Q. 15

Find the equation of the plane passing through the intersection of the planes
5x - y + z = 10 and x + y - z = 4 and parallel to the line with direction ratios
2, 1, 1. Find also the perpendicular distance of (1, 1, 1) from this plane.

Answer :

Given – A plane passes through the intersection of 5x - y + z = 10 and x + y - z = 4 and parallel to the line with direction ratios (2, 1, 1)


To find – Equation of the plane


Tip – If ax + by + cz + d = 0 and a’x + b’y + c’z + d’ = 0 be two planes, then the equation of the plane passing through their intersection will be given by


(ax + by + cz + d) + λ(a’x + b’y + c’z + d’) = 0, where λ is any scalar constant


So, the equation of the plane maybe written as


(5x - y + z - 10) + λ(x + y - z - 4) = 0


(5 + λ)x + (- 1 + λ)y + (1 - λ)z + (- 10 - 4λ) = 0


This is plane parallel to a line with direction ratios (2, 1, 1)


So, the normal of this line with direction ratios ((5 + λ), (- 1 + λ), (1 - λ)) will be perpendicular to the given line.


Hence,


2(5 + λ) + (- 1 + λ) + (1 - λ) = 0


λ = - 5


The equation of the plane will be


(5 + (- 5))x + (- 1 + (- 5))y + (1 - Χ(- 5))z + (- 10 - 4Χ(- 5)) = 0


- 6y + 6z + 10 = 0


3y - 3z = 5


To find – Perpendicular distance of point (1, 1, 1) from the plane


Formula to be used - If ax + by + c + d = 0 be a plane and (a’, b’, c’) be the point, then the distance between them is given by


The distance between the plane and the line




units


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