From a point $P (λ, λ, λ)$ , perpendiculars $P Q$ and $P R$ are drawn respectively on the lines $y = x, z = 1$ and $y = - x, z = - 1$ . If $P$ is such that $∠ Q P R$ is a right angle, then the possible value(s) of $λ$ is(are)

\sqrt{2}

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- \sqrt{2}

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Solution

The correct option is C $- 1$
Equation of the lines are:
$y = x, z = 1 L_{1} : \frac{x}{1} = \frac{y}{1} = \frac{z - 1}{0}$
Point on the line $Q = (a, a, 1)$

$y = - x, z = - 1 L_{2} : \frac{x}{- 1} = \frac{y}{1} = \frac{z + 1}{0}$
Point on the line $R = (- b, b, - 1)$
Point $P = (λ, λ, λ)$
So,
$- - \to P Q = (a - λ)^i + (a - λ)^j + (1 - λ)^k$
$P Q$ is perpendicular to line $L_{1}$ ,
$(a - λ) + (a - λ) + 0 = 0 \Rightarrow λ = a$
$- - \to P R = (- b - λ)^i + (b - λ)^j + (- 1 - λ)^k$
Similarly $P R$ is perpendicular to line $L_{2}$ ,
$- 1 \times (- b - λ) + (b - λ) + 0 = 0 \Rightarrow b = 0$
As, $P Q$ and $P R$ is perpendicular,
$(1 - a) \times (- 1 - a) = 0 \Rightarrow a = \pm 1$
But when $a = 1$ point $P$ and $Q$ become same, so
$λ = a = - 1$

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