Q. 145.0( 3 Votes )

# Find the equation of the plane passing through the intersection of the planes x – 2y + z = 1 and 2x + y + z = 8 and parallel to the line with direction ratios proportional to 1, 2, 1. Find also the perpendicular distance of (1, 1, 1) from this plane.

Answer :

We know that equation of plane passing through the intersection of planes a_{1}x + b_{1}y + c_{1}z + d_{1} = 0 and a_{2}x + b_{2}y + c_{2}z + d_{2} = 0 is given by

(a_{1}x + b_{1}y + c_{1}z + d_{1}) + k(a_{2}x + b_{2}y + c_{2}z + d_{2}) = 0

So, equation of plane passing through the intersection of planes

x – 2y + z – 1 = 0 and 2x + y + z – 8 = 0 is

(x – 2y + z – 1) + k(2x + y + z – 8) = 0 ……(1)

⇒ x(1 + 2k) + y(– 2 + k) + z(1 + k) + (– 1 – 8k) = 0

We know that line is parallel to plane a_{2}x + b_{2}y + c_{2}z + d_{2} = 0 if a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 0

Given the plane is parallel to line with direction ratios 1,2,1

1×(1 + 2k) + 2×(– 2 + k) + 1×(1 + k) = 0

⇒ 1 + 2k – 4 + 2k + 1 + k = 0

⇒ k =

Putting the value of k in equation (1)

⇒

⇒

⇒ 9x – 8y + 7z – 21 = 0

We know that the distance (D) of point (x_{1},y_{1},z_{1}) from plane ax + by + cz – d = 0 is given by

o, distance of point (1,1,1) from plane (1) is

⇒

⇒

⇒

Taking the mod value we have

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