Three Dimensional Geometry Question 14
Question 14 - 29 January - Shift 1
Let the equation of the plane $P$ containing the line $x+10=\frac{8-y}{2}=z$ be $ax+by+3 z=2(a+b)$ 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
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Answer: 355
Solution:
Formula: Direction Cosines And Direction Ratio, A plane and a point
The line $\frac{x+10}{1}=\frac{y-8}{-2}=\frac{z}{1}$ have a point $(-10,8,0)$
with d. r. $(1,-2,1)$
$\because$ The plane $a x+b y+3 z=2(a+b)$
$\Rightarrow b=2 a$
and dot product of d.r.’s is zero
$\therefore a-2 b+3=0$
$\therefore a=1 , b=2$
Distance from $(1,27,7)$ is
$c=\frac{1+54+21-6}{\sqrt{14}}=\frac{70}{\sqrt{14}}=5 \sqrt{14}$
$\therefore a^{2}+b^{2}+c^{2}=1+4+350$
$=355$
