6) be given points.
Direction ratios of RS are 3 - 0, 5 - 3, 6 - 2 i.e. 3, 2, 4.
∴ direction cosines of RS are
i.e., 
Projection of PQ on RS
∴ PQ is perpendicular to RS
Hence the result.
The given points are A (1, – 1, 0), B (2, 1, – 1), C (– 3, 2, 2) and D (0, – 2, – 1).
Direction ratios of CD are 0 + 3, – 2 – 2, – 1 – 2 i.e.. 3, – 4, 3
∴ Direction ratios of CD are 0 + 3, – 2 – 2, – 1 – 2 i.e.. 3, – 4, 3
∴ direction cosines of CD are
(in magnitude)
Let l, m, n be the direction-cosines of the line and let the length of the line segment be r.
∵ projections of the line segment on the axis are 12, 4, 3,
∴ l r = 12, m r = 4, n r = 3
Squaring and adding, we get,
(l2 + m2 + n2) r2 = 144 + 16 + 9
∴ (1) (r)2 = 169, ∴ r = 13
∴ length of the line = 13 units
and direction-cosines of line are 
The projections of a directed line segment on the co-ordinate axes are 6, -3, 2. Find its length and direction cosines.
Let l, m, n be the direction cosines of the line and let the length of the line segment be r.
∵ projection of the line segment on the axes are 6, - 3, 2
∴ l r = 6, m r = – 3, n r = 2
Squaring and adding, we get,
(l2 + m2 + n2) r2 = 36 + 9 + 4
∴ r2 = 49 ⇒ r = 7
length of line = 7 units.
and direction cosines of line are 
Let l, m, n be the direction cosines of the line PQ and let the length of the line segment be r.
∵ projections of the line segment on the axes are 6, 2, 3.
∴ l r = 6, m r = 2, n r = 3
Squaring and adding. we get,
(l2 + m2 + n2) r2 = 36 + 4 + 9
∵ r2 = 49 ⇒ r = 7
Find a vector equation for the line.
be the position vector of any point on the given line
The Cartesian equation of a line is 
Find the vector equation for the line.
be the position vector of any point on the given line. 
Find the cartesian equation of the line which passes through the point (-2, 4, -5) and is parallel to the line given by 
Direction-ratios of the line 
∴ equation of the line through (–2, 4, –5) and having direction ratios 3, 5, 6 are
Find the equations of a line which is parallel to the line 
and which passes through the point (3, 0, 5).
For the cartesian and vector equation of a line which passes through the point (1, 2, 3) and is parallel to the line 
Find the equation of line (vector and cartesian both) which is parallel to the vector 
and which passes through the point (5, -2, 4).
and parallel to the vector 
is 
...(1)
and 
...(2)
is the position vector of (x, y, z).
Here 
The equation of line is
or 
∴ direction ratios of the line are 1, 2, 3.
∴ line is parallel to the vector
∴ vector equation of line is
Sponsor Area
The equation of a line is given by 
. Write the direction cosines of a line parallel to the above line.
. Find the direction cosines of a line parallel to this line.
direction cosines are 
.
and has direction ratios 2, 1, -6.
vector equation of line is
be the position vector of any point on the line.
be the position vector of any point on the line.
the equation of line is
we get,
Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector 
be the position vector of any point on the line.
and passes through (1, 2, 3).
and parallel to the vector 
is 
...(1)
and passes through (2, 3, 1).
and parallel to the vector 
and passes through (1, ,1, 1).
...(1) where X is a parameter
at a distance 
from the point (1, 2, 3).
required point is 
are position vectors of A, B respectively.
be the position vector of any point on the given line. 
we get,
are position vectors of A and B respectively.
be the position vector of any point on the given line. 
where λ is a parameter.
are position vectors of A and B respectively.
be the position vector of any point on the given line. 
are position vectors of A, B respectively.
be the position vector of any point on the given line.
the equation of line is
are position vectors of A, B respectively.
be the position vector of any point on the given line. 
the equation of line is
Here, 
The vector equation of line is
or 
...(1)
are collinear. 
are collinear.
is 
will lie on it
for 
Show that the point whose position vectors are given by 
are collinear.
and 
are collinear.
are position vectors of A, B respectively.
be the position vector of any point on the given line.
...(1)
...(2)
...(3)
...(4)
point is 
...(1)
and parallel to the line joining the points with position vectors 
and 
. Also, find the cartesian equivalent of this equations.
Let 
be the position vectors of A, B, and C respectively.
Now, 
The vector equation of line through A 
and parallel to vector 
is
or 
Now taking 
the equation of line is
or 
Comparing the coefficients of x, y, z, we get,
which is cartesian equation of line.
be the position vector of any point on the line. 
Direction-ratios of AB are 2 – 4, 3 – 5, 4 – 10 i.e.. – 2, – 2, – 6 i.e., 1, 1, 3 and AB passes through A (4, 5, 10)
∴ its vector equation is
Also Cartesian equations of AB are
Direction ratios of BC are 1– 2, 2 – 3, – 1, – 4
i.e. – 1, – 1, 5 i.e. 1, 1, 5 and BC passes through B (2, 3, 4)
∴ its vector equation is
Also cartesian equations of BC are
Let (α, β, γ) be coordinates of D.
Now mid-points of AC is same as that of BD.
and is in direction 
Find equations for the line in vector and in cartesian form .
Here 
Let 
be the position vector of any point P (x, y, z) on the line
∴ the vector equation of line is 
or 
This equation can be written as
Comparing coefficients of 
we get,
Equationing values of λ, we get, 
which are cartesian equations of the line.
and parallel to the line joining the points with position vectors 
and 
Also, find the cartesian equivalent of this equation.
Let 
be the position vector of A and 
be position vectors of B and C.
The vector equation of line through 
and parallel to vector 
is
or 
Now taking 
, the equation of line is
or 
Comparing the coefficients of 
we get, x = λ + 1, y = 2 λ – 2, z = – (2 λ + 3)
or 
which is cartesian equation of line.
Let 
be vectors parallel to the two given lines.
and 
Let θ be angle between given lines.
∴ θ is angle between 
be vectors parallel to these lines
Sponsor Area
are prependicular to each other. 
are prependicular, then find the value of k.
Show that the lines x = ay + b, z = cy + d and x = a' y + b' , z = c' y + d' are perpendicular to each other, if aa' + cc' + 1 = 0.
Given points are A (1, 2, 3), B (4, 5, 7), C (– 4, 3, –6) and D (2, 9, 2).
Direction ratios of AB are 4 – 1, 5 – 2, 7 – 3 i.e. 3, 3, 4
Direction ratios of CD are 2 + 4, 9 – 3, 2 + 6 i.e. 6, 6, 8 i.e. 3, 3, 4
Since direction ratios of AB and CD are proportional
∴ AB is parallel to CD.
∴ angle between them is 0°.
...(1)
...(2)
...(3)
∴ from (3), the equations of line are
Let a, b, c be the direction ratios of the line passing through the point (1, 2, – 4)
∴ equation of line is
...(1)
Since this line is perpendicular to the lines
and 3 a + 8 b – 5 c = 0
Solving these equations, we get,
∴ from (1), the equations of line is
∴ line passes through the point (1, 2, – 4) with position vector 
is parallel to 
.
∴ equation of line is
or 
...(1)
...(2)
...(1)
...(2)
...(4)
...(5)
...(6)
Direction ratios of AD are 4 k + 1 – 1, 4 – 2, – 2 k + 6 – 1
i.e.. 4 k , 2, – 2 k + 5
Direction ratios of BC are 4, 0, – 2
Since AD is perpendicular to BC
⊥ distance of A from BC = distance AD
Let D be foot of perpendicular form A on BC.
Direction ratios of AD are
k – 1, – k – 1 – 8, – 2 k + 3 – 4
i.e. k – 1, – k – 9, –2 k – 1
Direction ratios of BC are
2 – 0, – 3 + 1, – 1 – 3 i.e. 2, –2, – 4
i.e. 1, –1, –2.
Since AD is perpendicular to BC
∴ (k – 1) (1) + (– k – 9) (–1) + (–2 k – 1) (– 2) = 0
Let the given line AB be
Any point M on this line is (5 r – 3, 2 r + 1, 3 r – 4)
Let this point M be the first of perpendicular from P (0, 2, 3) on AB.
Direction-ratios of PM are
5 r – 3 – 0, 2 r + 1 – 2, 3 r – 4 – 3
i.e., 5 r – 3, 2 r – 1, 3 r – 7
Direction -ratios of AB are 5, 2, 3.
Since PM ⊥ AB
∴ (5 r – 3) (5) + (2 r – 1) (2) + (3 r – 7) (3) = 0
∴ 25 r – 15 + 4 r – 2 + 9 r – 21 = 0
∴ 38 r – 38 = 0 ⇒ r – 1 = 0 ⇒ r = 1
∴ M is (5 – 3, 2 + 1, 3 – 4) i.e.. (2, 3, – 1),
which is required foot of perpendicular.
M is 
Find the length and foot of the perpendicular drawn from the point (3, 4, 5) on the line 
Any point M on this line is
(2 r + 2, 5 r + 3, 3 r + 1)
Let this point M be the foot of perpendicular from P (3, 4, 5) on AB
Direction-ratios of PM are
2 r + 2 – 3, 5 r + 3 – 4, 3 r + 1 – 5
i.e. 2 r – 1, 5 r – 1, 3 – 4
Direction-ratios of AB are 2, 5, 3
Since PM ⊥ AB
Required length of perpendicular = PM
Find the perpendicular distance of the point (1, 0, 0) form the line 
Any point M on this line is (2 r + 1, – 3 r – 1, 8 r – 10)
Let this point M be the foot of perpendicular form P( 1,0, 0) on AB.
Direction ratios of PM are
2 r + 1 – 1, –3 r – 1 – 0,
8 r – 10 – 0 i.e. 2 r, – 3 r – 1, 8 r – 10
Direction-ratios of AB are 2, –3, 8
Since PM ⊥ AB
∴ (2 r) (2) + (– 3r – 1) (–3) + (8 r – 10) (8) = 0
∴ 4 r + 9 r + 3 + 64 r – 80 = 0
∴ 77 r = 77 ⇒ r = 1
∴ M is (3, –4, –2)
Required distance = PM = 
Find the length of the perpendicular drawn from the point (1, ,2 3) on the line
Any point M on this line is
(3 r + 6, 2 r + 7, –2 r + 7)
Let this point M be the foot of perpendicular from P(1, 2, 3) on AB
Direction ratios of PM are
3 r + 6 – 1, 2 r + 7 – 2, – 2 r + 7 – 3 i.e. 3 r + 5, 2 r + 5, – 2 r + 4
Direction ratios of AB are 3, 2, –2
Since PM ⊥ AB
∴ (3 r + 5) (3) + (2 r + 5) (2) + (– 2 r + 4) (–2) = 0
∴ 9 r + 15 + 4 r + 10 + 4 r – 8 = 0
⇒ 17 r = –17 ⇒ r = – 1
∴ M is (3, 5, 9)
∴ required length of perpendicular = PM
Direction ratios of PM are
10 r + 11 – 2, – 4 r – 2 + 1, – 11 r – 8 – 5 i.e. 10 r + 9, – 4 r – 1, – 11 r – 13
Direction ratios of AB are 10, – 4, – 11.
Since PM ⊥ AB
∴ (10) (10 r + 9) + (–4) (–4 r – 1) + (–11) (– 11 r – 13) = 0
∴ 100 r + 90 + 16 r + 4 + 121 r + 143 = 0
∴ 237 r + 237 = 0 or 237 r = – 237 ⇒ r = – 1
∴ M is (– 10 + 11, 4 – 2, 11 – 8) i.e. (1, 2, 3)
which is foot of perpendicular.
Length of perpendicular = PM
Direction ratios of PM are
r – 5 – 2, 4 r – 3 – 4, – 9 r + 6 + 1 i.e. r – 7, 4 r – 7, – 9 r + 7
Direction ratios of AB are 1, 4, – 9
Since PM ⊥ AB
∴ (1) (r – 7) + (4) (4 r – 7) + (– 9) (–9 r + 7) = 0
∴ r – 7 + 16 r – 28 + 81 r – 63 = 0
∴ 98 r = 98 ⇒ r = 1
∴ Mis (1 – 5, 4 – 3, – 9 + 6) i.e. (– 4, 1, –3)
The equation of PM is
i.e., 
i.e., 
Any point M on this line is
(5 r + 4, 3 r + 2, 4 r + 3)
Let this point M be the foot of perpendicular from P (3, 2, 1) on AB.
Direction-ratios of PM are
5 r + 4 – 3, 3 r + 2 – 2, 4 r + 3 – 1 i.e. 5 r + 1, 3 r, 4 r + 2
Direction-ratios of AB are 5, 3, 4
Since PM ⊥ AB
∴ 5 (5 r + 1) + 3 (3 r) + 4 (4 + 2) = 0.
∴ 25 r + 5 + 9 r + 16 r + 8 = 0 ⇒ 50 r = – 13
∴ required length of perpendicular = PM
Any point D on it is
(4 k + 1, 4, – 2 k + 6)
Let D be foot of perpendicular from A on BC.
Direction ratios of AD are
4 k + 1–1, 4 – 2, – 2 k + 6 – 1 i.e., 4 k , 2, – 2 k + 5
Direction ratios of BC are 4, 0, – 2
Since AD is perpendicular to BC
∴ (4 k) (4) + (2) (0) + (– 2 k + 5) (–2) = 0
From P (1, 6, 3) , draw PM ⊥ AB and produce it to P' (α, β γ) such that M is mid-point of PP'. Then P' is image of P in line AB.
Any point M as line AB is
(r. 2 r + 1, 3 r + 2)
Direction ratios of AB are 1, 2, 3
Direction ratios of PM are r – 1 , 2 r + 1 – 6 , 3 r + 2 – 3
i.e. r – 1, 2 r – 5, 3 r – 1
∵ PM ⊥ AB∴ (1) (r – 1) + (2) (2 r – 5) + (3) (3 r – 1) = 0
∴ i + 4 r – 10 + 9 r – 3 = 0
∴ 14 r = 14 ⇒ r = 1
∴ M is (1, 3, 5)
Now M is mid-point of PP'
Direction-ratios of AB are 4 – 1, – 3 + 1, 1 + 3 i.e.. 3, – 2, 4 respectively.
Direction-ratios of AC are 3 – 1, –1 + 1, 2 + 3 i.e., 2, 0, 5 respectively.
Area of 
Find the area of the triangle whose vertices are (1, 2, 4), (-2, 1, 2), (2, 4, -3).
Tips: -
Note on parallelopiped and cube
(i) A parallelopipcd is a solid bounded by three pairs of parallel plane faces.
(ii) A rectangular parallelopiped is parallelopiped whose faces are all rectangles.
(iii) A cube is a parallelopiped whose faces are all squares.
Let OA = OB = OC = a, then the co-ordinates of O , A , B , C are (0, 0, 0), (a, 0, 0), (0, a, 0), (0, 0, a) respectively ; those of P, L, M, N are (a, a, a), (0, a, a), (a, 0, a), (a, a, 0) respectively.
The four diagonals are OP, AL, BM, CN. Direction cosines of OP are proportional to a – 0, a – 0, a – 0, i.e., a, a, a, i.e., 1,1,1.
Direction-cosines of AL are proportional to 0 – a, a – 0, a – 0 i.e., –a, a, a, i.e., – 1, 1, 1.
Direction-cosines of BM are proportional to a – 0, 0 – a, a – 0, i.e., a – a, a i.e., 1, – 1, 1.
Direction-cosines of CN are proportional to a – 0, a – 0, 0 – a i.e., a, a,– a i.e., 1, 1, – 1.
direction -cosines of OP are 
Directon-cosines of AL are 
Direction-cosines of BM are 
Direction-cosines of CN are 
Let l, m, n be direction-cosines of the line
∴ the line makes an angle α with OP.
or 
...(1)
Similarly 
...(2)
...(3)
Squaring and adding (1), (2), (3) and (4), we get,
cos2 α + cos2 β + cos2 γ + cos2 δ
Let ΔA = a, OB = b, OC = c, then the co-ordinates of O, A, B, C are (0, 0, 0), (a, 0, 0), (0, b, 0), (0, 0, c) respectively.
The co-ordinates of other points are shown in the figure.
The four diagonals are OP, AL, BM, CN
Direction-cosines of OP are a – 0, b – 0, c – 0 i.e., a, b. c respectively
Direction-cosines of AL are 0 – a, b – 0, c – 0 i.e., – a, b. c respectively
Direction-cosines of BM are a – 0, 0 – b, c – 0 i.e., a,– b, c respectively
Direction-cosines of CN are a – 0, b – 0, 0 – c i.e., a, b, – c respectively.
Let θ be the angle between OP and AL.
angle between OP and AL = 
Similarly angle between OP and BM = 
and angle between OP and CN = 
Proceeding in this way, we see that angles between four diagonals are given by 
Let OA = OB = OC = a, then the co-ordinates of O, A, B, C are (0, 0, 0), (a, 0, 0), (0, a, 0), (0, 0, a) respectively ; those of P, L, M, N are (a, a, a), (0, a, a), (a, 0, a), (a, a, 0) respectively.
The four diagonals are OP, AL, BM, CN. Direction cosines of OP are proportional to a – 0, a – 0, a – 0, i.e., a, a, a, i.e., 1, 1, 1.
Direction-cosines of AL are proportional to 0 – a, a – 0, a – 0 i.e., –a, a, a, i.e., – 1, 1, 1.
Direction-cosines of BM are proportional to a – 0, 0 – a, a – 0. i.e.. a – a, a i.e., 1, – 1, 1.
Direction-cosines of CN are proportional to a – 0, a – 0, 0 – a i.e., a, a, – a i.e., 1, 1, – 1.
Let θ be the angle between AL and BM
Similarly the angle between other two diagonals is also cos 
.
Direction-ratios of BC are
x3 – x2, y3 – y2, z3 – z2.
Directions-ratios of FE are
or 
or 
which are the same as that of BC
∴ FE || BC.
Also,
Hence the result.
Since < l, m, n > and < l + δl, m + δm, n + δn > are direction cosines of two lines
∴ l2 + m2 + n2 = 1 ...(1)
and (l + δl)2 + (m + δm)2 + (n2 + δn)2 = 1
or (l2 + m2 + n2) + 2 (l δl + m δm + n δn) + [(δl)2 + (δm)2 + (δn)2] = 1
or 1+2 (I δl + m δm + n δn) + [ (δl)2 + (δm)2 + (δn)2 ] = 1 [∵ of (1)]
or (δl)2 + (δm)2 + (δn)2 = – 2 (lδ l + m δm) + n δn) ....(2)
Now δθ is angle between two lines
∴ cos δθ = l (l + δl) + m (m + δm) + n (n + δn)
Find the angle between the two lines whose direction cosines are given by the equations:
l + m + n = 0, l2 + m2 – n2 = 0
The given equation are
l + m + n = 0 ...(1)
and l2 + m2 – n2 = 0 ....(2)
From (1), n = – (l + m) ....(3)
From (2) and (3), we get,
l2 + m2 – (l + m)2 = 0 or – 2 l m = 0
⇒ l m = 0
either l = 0
1. l + 0 .m + 0. n = 0
Also, l+m+n = 0
Solving,
or m = 0
0.l + l.m + 0.n = 0
Also, l + m + n = 0
Solving,
∴ direction ratios of the two lines are 0, – 1, 1 ; 1, 0, – 1
Let θ be the angle between the lines
∴ acute angle θ between the lines is given by
The given equation are
2 l – m + 2 n = 0 ...(1)
and m n + n l + l m = 0 ....(2)
From (1), m = 2 l + 2 n ....(3)
From (2) and (3), we get,
n (2 l + 2 n) + n l + l (2 l + 2 n) = 0
or 2 n l + 2 n2 + n l + 2 l2 + 2 n l = 0 or 2 l2 + 5 l n + 2 n2 = 0
⇒ (2 l + n) (l + 2 n) = 0
either 2l+ n = 0
i.e. 2l + 0 m + n = 0
Also, 2l - m + 2n = 0
Solving, we get,
or l + 2n = 0
i.e., l + 0 m + 2 n = 0
Also,
2l - m + 2n = 0
Solving, we get,
The given equations are
l + m + n = 0 ....(1)
and 2 l + 2 m – m n = 0 ...(2)
From (1), l = – (m + n) ...(3)
From (2) and (3), we get,
– 2 (m + n) + 2 m – m n = 0 or – 2 n – m n = 0
⇒ n (2 + m) = 0
Either n = 0
i.e., 0 l + 0 m + n = 0
Also,
l + m + n = 0
Solving, we get,
or 2 + m = 0 i.e., m = -2
from (1), l - 2 + n = 0
Now, l = 1, n = 1 satisfy it
Also, l = 1, m =-2, n = 1 satisfy (2)
we have
l = 1, m = -2, n = 1
∴ direction-ratios of two lines are 1, – 1, 0 and 1, –2, 1.
Let θ be the angle between the lines
...(3)
are the roots of the equation (3).
∴ l1 l2 = k (b w2 + c v2), m1 m2 = k (c u2 + a w2), n1 n2 = k (a v2 + b u2)
∴ l1 l2 + m1 m2 + n1 n2 = k [b w2'+ cv2 + c u2 + a w'2 + a v2 + b u2]
∴ lines are perpendicular
If k (b w2 + cv2 + c u2 + a w2 + a v2 + b u2] = 0 [∵ of (4)]
i.e., if b w2 + c v2 + c u2 + a w2 + a v2 + b u2 = 0
i.e., if u2 (b + c) + v2 (c + a) + w2 (a + b) = 0
(ii) The lines are parallel
if l1 = l2, m1 = m2, n1 = n2
i.e., if 
i.e., if equation (3) has equal roots
i.e., if disc = 0
i.e., if 4 c2 u2 v2 – 4 (a w2 + c u2) (b w2 + c v2) = 0
i.e., lf c2 u2 v2 – a bw4 – acv2 w2 –bcu2 W – c2 u2 v2= 0
i.e., if – a b w4 – a c v2 w2 – b c u2 w2 = 0
i.e., if a b w2 + a c v2 + b c u2 = 0 [Dividing by – w2]
i.e, if 
i.e., if 
The direction-cosines of the two lines are given by the equations
u l + v m + w n = 0 ....(1)
f m n + g n l + h l m = 0 ....(2)
From (1), w n = -(u l + v m), 
Putting this value of n in (2), we get,
∴ –f m (u l + v m)–g l(u l + v m) + h lm w = 0
∴ –u f l m – v f m2 – u g l2 – v g l m + h l m w = 0
∴ – u g l2 – (u f + v g – h w) Im – v f m2 = 0
or u g l2 + (u f + v g –h w )l m + v f m2 = 0
or 
...(3)
which is a quadratic is 
Let 
be its roots where l1 , m1 , n1 and l2, m2, n2 are the direction-cosines of the lines.
(i) The two lines will be perpendicular when
l1 l2 + m1 m2 + n1 n2 = 0 .....(4)
From (3), 
or 
(By symmetry)
Putting in (4), we see that lines are perpendicular
if 
i.e., if 
which is required condition.
(ii) The lines are parallel if l= l2, m1= m2, n1 = n2
i.e., if 
i.e., if the roots of (3) are equal
i.e., if disc, of (3) = 0
i.e., if (u f + v g – h w)2 – 4 (u g) (v f) = 0
i.e., if (u f + v g – h w)2 = 4 u v f g
i.e., if 
i.e., if 
i.e., if 
i.e., if 
i.e., if 
Find the shortest distance (S.D.) between the lines:
...(1)
...(2)
be point on line (2) with position vectors 
so that
be S.D. vector between given lines.
it is parallel to 
If 
is unit vector along 
, then
...(1)
...(2)
and T be point on line (2) with position vector 
so that
be the S.D. vector between given lines. 
is a unit vector along 
then
...(1)
...(2)
and T be the point on line (2) with position vector 
so that
be the S.D. vector between given lines.
is a unit vector along 
, then
Find the shortest distance between the lines:
and 
we get
...(1)
...(2)
and T be point on line (2) with position vector 
so that
be the S.D. vector between the given lines. Therefore, it is parallel to 
is a unit vector along 
,
The equations of two lines are
...(1)
and 
...(2)
Comparing these equations with
Let S be point on line (1) with position vector 
and T be point on line (2) with position vector 
so that
Now, 
Let 
be the S.D. vector between given lines. Therefore, it is parallel to 
If 
is a unit vector along 
, then
Now. S.D. = Projection of 
The equations of the two lines are
...(1)
and 
...(2)
Comparing these equations with
Let S be point on line (1) with position vector 
and T be point on line (2) with position vector 
so that
Now, 
Let 
be the S.D. vector between given lines. Therefore, it is parallel to 
If 
is a unit vector along 
, then
Now, S.D. = 
.
...(1)
...(2)
and T be point on line (2) with position vector 
so that
be the S.D. vector between given lines. Therefore, it is parallel to 
is a unit vector along 
, then 
= Projection of 
Sponsor Area
.
...(1)
...(2)
and T be the point on the line (2) with position vector 
so that
be the S.D. vector between given lines. Therefore, it is parallel to 
is a unit vector along 
then 
The equations of two lines are
and 
or 
Comparing these equations with
Now, 
S.D. = 
...(1)
...(2)
and T be point on line (2) with position vector 
so that
be the S.D. vector between given lines. Therefore, it is parallel to 
is a unit vector along 
, 
...(1)
...(2)
we get
and T be the point on the line (2) with position vector 
be the S.D. vector between given lines. Therefore, it is parallel to 
.
is a vector along 
then
= Projection of 
...(1)
...(2)
...(1)
...(2)
and line (2) passes through the point T with position vector 
Also both lines are parallel to the vector
and 
, we get,
...(1)
...(2)
and line (2) passes through the point T with position vector 
Also both lines are parallel to the vector 
...(1)
...(2)
and 
and T be point on limit (2) with position vector 
so that
be the S.D. vector between given lines. Therefore, it is parallel to 
is a unit vector along 
then
The equations of two lines are
...(1)
and 
...(2)
Comparing these equations with
Let S be points on line (1) with position vector 
and T be point on line (2) with position vector 
so that
Let 
be the S.D. vector between given lines. Therefore, it is parallel to 
If 
is a unit vector along 
, then
Now S.D. = Projection of 
on 
= Projection of 
on 
= 
∴ given lines interesect.
...(1)
...(2)
and T be point on line (2) with position vector 
so that
be the S.D. vector between given lines. Therefore, it is parallel to 
is a unit vector along 
, then
or 6x + 4y + 3z = 12The equation of plane is 2x + y – z = 5
or 
which is of intercept form 
where 
where 
Since the plane is parallel to plane ZOX i.e. y = 0
∴ its equation is of form
y = k ...(1)
This plane makes intercept 3 on y-axis i.e. this plane passes through (0, 3, 0)
∴ 3 = k
Putting k = 3 in (1), we get,
y = 3, which is required equation.
the direction-cosines of a normal to the given plane are 
The equation of plane is 3x – 4y + z + 5 = 0
or 3 x – 4 y + z = – 5
or – 3 x + 4 y – z = 5
Dividing both sides by 
which is of form lx + my + nz = p
where 
The direction-cosines of the normal to the plane are 
The equations of planes are
3x – 6y – 2z–7 = 0
and 2 x + y – k z – 5 = 0
Since two planes are perpendicular to each other.
∴ (3) (2) + (–6) (1) + (–2) (–k) = 0 [∵a1a2 + b1,b2 + c1c2 = 0]
∴ 6 – 6 + 2k = 0 ⇒ 2k = 0 ⇒ K = 0
The foot of the perpendicular drawn from the origin to the plane is (4, 3, 2). Find the equation of the plane. 
The equation of plane through M (4, 3, 2) is
a (x – 4) + b (Y–3) + c (z – 2) = 0 ...(1)
The direction-ratios of the line through the points O (0, 0. 0) and M (4, 3, 2) are
4 - 0, 3-0, 2-0 i.e. 4, 3, 2
∴ the line OM with direction-ratios 4, 3, 2 is normal to the plane (1)
∴ equation (1) of plane becomes
4 (X – 4) + 3 (y – 3) + 2 (z – 2) = 0 
or 4x – 16 + 3 Y – 9 + 2z – 4 = 0 or 4x + 3y + 2z – 29 = 0
The equation of plane through P (1, 2, –3) is
a (x – 1) + b (y – 2) +c (z +3) = 0 ...(1)
The direction ratios of the line through the points O (0, 0, 0) and P (1, 2, –3) are 1 - 0, 2 - 0, - 3 - 0.
i.e. ,1, 2,–3
∴ the line OP with direction ratios 1, 2, –3 is normal to the plane (1)
∴ equation (1) of plane becomes
1(x – 1) + 2 (y – 2) – 3 (z + 3) = 0
or x – 1 + 2y – 4-3z – 9 = 0 or x + 2y – 3z – 14 = 0.
The equation of plane through (–10, 5, 4) is
A(x + 10) + B (y – 5) + C (z – 4) = 0 ...(1)
The direction ratios of the line through the points (4, –1, 2) and (–10.5. 4) are –10 – 4, 5+1, 4-2. i.e.– 14, 6, 2 i.e. 7,–3,–1.
∵ the line with direction ratios 7, –3.–1 'is normal to the plane (1).
∴ equation (1) of plane becomes
or 
or 
which is the required equation of plane.
Here 
The vector equation of plane is 
or 
Take 
or 
or 
which is Cartesian equation of plane.
Here, 
The vector equation of plane is
or 
Take 
or 
which is the cartesian equation of the plane.
and perpendicular to the vector 
.
Here, 
This is cartesian equation of plane.
Find the equation of a plane which passes through (2, –3, 1) and is perpendicular to the line through the points (3, 4, –1) and (2,–1, 5).
The equation of plane through (2, –3, 1) is
A (x – 2) + B ( y + 3) + C (z – 1) = 0 ...(1)
The direction ratios of the line through the points (3, 4, –1) and (2, –1.5) are 2 – 3,–1– 4 ,5 + 1 i.e. -1,-5, 6 i.e. 1, 5,–6.
∴ the line with direction ratios 1, 5, –6 is normal to the plane (1)
∴ 1 (x – 2) + 5(y + 3)–6(z – 1) = 0 
or x – 2 + 5y + 15 – 6z + 6 = 0
or x + 5 y – 6 z + 19 = 0
which is required equation of plane.
The equation of plane is 
or 
or 
or 
which is cartesian equation of plane.
Find the angle between the planes
2x – y + z = 6 and x + y + 2z = 7
The equation of two planes are
2 x – y + z = 6
and x + y + 2 z = 7
Let θ be the angle between the planes
Equation of any plane parallel to the plane
2x – 3y + z + 9 = 0 is
2x – 3y + z + k = 0 ..(1)
∴ it passes through origin (0, 0, 0)
∴ 0 – 0 + 0 + k = 0 ⇒ k = 0
Putting k = 0 in (1), we get,
2x – 3y + z = 0
which is required equation of plane.
Equation of any plane parallel to the plane – 2x + y– 3 z = 0 is –2x + y –3 z = k .... (1)
∴ it passes through P (1,4, - 2)
∴ –2 + 4 + 6 = k ⇒ k = 8
Putting k = 8 in (1), we get, – 2x + y–3 z = 8 which is the required equation.
Find equation of the plane parallel to x + 3y – 2z + 7 = 0 and passing through the origin.
Equation of any plane parallel to the plane
x + 3y –2z + 7 = 0 is
x + 3y – 2z + k=0 ...(1)
∵ it passes through origin (0, 0, 0)
∴ 0 + 0– 0 + k = 0 ⇒ k = 0
Putting k = 0 in (1), we get,
x + 3y –2z = 0
which is required equation of plane.
Equation of any plane parallel to the plane
3x – 2y + z – 11 = 0 is
3x - 2y + z + k = 0 ...(1)
∴ it passes through origin (0, 0, 0)
0 - 0 + 0 + k = 0 ⇒ k = 0
Putting k = 0 in (1), we get,
3x –2y + z = 0
which is required equation of plane.
...(1)
satisfies (1)
The equation of any plane through (a, b, c) is
A (x – a) + B (y – b) + C (z – c) = 0 ...(1)
Consider the plane
∴ normal to the plane is the vector 
∴ normal to the plane has direction ratios 1, 1, 1.
∴ A = k, B = k, C = k
Putting values of A, B. C in (1), we get,
k (x – a) + k (y – b) + k (z – c) = 0
or x – a + y – b + z – c = 0
or x + y + z – a –b – c = 0
which is required equation of plane.
The equation of plane is
2x – 3y + 4z – 6 = 0 ...(1)
Direction ratios of the normal to the plane (1) are 2, –3, 4.
Dividing each by 
we get the direction cosines of the normal as 
∴ dividing (1) throughout by 
we get, 
This is of the form l x + m y + n z = p, where p is the distance of the plane from the origin.
distance of plane from the origin = 
.
Let (x1, y1, z1) be the coordinates of the foot of perpendicular drawn from origin 0(0, 0, 0) to the plane (1).
∴ direction ratios of OP are .x1– 0, y, –0, –0 i.e. x1, y1, z1
∴ Direction cosines of OP are
Since direction cosines and direction ratios of a line are proportional
Since P(x1, y1, z1) lies on plane (1)
Tips: -
If d is the distance from the origin and l, m, n are the direction cosines of the normal to the plane through the origin, then the foot of the perpendicular is (ld, md, nd).
we get the direction cosines of the normal as 
.
we get,
In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
z = 2
The equation of plane is
z = 2 or 0x + 0y + 1.z = 2
It is of form lx + my + nz = p
where l = 0, m = 0, n = 1, p = 2
∴ direction cosines of the normal to the plane are 0, 0, 1 and distance from origin = 2.
In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
x + y + z = 1
and distance from origin 
In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
2x + 3 y – z = 5
and distance from origin = 
In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
5y + 8 = 0
The equation of plane is
5y + 8 = 0 or 5y = – 8
or 0x – 5 y + 0z = 8
Dividing both sides by 
0x - y + 0 z = 
It is of the form lx + my + nz = p, where l = 0, m = -1, n = 0, 
∴ direction cosines of the normal to the plane are 0, – 1, 0 and distance from origin = 
In the following cases, find the coordinates of the foot of the perpendicular
drawn from the origin.
2x + 3y + 4z – 12 = 0
The equation of given plane is
2 x + 3 y + 4 z – 12 = 0 ...(1)
Dividing both sides by 
∴ direction cosines of the normal OP are 
where O is origin and P(x1,y1, z1) is foot of perpendicular.
Direction ratios of OP are x1 – 0, y1 – 0, – 0 i.e. x1, y1, z1.
Since direction cosines and direction ratios of a line are proportional.
Since P lies on plane (1)
The equation of plane is
0x + 3 y + 4 z = 6 ...(1)
Dividing both sides by 
It is of form lx + my + nz = p where l = 0, 
∴ direction cosines of the normal OP are 0, 
where O is origin and P (x1, y1, z1) is foot of perpendicular.
Direction ratios of OP are x1– 0, y1 – 0, z1 – 0 i.e. x1, y1, z1.
Since direction cosines and direction ratios of a line are proportional.
Since P lies on plane (1)
where O is origin and P (x1, y1, z1) is foot of perpendicular.
The equation of plane is
5 y + 8 = 0 or 5 y = – 8 or – 5 y = 8
Dividing both sides by 
...(1)
∴ direction ratios of the normal OP to the plane are 0, – 1, 0 where O is origin and P(x1, y1, z1) is foot of perpendicular.
Direction ratios of OP are x1– 0, y1 – 0, z1 – 0 i.e. x1, y1, z1.
Since direction cosines and direction ratios of a line are proportional.
Since P lies on plane (1)
...(1)
direction cosines of 
.
Let 
∴ required equation of plane is
or 
Here p = 7
and 
The vector equation of plane is
or 
from the origin and its normal vector from the origin is 
Also find its cartesian form.
Here 
and 
The required equation of plane is
or 
or 
Taking 
or 2x – 3 y + 4 z = 6, which is cartesian equation of plane.
...(1)
...(3)
Find also the distance of the plane from the origin.
which is perpendicular vector from the origin
are 
...(1)
...(1)
....(1)
we get,
which is the required equation. 
we get,
which is required vector equation of plane. Find the vector equation in scalar product form of the plane that contains the lines.
and 
...(1)
...(2)
.
as plane is perp. to line (2).
.
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#13 {main}</pre>](/application/zrc/images/qvar/MAEN12064762-3.png)
.Here, 
The vector equation of line is
or 
Take 
or 
or 
or 
or 
which is cartesian equation of plane.
.Here, 
The vector equation of plane is
or 
Take 
or 
or 
or 
or 
which is cartesian equation of plane.
Find also the point of intersection of this line and plane.
....(1)
and is parallel to the vector 
...(2)
The equations of given planes are
3 x – 6 + 2 z = 7
and 2 x + 2 y – 2 z = 5
∴ a1 = 3, b1 = – 6, c1 = 2
and a2 = 2, b2 = 2, c2 = – 2
Now,
The equations of two planes are
2x + y – 2 z = 5 and 3x – 6y –2 z = 7
These equations can be written as
and 
or 
Also, 
Let θ be the angle between the planes
The equations of given planes are
7x + 5y + 6z + 30 = 0 ...(1)
and 3x – y – 10z + 4 = 0 ...(2)
Direction ratios of normal to plane (1) are 7, 5, 6
Direction ratios of normal to plane (2) are 3, –1, –10.
Now 
∴ planes (1) and (2) are not parallel.
Again (7) (3) + (5) (– 1) + (6) (– 10) = 21 – 5 – 60 = – 44 ≠ 0
∴ planes (1) and (2) are not perpendicular to each other.
Let θ be angle between planes (1) and (2).
The equations of given planes are
2x + y + 3z – 2 = 0 ...(1)
and x – 2y + 5 = 0 ...(2)
Direction ratios of normal to plane (1) are 2, 1, 3
Direction ratios of normal to plane (2) are 1, – 2, 0.
Now, 
∴ planes (1) and (2) are not parallel.
Again (2) (1) + (1) (– 2) + (3) (0) = 2 – 2 + 0 = 0
∴ given planes (1) and (2) are perpendicular to each other.
The given planes are
2x – 2 y + 4 z + 5 = 0 ...(1)
and 3 x – 3 y + 6 z – 1 = 0 ...(2)
Direction ratios of normal to plane (1) are 2, –2, 4,
Direction ratios of normal to plane (2) are 3, – 3, 6.
Now, 
∴ given planes (1) and (2) are parallel.
The given planes are
2x – + 3z – 1 = 0 ...(1)
and 2x – y + 3z + 3 = 0 ...(2)
Direction ratios of normal to plane (1) are 2, –1, 3.
Direction ratios of normal to plane (2) are 2, – 1, 3.
Now, 
∴ given planes (1) and (2) are parallel.
The equations of given planes are
4x + 8 y + z – 8 = 0 ...(1)
and 0x + y + z – 4 = 0 ...(2)
Direction ratios of normal to plane (1) are 4, 8, 1.
Direction ratios of normal to plane (2) are 0, 1, 1.
Now, 
∴ given planes (1) and (2) are not parallel.
Again (4) (0) + (8) (1) + (1) (1) = 0 + 8 + 1 = 9 ≠ 0
∴ given planes (1) and (2) are not perpendicular to each other.
Let θ be angle between planes.
...(1)
The equation of any plane through the intersection of planes
x + y + z = 9 i.e., x + y + z – 9 = 0 and 2 x + 3 y + 4 z + 5 = 0 is
(x + y + z – 9) + k (2 x + 3 y + 4 z + 5) = 0 ...(1)
∴ it passes through the point (1, 1, 1)
∴ (1 + 1 + 1 – 9) + k (2 + 3 + 4 + 5) = 0
Putting this value of k in (1), we get,
or 
or 
or 
which is required equation of plane.
The equation of any plane through the intersection of planes
2x – 3 y + z – 9 = 0 and x – y + z – 4 = 0 is
(2 x – 3 y + z – 9) + k (x – y + z – 4) = 0 ...(1)
∴ it passes through origin (0, 0, 0,)
∴ (0 – 0 + 0 – 9) + k (0 – 0 + 0 – 4) = 0
Putting 
or 4 (2x – 3 y + z – 9) – 9 (x – y + z – 4) = 0
or 8 x – 12 y + 4 z – 36 – 9 x + 9 y – 9 z + 36 = 0
or – x – 3 y – 5 z = 0
or x + 3 y + 5 z = 0, which is required equation of plane.
Any plane passing through the intersection of planes
3x – y + 2 z – 4 = 0 and x + y + z – 2 = 0 is
(3x – y + 2 z – 4) + k (x + y + z – 2) = 0 ...(1)
∴ it passes through (2, 2, 1)
∴ (6 – 2 + 2 – 4) + k (2 + 2 + 1 – 2) = 0
Putting 
or 3 (3x – y + 2 z – 4) – 2 (x + y + z – 2) = 0
or 9x – 3 y + 6 z – 12 – 2x – 2 y – 2 z + 4 = 0
or 7x – 5 y + 4 z = 8, which is required equation of plane.
The equation of any plane through (– 1,–1, 2) is
a(x + 1) + b(y + 1) + c(z – 2) = 0 ...(1)
∴ it is perpendicular to the planes
2x + 3y – 3z = 2 and 5x – 4y + z = 6
∴ 2a + 3b – 3c = 0 ...(2)
and 5a – 4b + c = 0 ...(3)
Solving (2) and (3), we get,
∴ a = 9 k, 6 = 17 k, c = 23 k
Putting values of a, b, c in (1), we get
9 k (x + 1)+ 17 k (y + 1) + 23 k (z – 2) = 0
or 9 (x + 1) + 17 (+ 1) + 23 (z – 2) = 0
or 9 x + 9 + 17 y + 17 + 23 z – 46 = 0
or 9 x + 17 y + 23 z – 20 = 0
which is required equation of plane.
The equation of any plane through the intersection of planes
x + 2y + z – 3 = 0 and 2x – y – z – 5 = 0 is
(x + 2y + z – 3) + k (2x – y – z – 5) = 0 ...(1)
∴ it passes through (2, 1, 3)
∴ (2 + 2 + 3 – 3) + k (4 – 1– 3 – 5) = 0
Putting 
or 5 (x + 2 y + z – 3) + 4 (2 x – y – z – 3) = 0
or 5 x + 10 y + 5 z – 15 + 8 x – 4 y – 4 z – 12 = 0
or 13 x + 6 y + z – 27 = 0
which is equation of plane.
Direction ratios of normal to the plane are 13, 6, 1.
and the point (1, 1, 1).The equations of given planes are
...(1)
and 
...(2)
Any plane through the intersection of planes (1) and (2) is
or 
...(3)
It will pass through the point (1, 1, 1)
If 
Putting 
or 
or 
which is the required vector equation of the plane.
and the point (2, 1, 3).
...(1)
...(2)
in (3), we get,
and parallel to x-axis.
...(1)
...(2)
The equation of plane through the line of intersection of the planes
3x – 4y + 5z – 10 = 0 and 2x + 2y – 3z – 4 = 0 is
(3x – 4 y + 5 z – 10) + k (2x + 2y – 3z – 4) = 0 ...(1)
or (2k + 3) x + (2k – 4) y + (–3 k + 5) z – (4k + 10) = 0
Direction ratios of normal to the plane are
2k + 3, 2k – 4, –3k + 5
Consider the line x = 2y = 3z
or 
or 
Its direction ratios are 1, 
Since this line is parallel to plane (1).
∴ this line is perpendicular to the normal to the plane (1).
Putting 
or 3 (3x – 4y + 5z – 10) – 4 (2x + 2 y – 3z – 4) = 0
or 9x – 12y + 15z – 30 – 8x – 8y + 12z + 16 = 0
or x – 20y + 27z = 14
which is required equation of plane.
and which is perpendicular to the plane 
Any plane through the intersection of planes (1) and (2) is
(x + 2 y + 3 z – 4) + k (2 x + y – z + 5) = 0 ...(3)
i.e. (2 k + 1) x + (k + 2) y + (– k + 3) z + (5 k – 4) = 0
Direction ratios of the its normal are 2 k + 1, k + 2, – k + 3.
Again consider the plane
or 
or 
...(4)
Direction ratios of its normal are 5, 3, -6
Since plane (3) is perpendicular to plane (4)
∴ 5 (2 k + 1) + 3 (k + 2) + (–6) (– k + 3) = 0
∴ 10 k + 5 + 3 k + 6 + 6 k – 18 = 0
Putting 
or 19 (x + 2y + 3z – 4) + 7 (2 x + y – z + 5) = 0
or 19x+ 38y + 57z – 76 + 14x + 7y – 7z + 35 = 0
or 33x + 45y + 50z – 41 = 0
which is required equation of plane.
The equation of any plane through the line of intersection of he planes 2 x – y = 0 and 3 z – y = 0 or y – 3 z = 0 is
(2 x – y) + k (y – 3 z) = 0 ...(1)
or 2x – y + ky – 3 k z = 0
or 2x + (k – 1) y – 3 k z = 0
∴ this plane is perpendicular to the plane 4 x + 5 y – 3 z = 8.
∴ (2) (4) + (k – 1) (5) + (–3 k) (– 3) = 0 [∴ a1 a2 + b1 b2 + c1 c2 = 0]
∴ 8 + 5 k – 5 + 9 k = 0 or 14 k = –3
Putting this value of k in (1), we get 
or 14 (2 x – y) – 3 (y – 3 z) = 0
or 28 x – 14 y – 3 y + 9 z =0
or 28 x – 17 y + 9 z = 0
which is the required equation of the plane.
The equation of any plane through the line of intersection of the planes
x + y + z – 1 = 0 and 2x + 3 y + 4 z – 5 = 0 is
(x + y + z – 1) + k (2 x + 3 y + 4 z – 5) = 0 ...(1)
or (2 k + 1) x + (3 k + 1) y + (4 k + 1) z – (1 + 5 k) = 0
∴ this plane is perpendicular to the plane x – y + z = 0
∴ (2 k + 1) (1) + (3 k + 1) (– 1) + (4 k + 1) (1) = 0
∴ 2 k + 1 – 3 k – 1 + 4 k + 1= 0
Putting 
(x + y + z - 1) - 
or 3 (x + z –1) – (2 x + 3 y + 4 z – 5) = 0
or 3 x + 3 y + 3 z – 3 – 2 x – 3 y – 4 z + 5= 0
or x – z + 2 = 0
which is required equation of plane.
Any plane through the line of intersection of planes
x + 2 y + 3 z – 4 = 0
and 2 x + y – z + 5 = 0 is
(x + 2 y + 3 z – 4) + k (2 x + y – z + 5) = 0 ...(1)
i.e. (2 k + 1) x + (k + 2) y + (– k + 3) z + (5 k – 4) = 0
Direction ratios of its normal are 2 k + 1, k + 2, – k + 3
Again consider the plane
5 x + 3 y + 6 z + 8 = 0 ...(2)
Direction ratios of its normal are 5, 3, 6
Since plane (1) is perpendicular to plane (2)
∴ 5 (2 k + 1) + 3 (k + 2) + 6 (– k + 3) = 0
∴ 10 k + 5 + 3 k + 6 – 6 k + 18 = 0
Putting 
or 7 (x + 2 y + 3 z – 4) – 29 (2 x + y – z + 5) = 0
or 7 x + 14 y + 21 z – 28 – 58 x – 29 y + 29 z – 145 = 0
or – 51 x –15 y + 50 z – 173 = 0
or 51 x + 15 y – 50 z + 173 =0
which is required equation of plane.
Any plane through the line of intersection of planes
x + y – 2 z + 3 = 0 and 3 x – y – 2 z – 4 = 0 is
(x + y – 2 z + 3) + k (3 x – y – 2 z – 4) = 0 ...(1)
i.e. (3 k + 1) x + (– k + 1) y + (– 2 k – 2) z + (– 4 k + 3) = 0
Direction ratios of its normal are 3 k + 1, – k + 1, – 2 k – 2.
Again consider the plane
2 x + 3 y – z + 1 = 0 ...(2)
Direction ratios of its normal are 2, 3, – 1.
Since plane (1) is perpendicular to plane (2)
∴ (2) (3 k + 1) + (3) (– k + 1) + (– 1) (–2 k –2) = 0
∴ 6 k + 2 – 3 k + 3 + 2 k + 2 = 0
Putting 
or 5 (x + y – 2 z + 3) – 7 (3 x – y – 2 z – 4) = 0
or 5x + 5 y – 10 z + 15 – 21 x + 7 y + 14 z + 28 = 0
or – 16 x + 12 y + 4 z + 43 = 0
or 16 x – 12 y – 4z – 43 = 0
which is required equation of plane
The equation of plane passing through (0, – 1,–1) is
A (x – 0) + B(y + 1) + C(z + 1) = 0 ...(1)
∴ it passes through (4, 5, 1)
∴ A(4 – 0) + B(5 + 1) + C( 1 + 1) = 0
∴ 4A + 6B + 2C = 0 ⇒ 2A + 3B + C = 0 ...(2)
Again plane (1) passes through (3, 9, 4)
∴ A(3 – 0) + B(9 + 1) + C(4 + 1) = 0
∴ 3A + 10B + 5C = 0 ....(3)
Solving (2) and (3), we get,
Putting values of A, B, C in (1), we get,
5 k (x – 0) – 7 k (y + 1) + 11 k (z + 1) = 0
or 5 x – 7 y – 7 + 11 z + 11 = 0
or 5 x – 7 y + 11 z + 4 = 0
which is required equation of plane.
The equation of plane passing through (0, – 1, 0) is
A(x – 0) + B(y + 1) + C(z – 0) = 0 ...(1)
∴ it passes through (2, 1, – 1)
∴ A (2 – 0) + B(1 + 1) + C(– 1 – 0) = 0
∴ 2A + 2B – C = 0 ...(2)
Again plane (1) passes through (1, 1, 1)
∴ A( 1 – 0) + B(1 + 1) + C(1 – 0) = 0
∴ A + 2B + C = 0 ....(3)
Solving (2) and (3), we get,
Putting values of A, B, C in (1), we get,
4 k (x – 0) – 3 k (y + 1) + 2 k (z – 0) = 0
∴ 4 x – 3 y + 2 z = 0
∴ 4 x – 3 y + 2 z = 3
which is required equation of plane.
The equation of any plane through the points (0, – 1, 0) is
a (x – 0) + b (y + 1) + c (z – 0) = 0 ...(1)
∴ it passes through (1, 1, 1)
∴ a + 2 b + c = 0 ...(2)
Again plane passes through (3, 3, 0)
∴ 3 a + 4 b + 0 c = 0 ...(3)
Solve (2) and (3), we get,
Putting values of a, b,c in (1), we get,
4 k x – 3 k (y + 1) + 2 k z = 0
or 4 x – 3 y + 2 z = 3
which is required equation of plane.
The equation of any plane through (1, 1, 1) is
a (x – 1) + b (y – 1) + c (z – 1) = 0 ...(1)
∴ it passes through (1, –1, 1)
∴ a(1 – 1) + b (– 1 – 1) + c(1 – 1) = 0
∴ 0 a – 26 + 0 c = 0 ...(2)
Also (1) passes through (–7, –3, –5)
∴ a (–7–1) + b (–3–1) + c (–5 – 1) = 0
∴ – 8 a – 4 b – 6 c = 0
∴ 4 a + 2 b + 3 c = 0 ...(3)
From (2) and (3), we get,
∴ a = 3k, b = 0, c = – 4k
Putting values of a, b, c in (1), we get,
3k (x – 1) + 0 (y – 1) – 4k (z – 1) = 0
or 3 (x – 1) – 4(z – 1) = 0 or 3 x – 3 – 4z + 4 = 0
or 3x – 4 z + 1 = 0
which is required equation of plane.
Find the vector equation of the plane passing through the point A(2, 2, –1), B(3, 4, 2) and C (7, 0, 6). Also find the cartesian equation of the plane.
Here, 
The vector equations of plane is
or 
The Cartesian equation of plane is
or 
or (or – 2) (14 + 6) – (y – 2) (7 – 15) + (z + 1) (–2 – 10) = 0
or 20 (x – 2) + 8 (y – 2) – 12 (z + 1) = 0
or 5 (x – 2) + 2 (y – 2) – 3 (z + 1) = 0
or 5 x – 10 + 2y – 4 – 3 z – 3 = 0
or 5 x + 2 y – 3 z – 17 = 0
Here, 
where 
are position vectors of R, S, T respectively.
The vector equation of plane is
or 
The given points are (1,1, 0), (1, 2,1), (–2, 2, –1).
These points are not collinear.
The equation of plane passing through these points is
or 
Point P is (a, b, c)
PA ⊥ y z-plane and PB ⊥ z x-plane.
∴ A is (0, b, c) and B is (a, 0, c)
We are to find the equation of plane through (0, 0, 0), (0, b, c) and (a, 0, c).
The equation of plane through (0, 0, 0) is
λ (x – 0) + μ (y – 0) + v (z – 0) = 0
∴ λx + μ y + v z = 0 ...(1)
∴ it passes through (0, b, c) and (a, 0, c)
∴ 0 λ + b μ + c v = 0
and a λ + 0 μ + c v = 0
Solving these, we get,
Putting values of λ, μ,v in (1), we get,
k b c x + k c a y – k a b z = 0
or 
The equation of any plane through (2, 2, 1) is
a (x – 2) + b (y – 2) + c (z – 1) = 0 ...(1)
∴ it passes through (9, 3, 6)
∴ a (9 – 2) b (3 – 2) + c (6 – 1) = 0
∴ 7 a + b + 5 c = 0 ....(2)
Also plane (1) is perpendicular to the plane
2 x + 6 y + 6 z = 9
∴ a (2) + b (6) + c (6) = 0 [∴ a1 a2 + b1 b2 + c1 c2 = 0]
∴ 2 a + 6 b + 6 c = 0
or a + 3 b + 3 c = 0 ....(3)
From (2) and (3), we get,
Putting these values of a, b, c, in (1)
3 k (x – 2) + 4 k (y – 2) + (– 5 k) (z – 1) = 0
or 3 (x – 2) + 4 (y – 2) – 5 (z – 1) = 0
or 3 x – 6 + 4 y – 8 – 5 z + 5 = 0
or 3 x + 4 y – 5z – 9 = 0
which is the required equation of plane.
The equation of any plane through (2, 2, 1) is
a (x – 2) + b (y – 2) + c (z – 1) = 0 ....(1)
∴ it passes through (9, 3, 6)
∴ a (9 – 2) + b (3 – 2) + c (6 – 1) = 0
∴ 7 a + b + 5 c = 0 ...(2)
Also plane (1) is perpendicular to the plane 2 x + 6 y + 6 z = 9
∴ a (2) + b (6) + c (6) = 0 [∴ a1 a2 + b1 b2 + c1 c2 = 0]
∴ 2 a + 6 b + 6 c = 0
⇒ a + 3 b + 3 c = 0 ....(3)
From (2) and (3), we get, 
∴ a = 3 k, b = 4 k, c = 5 k
Putting these values of a , b, c, in (1),
3 k (x – 2) + 4 k (y – 2) + (– 5 k) (z – 1) – 0
or 3 (x – 2) + 4 (y – 2) – 5 (z – 1) = 0
or 3 x – 6 + 4 y – 8 – 5 z + 5 = 0
or 3 x + 4 y – 5 z – 9 = 0
which is the required equation of plane.
The equation of any plane through (1, – 1, 2) is
a (x – 1) + b (y + 1) + c (z – 2) = 0 ....(1)
∴ it passes through (2, – 2, 2)
∴ a (2 – 1) + b (– 2 + 1) + c (2 – 2) = 0
∴ a – b + 0 c = 0 ....(2)
Also plane (1) is perpendicular to the plane 6 x – 2 y + 2 c = 9
∴ 6 a – 2 b + 2 c = 0 ⇒ 3 a – b + c = 0 ....(3)
From (2) and (3), we get,
∴ a = k, b = k, c – 2 k
Putting these values of a, b, c in (1), we get,
k(x – 1) + k (y + 1) – 2 k (z – 2) = 0
or x – 1 + y + 1 – 2 z + 4 = 0
or x + y – 2 z + 4 = 0,
which is required equation of the plane.
The equation of plane through (2, 3, - 4) is
A (x – 2) + B (y – 3) + C (z + 4) = 0 ...(1)
∴ it passes through (1, –1, 3)
∴ A (1 – 2) + B (–1–3) + C (3 + 4) = 0
∴ – A– 4B + 7C = 0
∴ A + 4B – 7C = 0 ...(2)
Now plane (1) is parallel to x-axis.
∴ normal to the plane (1), with direction ratios A, B, C is perpendicular to x-axis with direction ratios 1, 0, 0.
∴ A (1) + B (0) + C (0) = 0
∴ A + 0B + 0C = 0 ...(3)
From (2) and (3), we get,
Putting values of A, B, C in (1), we get,
0 (x – 2) + 7k (y – 3) – 4k (z + 4) = 0
or 7 (y – 3) + 4 (z + 4) = 0
or 7 y – 21 + 4z + 16 = 0
or 7 y + 4 z – 5 = 0,
which is required equation of plane.
The equation of plane through (3, 4, 1) is
A(x – 3) + B(y – 4) + C(z – 1) = 0 ...(1)
Since it passes through (0, 1, 0)
∴ A(0 – 3) + B(1 – 4) + C(0 – 1) = 0
or –3A –3B –C = 0
∴ 3A + 3B + C = 0 ..... (2)
The equation of line is
Its direction ratios are 2, 7, 5
Since the line is parallel to plane (1) whose normal has direction ratios A, B, C.
∴ normal to plane (1) is perpendicular to line.
∴ 2A + 7B + 5C = 0 ...(3)
From (2) and (3), we get,
Putting these values of A, B, C in (1), we get,
8 k (x – 3) – 13 k (y – 4) + 15 k (z – 1) = 0
or 8 (x – 3) – 13(y – 4) + 15 (z – 1) = 0
or 8x – 24 – 13 y + 52 + 15 z – 15 = 0
or 8 x – 13 y + 15 z + 13 = 0,
which is required equation of plane.
The equation of any plane through (0, 0, 0) is
A (x – 0) + B (y – 0) + C (z – 0) = 0
or A x + B y + C z = 0 ...(1)
∴ it passes through (3, –1, 2)
∴ 3A – B + 2C = 0 ...(2)
Since plane (1) is parallel to the line 
∴ normal to the plane with direction ratios A, B, C is perpendicular to the line with direction ratios 1, – 4, 7.
∴ A (1) + B (– 4) + C(7) = 0 [∴ a1 a2 +b1 b2 +c1 c2 = 0]
∴ A – 4B + 7C = 0 ...(3)
Solving (2) and (3), we get,
Putting values of A, B, C in (1), we get,
k x – 19 k y – 11 k z = 0
or x – 19 y – 11 z = 0
which is required equation of plane.
The equation of any plane through (2, 1, 0) is
A(x – 2) + B(– 1) + C(z – 0) = 0 ...(1)
∴ it passes through (3, 2, 2)
∴ A(3 – 2) + B(2 – 1) + C(2 – 0) = 0
∴ A + B + 2C = 0 ...(2)
Since plane (1) is parallel to the line 
∴ normal to the plane with direction ratios A, B, C is perpendicular to the line with direction ratios 2, 3, 1.
∴ (2) (A) + (3) (B) + (1) (C) = 0
∴ 2A + 3B + C = 0 ...(3)
Solving (2) and (3), we get,
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∴ A = 5 k, B = – 3 k, C = – k
Putting these values of A, B, C in (1), we get,
5 k (x – 2) – 3 k (y – 1) – k (z – 0) = 0
or 5 (x – 2) – 3(y – 1) – z = 0
or 5 x – 10 – 3 y + 3 – z = 0
or 5 x – 3 y – z – 7 = 0
which is required equation of plane.
The equation of the plane through (1, 1, – 1) is
a (x – 1) + b (y – 1) + c (z + 1) = 0 ...(1)
∴ it is perp. to the planes
x + 2 y + 3 z = 7 and 2 x – 3 y + 4 z = 0
∴ a + 2 b + 3 c = 0 ...(2)
and 2 a – 3 b + 4 c = 0 ...(3)
solving (2) and (3), we get
∴ a = 17 k, b = 2 k, c = –7 k
Putting these values of a, b, c in (1), we get,
17 k (x – 1) + 2 k (y – 1) –7 k (z + 1) = 0
or 17 (x – 1) + 2(y – 1) –7(z + 1) = 0
or 17x – 17 + 2y – 2 – 7 z – 7 = 0
or 17 x + 2 y – 7 z – 26 = 0, which is required equation of plane.
The equation of any plane through (– 1, – 1, 2) is
a (x + 1) + b (y + 1) + c (z – 2) = 0 ....(1)
∴ it is perpendicular to the planes
3 x + 2 y – 3 z = 1 and 5 x – 4 y + z = 5
∴ 3 a + 2 b – 3 c = 0 ....(2)
and 5a – 4 b + c = 0 ...(3)
Solving (2) and (3), we get,
Putting values of a, b, c in (1), we get,
5 k (x + 1) + 9 k (y + 1) + 11 k (z – 2) = 0
or 5 (x + 1) + 9 (y + 1) + 11 (z – 2) = 0
∴ 5x + 9 y + 11 z – 8 = 0 which is required equation of plane.
The equation of any plane through (1, – 1, 2) is
a(x – 1) + b (y + 1) + c (z – 2) = 0 ...(1)
∴ it is perpendicular to the planes
2x + 3 y – 2 z = 5 and x + 2 y – 3 z = 8
∴ 2 a + 3 b – 2 c = 0 ...(2)
and a + 2 b – 3 c = 0 ...(3)
Solving (2) and (3), we get,
Putting values of a, b, c in (1), we get,
5 k (x – 1) – 4 k (y + 1) – k (z – 2) = 0
or 5 (x – 1) – 4 (y + 1) – 1 (z – 2) = 0
or 5 x – 5 – 4 y – 4 – z + 2 = 0
or 5 x – 4 y – z – 7, which is required equation of plane.
The equation of plane is
2x + y – 2z + 3 = 0 ...(1)
Direction ratios of the normal to the plane are 2, 1, – 2.
Let M be the foot of perpendicular from P(1, 2, 4) to the plane.
Now PM is a straight line which passes through P(1, 2, 4) and has direction ratios 2, 1,–2.
∴ its equations are
Any point M on line is. (2 r + 1, r + 2, – 2 r + 4)
∴ M lies on plane (1)
∴ 2 (2 r + 1) + (r + 2) – 2 (–2 r + 4) + 3 = 0
∴ 4 r + 2 + r + 2 + 4 r – 8 + 3 = 0
Length of perpendicular 
Find the coordinates of the image of the point (1, 3, 4) in the plane 2x – y +z + 3 = 0.
The equation of plane is
2x – y + z + 3 = 0 ...(1)
Direction ratios of normal to the plane are 2,–1, 1.
Let M be foot of perpendicular from P(l, 3, 4) to the plane.
Now PM is a straight line which passes through P( 1. 3. 4) and has direction ratios as 2, – 1, 1
∴ its equation is
Any point M on line is (2 r + 1, – r + 3, r + 4)
∴ M lies on plane (1)
∴ 2 (2 r + 1) – (– r + 3) + (r + 4) + 3 = 0
∴ 4r + 2 + r – 3 + r + 4 + 3 = 0.
∴ 6r = –6 ⇒ r = –1
∴ M is (– 2 + 1, 1 + 3, – 1 + 4) i.e. (–1,4, 3)
Let N (α, β,γ) be image of P in the plane (1) so that M is mid-point of PN.
Find the image of the point (1, 3, 4) in the plane x – y + z = 5.
From P (1, 3, 4) , draw PM ⊥ plane and produce it to P'such that M is mid-point of PP’ Then P' (α,β, γ) is image of P.
Direction ratios of PM are 1, – 1, 1
The equations of PM are
Any point on it is M (r + 1, – r + 3, r + 4)
∵ M lies on plane (1)
∴ (r + 1)–(–r + 3)+.(r + 4) = 5, ∴ r + 1 + r – 3 + r + 4 = 5
∴ 3r = 3 ⇒ r = 1
∴ M is (2, 2, 5)
Since M is mid-point of PP'
...(1)
...(2)
which is required locus.
.
...(1)
lies on the plane (1)
...(2)
...(3)
...(4)
Perpendicular
Parallel
passes through 
B.
Parallel
The equations of given planes are
2x – y + 4z = 5 and 5x –2.5y + 10z = 6
Now, 
∴ given planes are parallel
then find value of p.
Find the angle between the line
and the plane
Find the angle between the line 
and the plane 
Find the angle between the line 
and the plane 
and the plane 
Find the angle between the line 
and the plane x + 2y + 2 + 3 = 0
and the plane 2x – y + z= 1.
Find the angle between the line 
and the plane 2x + y – 3z + 4 = 0.
and the plane 10x + 2y – 11z = 3.
and the plane 3x + 4y + z + 5 = 0.
Find the angle between the line 
and the plane 
The equation of line is 
The equation of plane is 
Let θ be the angle between the line and plane
∴ line is parallel to the plane.
The equation of given planes are
2x – 3y + z + 3 = 0 .....(1)
and 4x –6y + 2z + 5 = 0 .....(2)
Let (x1.y1, z1) be any point on plane (1)
∴ 2x1 – 3y1 + z1 + 3 = 0
∴ 2x1 – 3y1 + z1 = –3 ...(3)
Length of perpendicular from (x1, b 1, b1) to plane (2) in
The equations of given planes are
2x – y + 3 z – 4 = 0 ... (1)
and 6 x – 3 y + 9 z + 13 = 0 ....(2)
Let (x1, y1 , z1 ) be any point on plane (1)
∴ 2 x1 –y1 + 3 z1 – 4 = 0 or 2x1– y, + 3 z1 =4 . (3)
Length of perpendicular from (x1, y1 , z1 ) to the plane (2) is
which is required distance between the planes.
and 
...(1)
...(2)
The equation of plane is
2x – 2y + 4 z + 5 = 0 ...(1)
Direction ratios of the normal to the plane are
2,–2, 4 i.e., 1,– 1, 2.
Let M be foot of perpendicular from P (1, 1, 2) to the plane.
Now PM is a straight line which passes through P (1. 1,2) and has direction ratios as 1, – 1, 2.
∴ its equations are
Any point M on line is (r + 1, – r + 1, 2 r + 2)
∵ M lies on plane (1)
∴ 2 (r + 1) – 2 (– r + 1) + 4 (2 r + 2) + 5 = 0.
∴ 2r + 2 + 2r – 2 + 8r + 8 + 5 = 0
The equation of plane is
3x – y – z = 7 ...(1)
Direction ratios of the normal to the plane are 3, – 1, – 1.
Let M be the foot of perpendicular from P (2, 2, 7) to the plane.
Now MP is a straight line which passes through P (2, 2, 7) and has direction ratios as 3, – 1, – 1.
its equations are 
Any point on line is M (3 r + 2, – r + 2, – r + 7)
∵ M lies on plane (1)
∵ 3 (3 r + 2) – (– r + 2) – (– r + 7) = 7
∴ 9r + 6 + r – 2 + r – 7 = 7
The equation of plane is
x + 2 y + 4 z = 38 ...(1)
Direction ratios of the normal to the plane are 1, 2, 4
Let M be foot of perpendicular from P(l, 2, 3) to the plane.
Now PM is a straight line which passes through P( 1, 2, 3) and has direction ratios as 1,2,4.
∴ its equations are
Any point M on line is (r + 1, 2 r + 2, 4 r + 3)
∵ M lies on plane (1)
∴ (r + 1) + 2 (2 r + 2) + 4 (4 r + 3) = 38
∴ r + 1 +4r + 4 + 16r + 12 = 38
∴ 21 R = 21 ⇒ r = 1
∴ M is (1 + 1,2 + 2,4 + 3) i.e. (2,4,7)
Let N(α, β, γ) be image of P in the plane (1) so that M is mid-point of PN.
∴ image is (3, 6, 11).
which are at unit distance from the origin.
...(2)
Let O be the origin and OA = a, OB = b, OC = c
∴ equation of plane passing through A, B and C is
or 
From the given condition, 
Now A, B, C are (a, 0, 0), (0, b, 0), (0, c, 0) respectively.
Let (x1 , y1 , z1 ) be the centroid of ΔABC.
locus of centroid 
is 
Let O be the origin and OA = a, OB.= b, OC = c.
∴ the equation of plane passing through A, B and C is
or 
From the given condition,
Now A, B, C are (a, 0,0), (0, b, 0), (0, 0, c) respectively.
Let (x1 ,y1 , z1) be the centroid of ΔABC.
Putting values of a, b, c in (1), we get,
or 
∴ locus of centroid (x1 ,y1, z1) is
The equation of plane through A (a, 0, 0) parallel to YZ-plane is x = a ... (2)
The equation of plane through B (0, b, 0) parallel to ZX-plane is y = b ... (3)
The equation of plane through C (0, 0, c) parallel to XY-plane is z = c ... (4)
Now to find the locus, we are to eliminate a, b, c.
Putting the values of a, b, c from (2), (3), (4) in (1), we get,
Distance between the two planes: 2x + 3y + 4z = 4 and 4x + 6y + 8 z = 12 is
2 units
4 units
8 units
D.
The equations of given planes
.
The equation of plane is
2x + 4 y – z = 1 ...(1)
The equation of line is
Any point on it is (2 r + 1, – 3 r + 2, 4 r – 3)
Let it lie on (1).
∴ 2 (2 r + 1) + 4 (–3 r + 2) – (4 r – 3) = 1
∴ 4 r + 2 – 12 r + 8 – 4r + 3 = 1
∴ – 12 r = – 12 ⇒ r = 1
∴ point is (2 + 1,–3 + 2,4 – 4) i.e. (3,–1,1).
meets the plane x + y + 4 z =6.
The equation of plane is
x + y+ 4 z = 6 ...(1)
The equation of line is
Any point on it is (2 r – 1, 3 r – 2, 4 r – 3)
Let it lie on (1)
∴ (2 r – 1) + (3 r – 2) + 4 (4 r – 3) = 6
∴ 2r – 1 + 3r – 2 +16r – 12 = 6
∴ 21 r = 21 ⇒ r = 1
∴ point is (2 – 1, 3 – 2, 4 – 3) i.e. (1, 1, 1).
meets the plane x + y + z = 2.
...(1)
The equation of given plane is
x – y + z = 5 ...(1)
The equation of the line through P (1,–2, 3) parallel to the line 
are 
Any point on it is Q (2 r + 1, 3 r –2, – 6 r + 3)
Let it lie on plane (1)
∴ 2r + 1 – 3r + 2 – 6r + 3 = 5
measured parallel to the plane 4 x + 12 y - 3z + 1 = 0.
The equation of line is 
or 
or 
Any point on this line is P(3λ + 2, 4λ –1, 2λ + 2)
It lies on the plane
Required distance = distance between (–1,–5,–10) and (2, –1, 2)
and the point of intersection of the line 
with the plane x – y + z = 5.
...(1)
Find the distance of the point (2, 3, 4) from the plane 3x + 2y + 2z + 5 = 0, measured parallel to the line 
The equation of given plane is
3x + 2y + 2z + 5 = 0 ...(1)
The equations of the line through P (2, 3, 4) parallel to the line
Any point on it is Q (3r + 2, 6 r + 3, 2r + 4)
Let it lie on plane (1)
∴ 3 (3 r + 2) + 2 (6 r + 3) + 2 (2 r + 4) + 5 = 0
or 9r + 6 + 12r + 6 + 4r + 8 + 5 = 0
or 25 r= – 25 or r = – 1
∴ point Q is ( – 3 + 2, – 6 + 3, –2 + 4) i.e. (–1,–3, 2)
...(1)
...(2)
intersect.The given lines are
...(1)
and 
...(2)
∴ lines (1) and (2) are coplanar
But
are not parallel
∴ lines (1) and (2) intersect.
...(1)
...(2)
Here, 
The vector equation of plane is
or 
Cartesian equation of plane is
or (x – 1) (24 – 18) – (y – 2) (16 + 12) + (z + 4) (6 + 6) = 0
or 6 (x – 1) – 28 (y – 2) + 12 (z + 4) = 0
or 3 (x – 1) – 14 (y – 2) + 6 (z + 4) = 0
or 3 x – 3 – l4y + 28 + 6z + 24 = 0
or 3 x – 14 y + 6 z + 49 = 0
and 
The vector equation of the plane is
or 
The cartesian equation of plane is
or 
or (x – 2) (10 + 10) – (y – 1) (5 – 15) + (z + 3) (–2 – 6) = 0
or 20 (x – 2) + 10 (y – 1) – 8 (z + 3) = 0
or 10 (x – 2) + 5 (y – 1) – 4 (z + 3) = 0
or 10x – 20 + 5y – 5 – 4z – 12 = 0
or 10x + 5y – 4 z = 37
The equation of any plane through (1, 2, 3) is
A(x – 1)+B(y – 2) + C(z – 3) = 0 ...(1)
∵ it passes through (0, – 1, 0)
∴ A (0 – 1) + B (– 1 – 2) + C (0 – 3) = 0
∴ – A – 3B – 3C = 0 ⇒ A + 3B + 3C = 0 ...(2)
Since plane (1) is parallel to the line 
∴ normal to the plane with direction ratios A, B, C is perpendicular to the line with direction ratios 2, 3, – 3.
∴ A(2) + B(3) + C(– 3) = 0 [∵ a1a2+ b1b2 + c1c2 = 0]
∴ 2A + 3B-3C = 0
Solving (2) and (3), we get,
Putting values of A, B, C in (1), we get,
6k(x – 1) + (–3 k)(y – 2) + k(z – 3) = 0 or 6 (x – 1) –3 (y – 2) + (z – 3) = 0
or 6x – 6 – 3y + 6 + z – 3 = 0
or 6x-3y + z=3
which is required equation of plane.
The equation of plane through (3, 2, 2) is
A (x – 3) + B (y – 2) + C (z – 2) = 0 ...(1)
Since it passes through (1, 0, –1)
∴ A (1 – 3) + B (0 – 2) + C (–1 – 2) = 0
or –2A – 2B – 3C = 0
∴ 2A + 2B + 3C = 0 .,.(2)
The equation of line is
Its direction ratios are 2, –2, 3
Since the line is parallel to plane (1) whose normal has direction ratios A, B, C.
∴ normal to plane (1) is perpendicular to line
∴ 2A – 2B + 3C = 0 ...(3)
From (2) and (3), we get,
Putting these values of A, B, C in (1), we get,
3k (x – 3) + 0 (y – 2) – 2k (z –2) = 0
or 3 (x – 3) – 2 (z – 2 ) = 0
or 3x – 9 –2z + 4 = 0 or 3x –2z – 5 = 0
The equation of any plane through (0, 0, 0) is
A (x – 0) + B (y – 0) + C (z – 0) = 0
or Ax + By + C z = 0 ...(1)
Since it is parallel to the vectors 
∴ normal to the plane with direction ratios A, B, C is perpendicular to the lines with direction ratios 1, 1,–1 and 3, 0,–1.
∴ A + B – C = 0 ...(2)
and 3A + 0B – C = 0 ...(3)
Solving (2) and (3), we get,
or 
Putting these values of A, B, C in (1), we get,
kx + 2ky + 3kz = 0
or x + 2 y + 3z = 0
which is required equation of plane.
Here, 
The vector equation of plane is
Cartesian equation of plane is
or 
or (–3 – 6) (x – 1) – (–2 – 6) (y – 2) + (2 – 3) (z + 4) = 0
or –9 (x– 1) + 8 (y – 2) – (z + 4) = 0
or –9x + 9 + 8y –16-z – 4 = 0
or –9 x + 8 y – z – 11 = 0 or 9 x – 8 y + z + 11 = 0.
intersect. Find the point of intersection also. 
...(1)
...(2)
...(3)
are coplanar.
and 
intersect each other.
The equations of lines are
...(1)
and 
...(2)
Any point on line (1) is (4r + 5, 4r + 7, - 5r -3), It lies on line (2)
if 
i.e, if 
...(3)
From the first and second members,
Substituting this value of r in (3), we get,
or 
∴ the lines intersect at (– 4 + 5, –4 + 7, 5 – 3) i.e. (1,3,2).
do not intersect each other.
...(1)
...(2)
...(3)
which is not true
are coplanar.
i.e. if 2(0) = 0 [Since C1, C2 are identical]
i.e. if 0 = 0, which is true.
Hence the result.
then find the value of k.
The equation of plane is 4x + 4y – kz = 0 ...(1)
It passes through the origin (0, 0, 0).
The equation of line is
...(2)
The line has direction ratios 2, 3, 4.
Since the line (2) lies in the plane (1)
∴ normal to the plane with direction ratios 4, 4, – k is perpendicular to the line (1).
∴ (2) (4) + (3) (4) + (4) (– k) = 0
∴ 8 + 12 – 4 k = 0 ⇒ 4 k = 20 ⇒ k = 5.
contains the line whose vector equation is 
...(1)
...(2)
and is parallel to the vector 
is a vector normal to the plane (1). It will be perpendicular to the line (2) if it is perpendicular to 
Show that the line 
lies in the plane 
...(1)
...(2)
and is parallel to the vector 
is a vector normal to the plane (1). It will be perpendicular to the line (2) if it is perpendicular 
lies in the plane 
whose vector equation is 
...(1)
...(2)
lies in the plane
is perpendicular to 
Let a, b, c the direction ratios of the line passing through the point (1, - 2, 3).
∴ equations of line are
...(1)
Since line (1) parallel to the plane x - y + 2z = 5 whose direction ratios of the normal are 1, -1, 2
∴ a(1) +b(– 1) + c(2) = 0
or a – b +2c=0 ... (2)
Again line (1) is parallel to the plane 3x + 2y – z = 6
∴ 3a + 2b – c = 0 ...(3)
From (2) and (3), we get
∴ from (1), the equations of line are
...(1)
...(2)
...(3)
Let a. b, c be the direction ratios of the line passing through the point (1, 2, – 4)
∴ equation of line is
...(1)
Since this line is perpendicular to the lines
and 
Solving these equations, we get,
∴ line passes through the point (1, 2,–4) with position vector 
is parallel to 
∴ equation of line is
or 
...(1)
are co-planar, find their common point and the equation of the plane in which way they lie. 
...(1)
...(2)
...(3)
we get
or – 1 = – 1 = – 1, which is true
∴ the lines intersect, and ∴ are coplanar
The point of intersection of lines is (–4 + 5, –4 + 7, 5 – 3) i.e. (1,3,2)
The equation of plane in which given lines lie is
or 
or (x – 5) (12+ 5)–(y–7) (12 + 25) + (z + 3) (4 – 28) = 0
or 17 (x – 5) – 47 (y – 7) – 24 (z + 3) = 0
or 17x – 47 y – 24 z + 172 = 0
which is required equation of plane.
Direction-ratios of first line are 2, 2, 1.
Direction-ratios of second line joining the points (3, 1, 4), (7, 2, 12) are 7 – 3, 2 – 1 12 – 4 i.e., 4, 1, 8
Let θ be the angle between the two lines
The direction-ratios of the two lines are 5, – 12, 13 and – 3, 4, 5.
Let θ be the angle between the lines.
Find the value of p, so that the lines:
are perpendicular to each other. Also find the equations of a line passing through a point (3, 2, -4) and parallel to line l1.
The equation of line L1:
Thus the equation of the line passing through the point (3, 2, -4) and parallel to the line L1 is:
Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x- y + z = 0. Also find the distance of the plane obtained above, from the origin.
Equation of the plane passing through the intersection of the planes x+y+z=1 and 2x+3y+4z=5 is:
Thus, the equation of the plane is:
A manufacturing company makes two types of teaching aids A and B of Mathematics for class XII. Each type of A requires 9 labour hours of fabricating and 1 labour hour for finishing. Each type of B requires 12 labour hours for fabricating and 3 labour hours for finishing. For fabricating and finishing, the maximum labour hours available per week are 180 and 30 respectively. The company makes a profit of 80 on each piece of type A and 120 on each piece of type B. How many pieces of type A and type B should be manufactured per week to get a maximum profit? Make it as an LPP and solve graphically. What is the maximum profit per week?
Let x be the number of pieces manufactured of type A and y be the number of pieces manufactured of type B. Let us summarize the data given in the problem as follows:
| Product | Time for Fabricating (in hours) | Time for Finishing (in hours) | Maximum labour hours available |
| Type A | 9 | 1 | 180 |
| Type B | 12 | 3 | 30 |
| Maximum Profit (in Rupees) | 80 | 120 |
| Points | Value of Z |
| A(12, 6) | Z = 80 x 12 + 120 x 6 = Rs. 1680 |
| B(0, 10) | Z = 80 x 0 +120 x 10 = Rs. 1200 |
| C(20, 0) | Z = 80 x 20 + 120 x 0 = Rs.1600 |
Find the Cartesian equation of the line passes through the point (-2, 4, -5) and is parallel to the line 
The equation of the given line is:
The required line is parallel to the given line. Therefore, direction ratios of the required line are same as the direction ratio of the given line. So, the direction ratios of the required line are 3, -5, and 6.
Thus, the equation of the straight line passing through (-2, 4, -5) and having direction ratios 3, -5, 6 is
Find the distance of the point (-1,-51-10) from the point of intersection of the line and the plane
Cartesian equation of line and plane,
Find the shortest distance between the following lines:
The vector form of this equation is:
The vector form of this equation is:
Now, the shortest distance between these two lines is given by:
Find the equation of the plane passing through the point (−1, − 1, 2) and perpendicular to each of the following planes: 2x + 3y – 3z = 2 and 5x – 4y + z = 6
The equation of the piane passing through the point ( -1, -1, 2 ) is:
a( x + 1 ) + b( y + 1 ) + c ( z - 1 )= 0 ............(1)
where a, b and c are the direction ratios of the normal to the plane
.
It is given that the plane (1) is perpendicularto the planes.
2x + 3y - 3z = 2 and 5x - 4y + z = 6
2a + 3b - 3c = 0 ..............(2)
5a - 4b + c = 0 ...............(3)
solving equations (2) and (3), we have:
So the direction ratios of the normal to the required plane are multiples of 9,
17, and 23.
Thus, the equation of the required plane is:
9(x + 1) + 17( y + 1) + 23( z - 2) = 0
or 9x + 17y + 23z = 20
Find the equation of the plane passing through the points (3, 4, 1) and (0, 1, 0) and parallel to the line
Equation of the plane passing through the point (3, 4, 1) is:
a ( x - 3 ) + b ( y - 4 ) c ( z - 1 ) = 0 .............(1)
Where a, b, c are the direction ratios of the normal to the plane
It is given that the plane (1) passes through the point 9 0, 1, 0 ).
a - 3 + b - 3 + c - 1 = 0
3a + 3b + c = 0 ...............(2)
It is also given that the plane (1) is parallel to the line
.
So, this line is perpendicular to the normal of the plane (1).
2a + 7b + 5c = 0 ................(3)
Solving equations (2) and (3), we have:
So, the direction ratios of the normal to the required plane are multiples of 8, -13, 15.
Therefore, equation (1) becomes:
8 ( x - 3 ) -13 ( y - 4 ) + 15 ( z - 1) = 0
8x - 13y + 15z + 13 = 0.
Which is the required equation of the plane.
Find the value of λ so that the lines, are perpendicular to each other.
Given lines are
let us rewrite the equations of the given lines as follows:
That is we have,
And
The lines are perpendicular so angle between them is
So, cos = 0
Here ( a1, b1, c1 ) = ( -3, , 2 )
and
( a2, b2, c2 ) = ( , 1, -7 )
For perpendicular lines
Find the equation of the plane passing through the point (-1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0.
Let the equation of the plane be,
A ( x - x1 ) + B ( y - y1 ) + C ( z - z1 ) = 0
Plane passes throughthe points ( -1, 3, 2 )
A ( x + 1 ) + B ( y - 3 ) + C ( z - 2 ) = 0 ........(i)
Now applying the condition of perpendicularity to the plane (i) with planes
x + 2y + 3z = 5 and 3x + 3y + z = 0, We have,
A + 2B + 3C = 0
3A + 3B + C = 0
Solving we get
A + 2B + 3C = 0
9A + 9B + 3C = 0
By cross multiplication, we have,
By substituting A and C in equation (i), we get,
Substituting the values of A, B and C in equation (i), we have,
What is the cosine of the angle which the vector makes with y-axis?
The y-axis can be represented in vector form by
Write the vector equation of the following line:
We get, x1 = 5, y1 = -4, z1 = 6, a = 3, b = 7, c = -2
Thus, the required line is parallel to the vector and passes through the point ( 5, -4, 6 ).
The vector form of the line can be written as where is a constant.
Thus, the required equation is
Find the Cartesian equation of the plane passing through the points A(0, 0, 0) and B(3, -1, 2) and parallel to the line
Let the equation of the plane be ax + by + cz + d = 0 ........(i)
Since the plane passes through the point A ( 0, 0, 0 ) and B ( 3, -1, 2),
we have
a x 0 + b x 0 + c x 0 + d = 0
d = 0 ................(ii)
Similarly for point B ( 3, -1, 2 ), a x 3 + b x ( - 1 ) + c x 2 + d = 0
3a - b + 2c = 0 ( Using , d = 0 ) ............(iii)
The required plane is parellel to the above line .
Therefore, a x 1 + b x ( - 4 ) + c x 7 = 0
a - 4b + 7c = 0 ............(iv)
Cross multiplying equations (iii) and (iv), we obtain:
Substituting the values of a, b and c in equation ( 1 ), we obtain the equation of plane as:
kx - 19ky - 11kz + d = 0
k ( x - 19y - 11z ) = 0 ..........( From equation (ii) )
x - 19y - 11z = 0
So, the equation of the required plane is x - 19y - 11z .
Write the vector equations of the following lines and hence determine the distance between them:
The vector form of the above equation is,
The vector form of this equation is
Since is same in equation (i) and (ii), the two lines are parallel.
Distance d, between the two parallel lines is given by the formula,
On substitution, we get
Write the intercept cut off by the plane 2x + y – z = 5 on x-axis.
2 x + y - z = 5
Dividing both sides by 5,
Thus, the intercept cut off by the given plane on the x-axis is .
Find the angle between the following pair of lines:
And check whether the lines are parallel or perpendicular.
= - 2 + 14 - 12
= 0
The angle between the given pair of lines is given by the relation,
Thus, the given lines are perpendicular to each other and the angle
between them is .
Find the equation of the plane which contains the line of intersection of the planes
and which is perpendicular to
the plane .
The equations of the given planes are
The equation of the plane passing through the line of intersection of the given
planes is
This is the vector equation of the required plane.
The lines p(p2+ 1) x – y + q = 0 and (p2+ 1)2x + (p2+ 1) y + 2q = 0 are
perpendicular to a common line for
no value of p
exactly one value of p
exactly two values of p
more than two values of p
B.
exactly one value of p
Let the line 
lie in the plane x + 3y – αz + β = 0. Then (α, β) equals
(6, – 17)
(–6, 7)
(5, –15)
(–5, 5)
B.
(–6, 7)
2 + 3 × 1 – α (–2) + β = 0
2α + β = –5 ... (i)
3 – 15 – 2α = 0
2α = –12
B = –5 + 12 = 7
(α, β) ≡ (–6, 7)
If P and Q are the points of intersection of the circles x2+ y2+ 3x + 7y + 2p – 5 = 0 and x2+ y2+ 2x + 2y – p2 = 0, then there is a circle passing through P, Q and (1, 1) for
all values of p
all except one value of p
all except two values of p
exactly one value of p
B.
all except one value of p
Radical axis is x + 5y + p2 + 2p – 5 =0
Equation of circle is
x2 + y2 + 3x + 7y + 2p – 5 + λ [x + 5y + p2 + 2p – 5 ] = 0 …. (i)
(i) passes through (1, 1)
If the straight lines 
intersect at a point, then the integer k is equal to
-5
2
5
-5
A.
-5
Let L be the line of intersection of the planes 2x + 3y + z = 1 and x + 3y + 2z = 2. If L makes an angle α with the positive x-axis, then cos α equals-
1/2
1
A.
If direction cosines of L be l, m, n, then
2l + 3m + n = 0
l + 3m + 2n = 0
The resultant of two forces P N and 3 N is a force of 7 N. If the direction of the 3 N force were reversed, the resultant would be 
N. The value of P is
5N
6N
4N
3N
A.
5N
If a line makes an angle of π/4 with the positive directions of each of x-axis and y-axis, then the angle that the line makes with the positive direction of the z-axis is
π/6
π/3
π/4
π/2
D.
π/2
l = cos π/4 , m = cos π/4
we know l2 + m2 + n2 = 1
Hence angle with the positive direction of z-axis is π/2 .
If (2, 3, 5) is one end of a diameter of the sphere x2+ y2+ z2 − 6x − 12y − 2z + 20 = 0, then the coordinates of the other end of the diameter are
(4, 9, –3)
(4, –3, 3)
(4, 3, 5)
(4, 3, –3)
A.
(4, 9, –3)
Coordinates of centre (3, 6, 1)
Let the coordinates of the other end of diameter are (α, β, γ)
Hence α = 4, β = 9 and γ = −3.
If one of the lines of my2+ (1 − m2)xy − mx2 = 0 is a bisector of the angle between the lines xy = 0, then m is
−1/2
-2
1
2
C.
1
Equation of bisectors of lines xy = 0 are y = ± x
put y = ± x in my2+ (1 − m2)xy − m = 0,
we get (1 − m2) x2 = 0
⇒ m = ± 1
The normal to the curve x = a(cosθ + θ sinθ), y = a( sinθ - θ cosθ) at any point ‘θ’ is such that
it passes through the origin
it makes angle π/2 + θ with the x-axis
it passes through (aπ/2 ,-a)
it is at a constant distance from the origin
D.
it is at a constant distance from the origin
Clearly dy/dx = an θ
⇒ slope of normal = - cot θ
Equation of normal at ‘θ’ is
y – a(sin θ - θ cos θ) = - cot θ(x – a(cos θ + θ sin θ)
⇒ y sin θ - a sin2 θ + a θ cos θ sin θ = -x cos θ + a cos2 θ + a θ sin θ cos θ
⇒ x cos θ + y sin θ = a
Clearly this is an equation of straight line which is at a constant distance ‘a’ from origin.
The line parallel to the x−axis and passing through the intersection of the lines ax + 2by + 3b = 0 and bx − 2ay − 3a = 0, where (a, b) ≠ (0, 0) is
below the x−axis at a distance of 3/2 from it
below the x−axis at a distance of 2 /3 from it
above the x−axis at a distance of 3/ 2 from it
above the x−axis at a distance of 2/ 3 from it
A.
below the x−axis at a distance of 3/2 from it
ax + 2by + 3b + λ(bx – 2ay – 3a) = 0
⇒ (a + bλ)x + (2b – 2aλ)y + 3b - 3λa = 0
a + bλ = 0
⇒ λ = -a/b
If the angle θ between the line 
and the plane 
is such of sin θ = 1/3 the value of λ is
5/3
-3/5
3/4
-4/3
A.
5/3
Angle between line and normal to plane is 
where θ is the angle between line & plane
The angle between the lines 2x = 3y = − z and 6x = − y = − 4z is
0o
90o
45o
30o
B.
90o
Angle between the lines 2x = 3y = - z & 6x = -y = -4z is 90°
Since a1a2 + b1b2 + c1c2 = 0
If the plane 2ax − 3ay + 4az + 6 = 0 passes through the midpoint of the line joining the centres of the spheres
x2 + y2 + z2 + 6x − 8y − 2z = 13 and x2 + y2 + z2 − 10x + 4y − 2z = 8, then a equals
-1
1
-2
2
C.
-2
Plane 2ax – 3ay + 4az + 6 = 0 passes through the mid point of the centre of spheres x2 + y2 + z2 + 6x – 8y – 2z = 13 and x2 + y2 + z2 – 10x + 4y – 2z = 8 respectively centre of spheres are (-3, 4, 1) & (5, - 2, 1) Mid point of centre is (1, 1, 1) Satisfying this in the equation of plane,
we get 2a – 3a + 4a + 6 = 0
⇒ a = -2.
The distance between the line 
and the plane 
10/9
3/10
10/3
B.
Distance between the line 
equation of plane is x + 5y + z = 5 ∴ Distance of line from this plane = perpendicular distance of point (2, -2, 3) from the plane
If a vertex of a triangle is (1, 1) and the mid-points of two sides through this vertex are (-1, 2) and (3, 2), then the centroid of the triangle is
(-1, 7/3)
(-1/3, 7/3)
(1, 7/3)
(1/3, 7/3)
C.
(1, 7/3)
Vertex of triangle is (1, 1) and midpoint of sides through this vertex is (-1, 2) and (3, 2) ⇒ vertex B and C come out to be (-3, 3) and (5, 3)
therefore centroid is 
The plane x + 2y – z = 4 cuts the sphere x2 + y2 + z2 – x + z – 2 = 0 in a circle of radius
3
1
2
B.
1
Perpendicular distance of centre 
from x + 2y – 2 = 4
Let A (2, –3) and B(–2, 1) be vertices of a triangle ABC. If the centroid of this triangle moves on the line 2x + 3y = 1, then the locus of the vertex C is the line
2x + 3y = 9
2x – 3y = 7
3x + 2y = 5
3x – 2y = 3
A.
2x + 3y = 9
If C be (h, k) then centroid is (h/3, (k – 2)/3) it lies on 2x + 3y = 1.
⇒ locus is 2x + 3y = 9.
A line makes the same angle θ, with each of the x and z-axis. If the angle β, which it makes with y-axis, is such that sin2β = 3sin2θ , then cos2θ equals
2/3
1/5
3/5
2/3
C.
3/5
A line makes angle θ with x-axis and z-axis and β with y-axis.
∴ l = cosθ, m = cosβ,n = cosθ
We know that, l2+ m2+ n2= 1
cos2θ + cos2β +cos2θ =1
2cos2θ = 1- cos2β
2cos2θ = sin2β
But sin2β = 3 sin2θ
therefore from equation (i) and (ii)
3sin2θ = 2cos2θ
3(1-cos2θ) = 2cos2θ
3-3cos2θ = 2cos2θ
3 = 5cos2θ
Distance between two parallel planes 2x + y + 2z = 8 and 4x + 2y + 4z + 5 = 0 is
3/2
5/2
7/2
9/2
C.
7/2
The distance between 4x + 2y + 4z - 16 = 0 and 4x + 2y + 4z + 5 = 0 is
A line with direction cosines proportional to 2, 1, 2 meets each of the lines x = y + a = z and x + a = 2y = 2z. The co-ordinates of each of the point of intersection are given by
(3a, 3a, 3a), (a, a, a)
(3a, 2a, 3a), (a, a, a)
(3a, 2a, 3a), (a, a, 2a)
(2a, 3a, 3a), (2a, a, a)
B.
(3a, 2a, 3a), (a, a, a)
Any point on the line 
and any point on the line 
Now direction cosine of the lines intersecting the above lines is proportional to (2t2 – a – t1, t2 – t1 + a, t2 – t1).
Hence 2t2 – a – t1 = 2k , t2 – t1 + a = k and t2 – t1 = 2k
On solving these, we get t1 = 3a , t2 = a. Hence points are (3a, 2a, 3a) and (a, a, a).
If the straight lines x = 1 + s, y = –3 – λs, z = 1 + λs and x = t/ 2 , y = 1 + t, z = 2 – t with parameters s and t respectively, are co-planar then λ equals
–2
–1
-1/2
0
A.
–2
Given 
are coplanar then plan passing through these lines has normal perpendicular to these lines
⇒ a - bλ + cλ = 0 and a/2 +b -c =0 (where a, b, c are direction ratios of the normal to the plan) On solving, we get λ = -2.
The intersection of the spheres x2 +y2 +z2 + 7x -2y-z =13 and x2 +y2 +z2 -3x +3y +4z = 8 is the same as the intersection of one of the sphere and the plane
x-y-z =1
x-2y-z =1
x-y-2z=1
2x-y-z =1
D.
2x-y-z =1
Required plane is S1 – S2 = 0 where
S1 = x2 + y2 + z2 + 7x – 2y – z – 13 = 0
and S2 = x2 + y2 + z2 – 3x + 3y + 4z – 8 = 0
⇒ 2x – y – z = 1.
If the straight line y = mx + c (m > 0) touches the parabola y2 = 8(x + 2), then the minimum value taken by c is
12
8
4
6
C.
4
The tangent of slope m must be of the form y = m(x + 2) + a/m
The equation of the plane which contains the line of intersection of the planes x + y + z – 6 = 0 and 2x + 3y + z + 5 = 0 and perpendicular to the xy plane is:
x – 2y + 11 = 0
x + 2y + 11 = 0
x + 2y – 11 = 0
x – 2y – 11 = 0
B.
x + 2y + 11 = 0
Equation of the required plane is
(x + y + z – 6) + λ(2x + 3y + z + 5) = 0
i.e. (1 + 2λ)x + (1 + 3λ)y + (1 + λ)z + (–6 + 5λ) = 0
This plane is perpendicular to xy plane whose
equation is z = 0
i.e. 0 . x + 0 . y + z = 0
∴ By condition of perpendicularity
0 .(1 + 2λ) + 0. (1 + 3λ) + (1 + λ) .1 = 0
i.e. λ = –1
∴ Equation of required plane is
(1 – 2)x + (1 – 3)y + (1 – 1)z + (–6 – 5) =0
or x + 2y + 11 = 0
The curve y = (λ + 1)x2 + 2 intersects the curve y = λx + 3 in exactly one point, if λ equals -
{–2, 2}
{1}
{-2}
{2}
C.
{-2}
for point of intersection at exactly one point
λx + 3 = (λ + 1)x2 + 2
(λ + 1)x2 – λx – 1 = 0
Δ = 0
λ2 + 4(λ + 1) = 0
λ2 + 4λ + 4 = 0
(λ + 2)2 = 0 λ = – 2
Sponsor Area
Sponsor Area
. Then
