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Question

# Find the vector 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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Solution

## $\text{The equation of the plane passing through the line of intersection of the given planes is}\phantom{\rule{0ex}{0ex}}x+y+z-1+\lambda \left(2x+3y+4z-5\right)=0\phantom{\rule{0ex}{0ex}}\left(1+2\lambda \right)x+\left(1+3\lambda \right)y+\left(1+4\lambda \right)z-1-5\lambda =0...\left(1\right)\phantom{\rule{0ex}{0ex}}\text{This plane is perpendicular to}x-y+z=0.\text{So,}\phantom{\rule{0ex}{0ex}}1+2\lambda -1\left(1+3\lambda \right)+1+4\lambda =0\text{(Because}{a}_{1}{a}_{2}+{b}_{1}{b}_{2}+{c}_{1}{c}_{2}=0\text{)}\phantom{\rule{0ex}{0ex}}⇒1+2\lambda -1-3\lambda +1+4\lambda =0\phantom{\rule{0ex}{0ex}}⇒3\lambda +1=0\phantom{\rule{0ex}{0ex}}⇒\lambda =\frac{-1}{3}\phantom{\rule{0ex}{0ex}}\text{Substituting this in (1), we get}\phantom{\rule{0ex}{0ex}}\left(1+2\left(\frac{-1}{3}\right)\right)x+\left(1+3\left(\frac{-1}{3}\right)\right)y+\left(1+4\left(\frac{-1}{3}\right)\right)z-1-5\left(\frac{-1}{3}\right)=0\phantom{\rule{0ex}{0ex}}⇒x-z+2=0$

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