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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.

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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.

Posted by

Divya Prakash Singh

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