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A bisector of the angle between the normals of the planes $4 x+3 y=5$ and $x+2 y+2 z=4$ is along the vector
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1987 Upvotes
Verified Answer
The correct answer is:
$(7 \hat{i}-\hat{j}-10 \hat{k})$
Given planes are $4 x+3 y-5=0 \& x+2 y+2 z-4=0$. Equation of plane which bisects the angle between the given planes.
$\begin{aligned}
& \frac{4 x+3 y-5}{\sqrt{16+9}}= \pm \frac{x+2 y+2 z-4}{\sqrt{1+4+4}} \\
& \frac{4 x+3 y-5}{5}= \pm \frac{x+2 y+2 z-4}{3}
\end{aligned}$
Take positive,
$\begin{aligned}
& 12 x+9 y-15=5 x+10 y+10 z-20 \\
& 7 x-y-10 z+5=0 \\
& \text { So, }(x \hat{i}+y \hat{j}+7 \hat{k})(7 \hat{i}-\hat{j}-10 \hat{k})+5=0
\end{aligned}$
Take negative,
$\begin{aligned}
& 12 x+9 y-15=-5 x-10 y-10 z+20 \\
& 17 x+19 y+10 z-35=0
\end{aligned}$
Therefore, required normal vector is $7 \hat{i}-\hat{j}-10 \hat{k}$
$\begin{aligned}
& \frac{4 x+3 y-5}{\sqrt{16+9}}= \pm \frac{x+2 y+2 z-4}{\sqrt{1+4+4}} \\
& \frac{4 x+3 y-5}{5}= \pm \frac{x+2 y+2 z-4}{3}
\end{aligned}$
Take positive,
$\begin{aligned}
& 12 x+9 y-15=5 x+10 y+10 z-20 \\
& 7 x-y-10 z+5=0 \\
& \text { So, }(x \hat{i}+y \hat{j}+7 \hat{k})(7 \hat{i}-\hat{j}-10 \hat{k})+5=0
\end{aligned}$
Take negative,
$\begin{aligned}
& 12 x+9 y-15=-5 x-10 y-10 z+20 \\
& 17 x+19 y+10 z-35=0
\end{aligned}$
Therefore, required normal vector is $7 \hat{i}-\hat{j}-10 \hat{k}$
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