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UPSC NDA I 2015 Mathematics Solved Paper

For the next two (2) items that follow:

From the point

P(3,-1,11)

,a perpendicular is drawn

on the line L given by the equation

\frac{x}{2}=\frac{y-2}{3}=\frac{z-3}{4}

Let Q be the foot of the perpendicular

[NDA I 2015]

Question : 55 of 120

Marks: +1, -0

Solution:

The vector joining P and Q will be perpendicular to the direction vector of the line.

Hence, the dot product of the vector joining P and Q and the direction vector of the line will be zero.

Let

\frac{x}{2}=\frac{y-2}{3}=\frac{z-3}{4}=t

x=2t

y=3t+2

z=4t+3

Hence

\vec{PQ}=(2t-3)\hat{i}+(3t+3)\hat{j}+(4t-8)\hat{k}

\vec{PQ}\cdot\vec{n}=((2t-3)\hat{i}+(3t+3)\hat{j}+(4t-8)\hat{k})\cdot(2\hat{i}+3\hat{j}+4\hat{k})=0

\therefore 4t-6+9t+9+16t-32=0

29t-29=0

t=1

\therefore \vec{PQ}=(2t-3)\hat{i}=(3t+3)\hat{j}+(4t-8)\hat{k}

=(-1)\hat{i}+6\hat{j}-4\hat{k}

Therefore, the direction ratios of the segment PQ will be

Dr's

=\langle -1,6,-4\rangle

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