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Let $P(1,-2,5)$ be the foot of the perpendicular drawn from the origin to the plane $\pi_1$ and the same $P$ be the foot of the perpendicular from $(1,2,-1)$ to the plane $\pi_2$. Then the acute angle between the planes $\pi_1$ and $\pi_2$ is
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Verified Answer
The correct answer is:
$\cos ^{-1}\left(\frac{19}{\sqrt{390}}\right)$
According to given informations, the direction
ratios of normal to the plane $\pi_1$ are $1-0,-2-0$, $5-0$ i.e. $1,-2,5$.
And similarly direction ratios of normal to the plane $\pi_2$ are
$1-1,2-(-2),-1-5$ i.e. $0,4,-6$
So, the acute angle between planes $\pi_1$ and $\pi_2$ is
$\cos ^{-1}\left|\left(\frac{(1)(0)+(-2)(4)+(5)(-6)}{\sqrt{1+4+25} \sqrt{0+16+36}}\right)\right|$
$\begin{aligned}=\cos ^{-1}\left|\frac{-8-30}{\sqrt{30} \sqrt{52}}\right| & =\cos ^{-1}\left(\frac{38}{2 \sqrt{30 \times 13}}\right) \\ & =\cos ^{-1} \frac{19}{\sqrt{390}}\end{aligned}$
Hence, option (a) is correct.
ratios of normal to the plane $\pi_1$ are $1-0,-2-0$, $5-0$ i.e. $1,-2,5$.
And similarly direction ratios of normal to the plane $\pi_2$ are
$1-1,2-(-2),-1-5$ i.e. $0,4,-6$
So, the acute angle between planes $\pi_1$ and $\pi_2$ is
$\cos ^{-1}\left|\left(\frac{(1)(0)+(-2)(4)+(5)(-6)}{\sqrt{1+4+25} \sqrt{0+16+36}}\right)\right|$
$\begin{aligned}=\cos ^{-1}\left|\frac{-8-30}{\sqrt{30} \sqrt{52}}\right| & =\cos ^{-1}\left(\frac{38}{2 \sqrt{30 \times 13}}\right) \\ & =\cos ^{-1} \frac{19}{\sqrt{390}}\end{aligned}$
Hence, option (a) is correct.
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