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Angle between the planes $x+y+2 z=6$ and $2 x-y+z=9$ is
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Verified Answer
The correct answer is:
$\frac{\pi}{3}$
Equations of planes are
$x+y+2 z-6=0$
and $\quad 2 x-y+z-9=0$
$\cos \theta=\frac{a_{1} a_{2}+b_{1}b_{2}+c
_{1}c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}$
$=\frac{1 \times 2+1 \times(-1)+2 \times 1}{\sqrt{1+1+4} \sqrt{4+1+1}}$
$=\frac{2-1+2}{\sqrt{6} \times \sqrt{6}}=\frac{3}{6}=\frac{1}{2}$
$\Rightarrow \quad \cos \theta=\cos \left(\frac{\pi}{3}\right)$
$\therefore \quad \theta=\frac{\pi}{3}$
$x+y+2 z-6=0$
and $\quad 2 x-y+z-9=0$
$\cos \theta=\frac{a_{1} a_{2}+b_{1}b_{2}+c
_{1}c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}$
$=\frac{1 \times 2+1 \times(-1)+2 \times 1}{\sqrt{1+1+4} \sqrt{4+1+1}}$
$=\frac{2-1+2}{\sqrt{6} \times \sqrt{6}}=\frac{3}{6}=\frac{1}{2}$
$\Rightarrow \quad \cos \theta=\cos \left(\frac{\pi}{3}\right)$
$\therefore \quad \theta=\frac{\pi}{3}$
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