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

equation of the plane passing through the intersection of the plane 4x-y+z=10 and the plane x+y-z=4 is given by: since  the plane (1) is  parallel to the line with the direction ratios proportional  to (2,1,1)

therefore substituting the value of λ in (1)

-5y+5z+6=0 i.e. 5y-5z-6=0...(1a)

(ii) let perpendicular distance of (1,1,1) from plane 1(a) be l. • 20

The equation of the plane passing through

the intersection of the two planes is

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

From this you get the dr's of the plane as

(4+k):(-1+k):(1-k)

Now using a1a2+b1b2+c1c2=0 with

the point (2,1,1),

you get k value as -4...

You substitute it in eqn---1..

So u get the eqn of the plane..

Now with its drs and the given point,

distance can be foundout using distance formula...

Got it?

• -7

Equation of plane::  5Z - 5Y + 6.

dist. of point (1,1,1) from plane = (6/5)(1/2)1/2

• -4
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