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MHT CET · Maths · Three Dimensional Geometry

Let \(M\) and \(N\) be foots of the perpendiculars drawn from the point \(\mathrm{P}(a, a, a)\) on the lines \(x-\mathrm{y}=0, \mathrm{z}=1\) and \(x+\mathrm{y}=0, \mathrm{z}=-1\) respectively and if \(\angle \mathrm{MPN}=90^{\circ}\) then \(a^2=\)

  1. A 1
  2. B 4
  3. C 6
  4. D 9
Verified Solution

Answer & Solution

Correct Answer

(A) 1

Step-by-step Solution

Detailed explanation

\(\mathrm{M}\) is the foot of the perpendicular from \(\mathrm{P}(a, a, a)\) on \(x-y=0, z=1\). Line is \((\lambda, \lambda, 1)\). \(\vec{PM} = (\lambda-a, \lambda-a, 1-a)\). Direction vector \((1,1,0)\). \(\vec{PM} \cdot (1,1,0) = 0 \Rightarrow (\lambda-a) \cdot 1 + (\lambda-a) \cdot 1 + (1-a) \cdot 0 = 0 \Rightarrow 2(\lambda-a)=0 \Rightarrow \lambda=a\). So, \(\mathrm{M}(a, a, 1)\).