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What is the cosine of angle between the planes $\mathrm{x}+\mathrm{y}+\mathrm{z}+\mathrm{l}=0$ and $2 \mathrm{x}-2 \mathrm{y}+2 \mathrm{z}+1=0 ?$
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$1 / 3$
If $\theta$ be the angle between the planes $a_{1} x+b_{1} y+c_{1} z+d_{1}=0$ and $a_{2} x+b_{2} y+c_{2} z+d_{2}=0$
then $\cos \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \cdot \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}$
$\begin{aligned} \cos \theta &=\frac{1(2)+1(-2)+1(2)}{\sqrt{1^{2}+1^{2}+1^{2}}} \cdot \sqrt{2^{2}+(-2)^{2}+2^{2}} \\ &=\frac{2}{\sqrt{3} 2 \sqrt{3}}=\frac{1}{3} \end{aligned}$
then $\cos \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \cdot \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}$
$\begin{aligned} \cos \theta &=\frac{1(2)+1(-2)+1(2)}{\sqrt{1^{2}+1^{2}+1^{2}}} \cdot \sqrt{2^{2}+(-2)^{2}+2^{2}} \\ &=\frac{2}{\sqrt{3} 2 \sqrt{3}}=\frac{1}{3} \end{aligned}$
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