Question

If O be the origin and the coordinates of P be (1, 2, −3), then find the equation of the plane passing through P and perpendicular to OP.

Answer 1

The coordinates of point P are ( 1,2,−3 ).

The direction ratios of OP are given by,

1−0=1 2−0=2 −3−0=−3

The equation of plane having direction ratios of normal are a,b,c, passing through the point ( x 1 , y 1 , z 1 ) is given by,

a( x− x 1 )+b( y− y 1 )+c( z− z 1 )=0

Substitute the values of ( x 1 , y 1 , z 1 ) as ( 1,2,−3 ) and a,b,c as 1,2,−3 respectively,

1( x−1 )+2( y−2 )−3( z+3 )=0 x+2y−3z−14=0

Thus, the equation of plane passing through the origin andpoint P ( 1,2,−3 ) is x+2y−3z−14=0.