Let the foot of the perpendicular from the point (1, 2, 4) on the line $${{x + 2} \over 4} = {{y - 1} \over 2} = {{z + 1} \over 3}$$ be P. Then the distance of P from the plane $$3x + 4y + 12z + 23 = 0$$ is :
JEE · Math · previous-year question
- A.5correct
- B.$${{50} \over {13}}$$
- C.4
- D.$${{63} \over {13}}$$
Answer
A. 5
Explanation
$$L:{{x + 2} \over 4} = {{y - 1} \over 2} = {{z + 1} \over 3} = t$$ Let P = (4t $$-$$ 2, 2t + 1, 3t $$-$$ 1) $$\because$$ P is the foot of perpendicular of (1, 2, 4) $$\therefore$$ $$4(4t - 3) + 2(2t - 1) + 3(3t - 5) = 0$$ $$ \Rightarrow 29t = 29 \Rightarrow t = 1$$ $$\therefore$$ P = (2, 3, 2) Now, distance of P from the plane $$3x + 4y + 12z + 23 = 0$$, is $$\left| {{{6 + 12 + 24 + 23} \over {\sqrt {9 + 16 + 144} }}} \right| = {{65} \over {13}} = 5$$
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