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Three Dimensional Geometry

Question
CBSEENMA12035914

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  x - 41 = y + 3-4 = z + 17

Solution

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)

Given equation of the line is  x - 41 = y + 3- 4 = z + 17We can also write the above equation as  x - 41 = y - ( -  3 )- 4 = z - ( -1 )7

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:

a( - 1 ) x 7 - ( - 4 ) x 2 = b2 x 1 - 3 x 7 = c3 x ( -4 ) - 1 x ( - 1 ) a- 7 + 8 = b2 - 21 = c- 12 + 1 a1 = b- 19 = c- 11 = k a = k,  b = - 19 k,  c = - 11 k

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 .

Some More Questions From Three Dimensional Geometry Chapter

Find the direction cosines of x, y and z-axis.