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Let $\pi_1$ be a plane passing through the point $\bar{i}+\bar{j}+\bar{k}$ and perpendicular to the vector $-\bar{j}+2 \bar{k}$. Let the line $\mathrm{L}$ passing through the points $3 \bar{i}-2 \bar{j}+\bar{k}$ and $-\bar{i}+3 \bar{j}+\bar{k}$ be a normal to the plane $\pi_2$. If the angle between the planes $\pi_1$ and $\pi_2$ is $\theta$ then $\cos \theta=$
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Verified Answer
The correct answer is:
$\sqrt{\frac{5}{41}}$
We have to find angle between two plans $\pi_1$ and $\pi_2$.
for plane $\pi_1, \pi_2=\pi_1, \bar{\pi}_1=-\hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}}$
and for plane $\pi_2, \bar{\pi}_2=4 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}$
hence $\cos \theta=\frac{|5|}{\sqrt{5} \cdot \sqrt{41}}$
$=\sqrt{\frac{5}{41}}$
for plane $\pi_1, \pi_2=\pi_1, \bar{\pi}_1=-\hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}}$
and for plane $\pi_2, \bar{\pi}_2=4 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}$
hence $\cos \theta=\frac{|5|}{\sqrt{5} \cdot \sqrt{41}}$
$=\sqrt{\frac{5}{41}}$
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