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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
If the image of the point $P(1, 2, a)$ in the line
$\frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2}$
is $Q(5, b, c)$, then $a^2 + b^2 + c^2$ is equal to
Solution
Point $M = \left(3,; \frac{b+1}{2},; \frac{c+a}{2}\right)$ satisfies the line
$\frac{3-6}{3} = \frac{\frac{b+1}{2}-7}{2} = \frac{\frac{c+a}{2}-7}{-2}$
$\Rightarrow -1 = \frac{b-12}{4} = \frac{c+a-14}{-4}$
$\Rightarrow b = 8 \quad ...(1)$
$c + a = 18 \quad ...(2)$
Now $PQ \perp L$
$\Rightarrow (4\hat{i} + (b-2)\hat{j} + (c-a)\hat{k}) \cdot (3\hat{i} + 2\hat{j} - 2\hat{k}) = 0$
$\Rightarrow 12 + 2(b-2) - 2(c-a) = 0$
$\Rightarrow 6 + (b-2) - (c-a) = 0$
$\Rightarrow b - c + a + 4 = 0$
$\Rightarrow 8 - c + a + 4 = 0$
$\Rightarrow c - a = 12 \quad ...(3)$
From (2) & (3):
$a = 3,; c = 15$
$\Rightarrow a^2 + b^2 + c^2 = 9 + 64 + 225 = 298$
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