Discussion
From a point \(P(\lambda, \lambda, \lambda)\), perpendiculars \(PQ\) and \(PR\) are drawn respectively on the lines \(y = x, z = 1\) and \(y = -x, z = -1\). If \(P\) is such that \(\angle QPR\) is a right angle, then the possible value(s) of \(\lambda\) is(are)
Answer: C
The only possible value is \(\lambda = -1\).
Line 1 is the set of points \((t, t, 1)\). The foot of the perpendicular from \(P(\lambda, \lambda, \lambda)\) minimises \(2(\lambda - t)^{2} + (\lambda - 1)^{2}\), so \(t = \lambda\) and \(Q = (\lambda, \lambda, 1)\).
Line 2 is the set of points \((s, -s, -1)\). Minimising \((\lambda - s)^{2} + (\lambda + s)^{2} + (\lambda + 1)^{2}\) gives \(s = 0\), so \(R = (0, 0, -1)\).
\[ \begin{aligned} \vec{PQ} &= (0, 0, 1 - \lambda), \quad \vec{PR} = (-\lambda, -\lambda, -1 - \lambda) \\ \vec{PQ} \cdot \vec{PR} &= -(1 - \lambda)(1 + \lambda) = \lambda^{2} - 1 = 0 \end{aligned} \]
So \(\lambda = \pm 1\). But for \(\lambda = 1\), P = (1, 1, 1) lies on line 1 itself, so Q = P and there is no angle QPR. Hence \(\lambda = -1\). The values \(\pm\sqrt{2}\) do not satisfy \(\lambda^{2} = 1\).
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