11  Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x – y + z = 0.

Answers (1)

The equation of the plane through the intersection of the given two planes, x+y+z =1  and  2x+3y+4z =5 is given in Cartesian form as;

(x+y+z-1) +\lambda(2x+3y+4z -5) = 0

or (1+2\lambda)x(1+3\lambda)y+(1+4\lambda)z-(1+5\lambda) = 0             ..................(1)

So, the direction ratios of (1) plane are a_{1},b_{1},c_{1} which are (1+2\lambda),(1+3\lambda),\ and\ (1+4\lambda).

Then, the plane in equation (1) is perpendicular to x-y+z= 0 whose direction ratios a_{2},b_{2},c_{2} are 1,-1,\ and\ 1.

As planes are perpendicular then,

a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} = 0

we get,

(1+2\lambda) -(1+3\lambda)+(1+4\lambda) = 0

or  1+3\lambda = 0

or  \lambda = -\frac{1}{3}

Then we will substitute the values of \lambda in the equation (1), we get

\frac{1}{3}x-\frac{1}{3}z+\frac{2}{3} = 0

or   x-z+2=0

This is the required equation of the plane.

 

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