Sponsor Area
The x-axis and y-axis taken together determine a plane known as ____________.
Name the octants in which the following points lie:
A (2, 3, 4), B (6, -3, 3), C (2, -1, -6), D (2, 2, -3), E (-1, 3, -6), F (-1, 3, 3), G (-3, -2,5) and H (-1,-2,-5).
A(2, 3, 4) : x>0, y>0, z>0
The point is in octant XOYZ or I
B(6, -3, 3): x>0, y<0, z>0
The point is in octant XOY'Z or IV
C(2, -1, -6) : octant XOY'Z' or VIII
D(2, 2, -3) : octant XOYZ' or V
E(-1, 3, -6): octant X'OYZ' or VI
F(-1, 3, 3): octant X'OYZ or II
G(-3, -2, 5): octant X'OY'Z' or III
H(-1, -2, -5): octant X'OY'Z or VII
Find the perpendicular distance of:
(i) A (2, -3, 4) from XY-plane
(ii) B (4, -3, 1) from ZX-plane
(iii) C (3, -2, -1) from YZ-plane
(i) Distance of P(a, b, c) from XY-plane =
∴ Distance of A(2, -3, 4) from XY - plane =
(ii) Distance of P(a, b, c) from ZX-plane =
∴ Distance of B(4, -3, 1) from ZX-plance =
(iii) Distance of P(a, b, c) from YZ-plane =
∴ Distance of C (3, -2, -1) from YZ-plane =
The given point is A (2, -3, 6) and has y-co-ordinate negative.
∴ The seven other points are:
(i) B for which all x, y, z are positive
∴ B is (2, 3, 6)
(ii) C for which x is negative, y and z are positive.
∴ C is (-2, 3, 6)
(iii) D for which x, y are positive and z is negative.
∴ D is (2, 3, -6)
(iv) E for which x, y are negative and z is positive.
∴ E is (-2, -3, 6)
(v) F for which x, z are negative and y is positive.
∴ F is (-2, 3, -6)
(vi) G for which x is positive and y, z are negative.
∴ G is (2, -3, -6)
(vii) H for which x, y, z are all negative.
∴ H is (-2, -3, -6)
Perpendicular distance of point P from XY-plane = (∵ c>0)
Similarly, the perpendicular distances of P(a, b, c) from ZX-plane and YZ-plane are b and a respectively. According to the question, we have
∴ a + b - 2c = 0
Which is the required relation between a, b and c
The point is P(x, y, z), where x, y, z>0
Distance from ZX-plane =
Distance from XY-plane =
Distance from YZ-plane =
According to the question, we have
2y - z - x = 4
Hence, the locus is x - 2y + z + 4 =0
Find the octant in which the following points lie:
A (-3,1, 2), B (-3, 1, -2), C (-3, -1,-2), D (3, -1, -2) and E (3,1, 2).
Solution not provided.
Ans. A : X'OYZ or II; B : X'OYZ' or VI; C : X'OY'Z' or VII; D : XOY'Z' or VIII; E: XOYZ or I
Solution not provided.
Ans. (-5, 0, 0)
Solution not provided.
Ans. (0, 0, 3)
Solution not provided.
Ans. (0, -4, 0)
Solution not provided.
Ans. (4, 3, 0)
Solution not provided.
Ans. (0, 4, -5)
Sponsor Area
Solution not provided.
Ans. (-3, 0, -5)
Find the perpendicular distance of point A (-3, 2, -4) from:
(a) ZX-plane (b)XY-plane (c) YZ-plane.
Solution not provided.
Ans. (a) 2 (b) 4 (c) 3
Solution not provided.
Ans. B(2, 1, 7), C(-2, 1, 7), (2, -1, 7), D(2, 1, -7) , F(-2, -1, 7), G(2, -1, -7), H(-2, -1, -7)
The co-ordinates of a point H are (-2, 3, -4). Find
(i) The co-ordinates of the foot of perpendicular from this point to ZX-plane.
(ii) The co-ordinates of the foot of perpendicular from this point to the z-axis.
(iii) The length of perpendicular from this point to XY-plane.
Solution not provided.
Ans. (i) (-2, 0, -4)
(ii) (0, 0, -4)
(iii) 4
Given points are : A(-2, 3, 5), B(1, 2, 3) and C(7, 0, -1)
Using distance formula, , we have
Now, AB + BC = = CA
Hence, points A, B, C are collinear.
Verify the following:
(0, 7, 10), (-1, 6, 6) and (-4, 9, 6) are the vertices of a right angled triangle.
Show that the points:
P (-1,-6, 10), Q (1, -3, 4), R (-5, -1, 1) and S (-7, -4, 7) are the vertices of a rhombus.
Sponsor Area
Let P (x, y, z) be the required point and
O (0, 0, 0), A (a, 0, 0), B (0, b, 0) and C (0, 0, c) be the given points.
∴
Also,
Hence, the required point is
The given points are A (6, 5, -3) and B (x, -7, 0)
Given that
Squaring both sides, we get
x = 2, x = 10
Hence, the value of x is 2 or 10.
Let the point on z-axis be P (0,0, z).
It is equidistant from A (3, 2, 1) and B (5, 2, 5)
Using distance formula, we have
Hence, the required point is P (0, 0, 5).
Find points on x - axis which are at a distance of units from point A (1, 2, 3).
Let the required point on x-axis be P (x, 0, 0)
It is at a distance of units from A (1, 2, 3)
Hence, the required points on x-axis are (5, 0, 0) or (-3, 0, 0).
Let one of the points be P(x, y, z).
A (1, 2, 3) and B (3, 2, -1) are two given points
Now,
Using distance formula, we have
Hence, the equation of set of points P (or the locus) is x - 2z = 0
Let one of the points be P (x, y, z).
A (4, 0, 0) and B (-4, 0, 0) be given points.
Given condition is :
Using distance formula, we have
Squaring both sides, we get
Again, squaring both sides, we get
Hence, the required locus is .
Let one of the points be P(x, y, z)
A (3, 4 5) and B (-1, 3, -7) be two given points
The given condition is: ...(i)
Using distance formula, we have
∴ From (i), we have
Hence, the equation of set of points P is
Let point P be (x, y, z) and A (1, 2, 3) be the given point. The given condition for locus is:
Squaring both sides, we get
Hence, locus of P is
Solution not provided.
Ans. (a) (b)
Solution not provided.
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Find the co-ordinates of a point equidistant from four points.
O (0, 0, 0), A (2p, 0, 0), B (0, 2q, 0) and C (0, 0, 2r). Find also the distance.
Solution not provided.
Ans. (p, q, r),
Solution not provided.
Ans. k = 9, -15
Solution not provided.
Solution not provided.
Solution not provided.
Solution not provided.
Solution not provided.
Ans. (0, 7, 0), (0, -5, 0)
Solution not provided.
Solution not provided.
Ans. (2, 0, 0)
Solution not provided.
Ans.
Solution not provided.
Ans. 6x - 8y + 16z - 55 = 0
Solution not provided.
Ans. 10x + 4y - 10z - 14 = 0
The ratio is 2 : 3 externally.
∴ Point Q is (-8, 17, 3)
Let point B (2, 4, 6) divides the join of A (-4, 6, 10), C (14, 0, -2) in the ratio k : 1
∴ The co-ordinates of point B are
On putting
When
∴ Point B (2, 4, 6) divides the join of A (-4, 6, 10) and C (14, 0, -2) in the ratio internally. Hence, A, B, C are collinear and ratio of division at B is 1 : 2 internally.
Let R and S be the two points of trisection:
Ratio at R is 1 : 2
∴ Point R is
Also, ratio at S is 2 : 1
∴ Point S is
Find the ratio in which the YZ-plane divides the line segment formed by joining the points (–2, 4, 7) and (3, –5, 8).
Let YZ-plane divides the join of A (-2, 4 7) and B (3, -5, 8) in ratio k : 1 at point P.
Since point P lies on YZ-plane x co-ordinate = 0
Hence, the ratio is internally.
Find the ratio in which the join of P (2, 1, -1) and Q (3, 2, 4) is divided by XY-plane.
Let XY-plane divides PQ in ratio k : 1 at point R.
Point R is
Since R lies on XY-plane.
∴ z co-ordinate = 0
Hence, the ratio is internally.
Let point R (4, y, z) divide the join of P (2, - 3, 4) and Q (8, 0, 10) in the ratio k:1. R is
∴
∴
Hence, point R is (4, -2, 6).
AD is bisector of Ratio at D is c : b, where
∴ For point D:
Hence, point D is
Sponsor Area
Let the co-ordinates of the other end B be
∴ co-ordinates of mid-point of line segment
AB are
Hence, point B is (-4, -5, 5)
Here, A ((3, -5, 7)) and B (-1, 7, -6) be the given vertices of a
Let C be (x, y, z) Centroid of triangle ABC is
(Given)
Hence, vertex C is (1, 1, 2).
Here A (2a, 2, 6), B (-4, 3b, -10) and C (8, 14, 2c) be the given vertices of a
∴ Centroid of is
But, centroid is given to be (0, 0, 0)
∴
Also,
and
Hence,
Three vertices of parllelogram ABCD are A (3, -1, 2), B (1, 2, -4), and C (-1, 1, 2).
Let the fourth vertex be D (x, y, z). ABCD is a parallelogram. Diagonals AC and BD bisect each other
Mid-point of AC and BD are coincident.
x = 1, y = -2, z = 8
Hence, vertex D is (1, -2, 8)
Using distance formula, we have
∴ ABCD is not a rectangle.
Let the co-ordinates of the vertices A, B and C respectively be
(2, 1, -1) is mid-point of BC.
∴
...(i)
...(ii)
....(iii)
(3, 2, 4) is mid-point of CA.
∴
...(iv)
...(v)
...(vi)
(1, -2, 3) is mid-point of AB.
∴
...(vii)
...(viii)
...(ix)
Adding (i), (iv) and (vii), we get
...(x)
Using (i) in (x), we get
Using (iv) in (x), we get
Using (vii) in (x), we get
Adding (ii), (v) and (viii), we get
...(xi)
Using (ii) in (xi), we get
Using (v) in (xi), we get
Using (viii) in (xi), we get
Adding (iii), (vi) and (ix), we get
...(xii)
Using (iii) in (xii), we get
Using (vi) in (xii), we get
Using (ix) in (xii), we get
Hence, the three vertices A, B and C are (2, -1, 8), (0, -3, -2), (4, 5, 0) respectively.
Solution not provided.
Ans. (1, 1, -4)
Solution not provided.
Ans. (4, -11, 5)
Solution not provided.
Ans. 3:2
Solution not provided.
Ans. 3:5
Solution not provided.
Ans. B (-4, 1, 2)
Solution not provided.
Ans. (6, -4, -2), (8, -10, 2)
Solution not provided.
Ans. x = 4, y = -9, z = 10
Solution not provided.
Ans. No
Solution not provided.
Ans. D (0, 6, -3)
Solution not provided.
Ans. C (15, -10, 16)
Using section formula, show that points A (2, -3, 4), B (-1, 2, 1) and are collinear.
Solution not provided.
Solution not provided.
Ans. 3 :2 externally ;
Solution not provided.
Ans. 1 : 3 externally ;
Solution not provided.
Ans.
Find the octant in which following points lie:
(i) (1, 4, -2) (ii) (2, 1, 4) (iii) (-2, -1, 5) (iv) (1, -3, -1) (v) (2, -4, 3)
Solution not provided.
Ans. (i) XOYZ' or V (ii) XOYZ or I (iii) X'OY'Z or III (iv) XOY'Z' or VIII (v) XOY'Z or IV
Write down the co-ordinates of the foot of perpendicular from point P (2, - 3, 4) upon (i) x-axis (ii)y-axis (iii) z-axis.
Solution not provided.
Ans. (i) (2, 0, 0) (ii) (0, -3, 0) (iii) (0, 0, 4)
Write the co-ordinates of the foot of perpendiculars from point P (2, 4, 5) to (i) the XY-plane (ii) the YZ-plane (iii) the ZX-plane.
Solution not provided.
Ans. (i) (2, 4, 0) (ii) (0, 4, 5) (iii) (2, 0, 5)
Solution not provided.
Solution not provided.
Solution not provided.
Ans. 10x + 6y -18z - 29 = 0
Find the co-ordinates of a point on y-axis which is at distance of units from point P(3, -2, 5).
Solution not provided.
Ans. (0, 2, 0) or (0, -6, 0)
Solution not provided.
Ans. 2 : 3 external
Solution not provided.
Ans.
Solution not provided.
Ans. x+5y-5z+23=0
Solution not provided.
Ans. (3, -4, 7)
Show that point A (5, 4, 2), B (-1, -2, 4) and are collinear. Find also the ratio in which C divides the join of A and B.
Solution not provided.
Ans. 2:3
Solution not provided.
Ans. x = 1, y = -11, z = 8
Sponsor Area
Solution not provided.
Ans. (3, 4, -5)
O is the point of concurrence of the co-ordinate axes. O is (0, 0, 0)
A lies on x-axis A is (a, 0, 0)
B lies on y-axis B is (0, b 0)
C lies on z-axis C is (0, 0, c)
D lies on plane XOY i.e. z = 0 D is (a, b, 0)
E lies on plane YOZ i.e. x = 0 E is (0, b, c)
F lies on plane ZOX i.e. y = 0 F is (a, 0, c)
Solution not provided.
Ans. O(0, 0, 0), A (6, 0, 0), B (0, 4, 0), C (0, 0, 5), D (6, 4, 0), E (0, 4, 5), F (6, 0, 5)
If are the vertices of triangle ABC, then prove that its centroid G is
The number of common tangent to the circles x2+y2-4x-6y-12=0 and x2+y2+6x+18y+26 = 0 is
1
2
3
4
C.
3
Number of common tangents depend on the position of the circle with respect to the each other.
(i) If circles touch externally ⇒C1C2 = r1+ r2,3 common tangents
(ii) If circles touch internally ⇒ C1C2 = r2-r1, 1 common tangents
(iii) If circles do not touch each other, 4 common tangents
Given equations of circles are
x2 +y2-4x-6y-12 = 0 .. (i)
x2+y2+6x+18y+26 =0 ... (ii)
Centre of circle (i) is C1 (2,3) and radius
=
Centre of circle (ii) is C2(-3,-9) and radius
Thus, both circles touch each other externally. Hence, there are three common tangents.
The angle between the lines whose direction cosines satisfy the equations l +m+n=0 and l2 = m2+n2 is
π/3
π/4
π/6
π/2
A.
π/3
We know that angle between two lines is
l +m +n= 0
⇒ l = - (m+n)
⇒ (m+n)2 = l2
⇒ m2 +n2 +2mn = m2 +n2
[∵ l2 = m2 +n2, given]
⇒ 2mn = 0
when m = 0 ⇒ l =-n
Hence, (l, m, n) is (1,0-1)
When n =0, then l =-m
Hence, (l,m,n) is (1,0-1)
If PS is the median of the triangle with vertices P(2,1), Q(6,-1) and R (7,3), then equation of the line passing through (1,-1) and parallel to PS is
4x-7y - 11 =0
2x+9y+7=0
4x+7y+3 = 0
2x-9y-11 =0
B.
2x+9y+7=0
Coordinate of S =
[∵ S is mid-point of line QR]
A slope of the line PS is -2/9.
Required equation passes through (1,-1) and parallel to PS is
Let a,b,c and d be non-zero numbers. If the point of intersection of the lines 4ax +2ay +c= 0 and 5bx +2by +d = 0 lies in the fourth quadrant and is equidistant from the two axes, then
2bc-3ad =0
2bc+3ad =0
2ad-3bc =0
3bc+2ad=0
C.
2ad-3bc =0
Let coordinate of the intersection point in the fourth quadrant be (α, -α) lies on both lines 4ax +2ay +c =0 and 5bx +2by +d =0
The x-coordinate of the incentre of the triangle that has the coordinates of mid points of its sides as (0, 1) (1, 1) and (1, 0) is
B.
Given mid points of sides of a triangle are (0,1) (1,1) and (1,0). Plotting these points on a graph and make a triangle.
Twenty meters of wire is available for fencing off a flower-bed in the form of a circular sector.Then the maximum area (in sq. m) of the flower-bed, is
30
12.5
10
25
D.
25
Total length = r + r + rθ = 20
Statement 1: The variance of first n even natural numbers is
Statement 2: The sum of first n natural numbers is and the sum of squares of first n natural numbers is
Statement–1 is true, Statement–2 is true, Statement–2 is a correct explanation for statement–1
Statement–1 is true, Statement–2 is true; Statement–2 is not a correct explanation for statement–1.
Statement–1 is true, statement–2 is false.
Statement–1 is false, Statement–2 is true
D.
Statement–1 is false, Statement–2 is true
If the letters of word SACHIN are arranged in all possible ways and these words are written out as in dictionary, then the word SACHIN appears at serial number
601
600
602
603
A.
601
Alphabetical order is A, C, H, I, N, S
No. of words starting with A – 5!
No. of words starting with C – 5!
No. of words starting with H – 5!
No. of words starting with I – 5!
No. of words starting with N – 5!
SACHIN – 1 601.
If non-zero numbers a, b, c are in H.P., then the straight line always passes through a fixed point. That point is
(-1, 2)
(-1, -2)
(1, -2)
(1, 1/2)
C.
(1, -2)
Let two numbers have arithmetic mean 9 and geometric mean 4. Then these numbers are the roots of the quadratic equation
x2 + 18x +16 = 0
x2-18x-16 = 0
x2+18x-16 =0
x2-18x +16 =0
D.
x2-18x +16 =0
Let α and β be two numbers whose arithmetic mean is 9 and geometric mean is 4.
∴ α + β = 18 ........... (i)
and αβ =16 ........... (ii)
∴ Required equation is x2 - (α + β)x + (αβ) = 0 ⇒ x2 - 18x + 16 = 0 [using equation (i) and equation (ii)]
The sum of the first n terms of the series when n is even. When n is odd the sum is
B.
19. The sum of n terms of given series = if n is even.
Let n is odd i.e. n = 2m + 1
Then, S2m+1 = S2m + (2m + 1)th term
Suppose four distinct positive numbers a1, a2, a3, a4 are in G.P. Let b1 = a1, b2 = b1 + a2, b3 = b2 + a3 and b4 = b3 + a4.
Statement-I: The numbers b1, b2, b3, b4are neither in A.P. nor in G.P. Statement-II: The numbers b1, b2, b3, b4 are in H.P.
Both statement-I and statement-II are true but statement-II is not the correct explanation of statement-I
Both statement-I and statement-II are true, and statement-II is correct explanation of Statement-I
Statement-I is true but statement-II is false.
Statement-I is false but statement-II is true
C.
Statement-I is true but statement-II is false.
a1 = 1
a2 = 2
a3 = 4
a4 = 8
So, b1 = 1
b2 = 1 + 2 = 3
b3 = 3 + 4 = 7
b4 = 7 + 8 = 15
The numbers b1, b2, b3 , b4 are not in G.P. and A.P. Statement-I is correct but Statement-II wrong.
Let a1, a2, a3, ...., a49 be in A.P. such that
33
66
68
34
D.
34
Let a1 = a and common difference = d
Given, a1 + a5 + a9 + ..... + a49 = 416
= a + 24d = 32 ...(i)
Also, a9 + a43 = 66
= a + 25d = 33 ...(ii)
Solving (i) & (ii),
We get d = 1, a = 8
Sponsor Area
Sponsor Area