Let Q and R be the feet of perpendiculars from the point P(a, a, a) on the lines x = y, z = 1 and x = –y, z = –1 respectively. If is a right angle, then is equal to __________. [2024]

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Solution

Q.
Let Q and R be the feet of perpendiculars from the point P(a, a, a) on the lines x = y, z = 1 and x = –y, z = –1 respectively. If $∠ Q P R$ is a right angle, then $12 a^{2}$ is equal to __________. [2024]

Ans.
(12)

Line $L_{1}$ is given by y = x, z = 1 can be expressed as

$L_{1} : \frac{x}{1} = \frac{y}{1} = \frac{z - 1}{0} = α$

$\Rightarrow x = α, y = α, z = 1$

Let the coordinate of Q on $L_{1}$ be $(α, α, 1)$

Line $L_{2}$ given by y = –x, z = –1 can be expressed as

$L_{2} : \frac{x}{1} = \frac{y}{- 1} = \frac{z + 1}{0} = β$ (say)

$\Rightarrow x = β, y = - β, z = - 1$

Let the coordinates of R on $L_{2}$ be $(β, - β, - 1)$

Direction ratios of PQ are $(a - α, a - α, a - 1)$ .

Now $P Q ⊥ L_{1}$

$∴ 1 (a - α) + 1 (a - α) + 0 (a - 1) = 0 \Rightarrow a = α$

Hence Q(a, a, 1)

Direction ratios of PR are $a - β, a + β, a + 1$

Now $P R ⊥ L_{2}$

$∴ 1 (a - β) + (- 1) (a + β) + 0 (a + 1) = 0 \Rightarrow β = 0$

Hence R(0, 0, —1)

Now, as $∠ Q P R = 90 °$

$\Rightarrow$ (a – a)(a – 0) + (a – a)(a – 0) + (a – 1)(a + 1) = 0

$\Rightarrow$ (a –1)(a + 1) = 0 $\Rightarrow$ a = 1 or a = –1

a = 1, rejected as P and Q are different points

$\Rightarrow$ a = –1, then $12 a^{2} = 12 \times {(- 1)}^{2} = 12$ .

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