Let the equation of the plane P containing the line $x+10=\frac{8-y}{2}=z$ be $ax+by+3z=2(a+b)$ and the distance of the plane $P$ from the point (1, 27, 7) be $c$. Then $a^2+b^2+c^2$ is equal to __________.
Answer (integer)
355
Solution
The line $\frac{x+10}{1}=\frac{y-8}{-2}=\frac{z}{1}$ have a point $(-10,8,0)$ with d. r. $(1,-2,1)$
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$\because$ the plane $a x+b y+3 z=2(a+b)$
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$\Rightarrow \mathrm{b}=2 \mathrm{a}$
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$\&$ dot product of d.r.'s is zero
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$\therefore \mathrm{a}-2 \mathrm{~b}+3=0$ $\therefore \mathrm{a}=1 \& \mathrm{~b}=2$
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Distance from $(1,27,7)$ is
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$c=\frac{1+54+21-6}{\sqrt{14}}=\frac{70}{\sqrt{14}}=5 \sqrt{14}$
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$\therefore \mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}=1+4+350$
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$=355$
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
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