JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let \(\mathrm{Q}\) and \(\mathrm{R}\) be the feet of perpendiculars from the point \(\mathrm{P}(\mathrm{a}, \mathrm{a}, \mathrm{a})\) on the lines \(\mathrm{x}=\mathrm{y}, \mathrm{z}=1\) and \(\mathrm{x}=-\mathrm{y}\), \(\mathrm{z}=-1\) respectively. If \(\angle \mathrm{QPR}\) is a right angle, then \(12 \mathrm{a}^2\) is equal to
- A \(13\)
- B \(14\)
- C \(10\)
- D \(12\)
Answer & Solution
Correct Answer
(D) \(12\)
Step-by-step Solution
Detailed explanation
\({x}{1}=\frac{y}{1}=\frac{z-1}{0}=r \rightarrow Q(r, r, 1) \) \(\frac{x}{1}=\frac{y}{-1}=\frac{z+1}{0}=k \rightarrow R(k,-k,-1) \) \(\overline{P Q}=(a-r) \hat{i}+(a-r) \hat{j}+(a-1) \hat{k} \) \(a=r+a-r=0 \) \(2 a=2 r \rightarrow a=r \)…
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