JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let \(\mathrm{P}(\alpha, \beta, \gamma)\) be the image of the point \(\mathrm{Q}(3,-3,1)\) in the line \(\frac{x-0}{1}=\frac{y-3}{1}=\frac{z-1}{-1}\) and \(R\) be the point \((2,5,-1)\). If the area of the triangle \(\mathrm{PQR}\) is \(\lambda\) and \(\lambda^2=14 \mathrm{~K}\), then \(\mathrm{K}\) is equal to :
- A \(36\)
- B \(72\)
- C \(18\)
- D \(81\)
Answer & Solution
Correct Answer
(D) \(81\)
Step-by-step Solution
Detailed explanation
\( \mathrm{RQ}=\sqrt{1+64+4}=\sqrt{69} \) \( \overrightarrow{\mathrm{RQ}}=\hat{\ell}-8 \hat{\mathrm{j}}+2 \hat{\mathrm{k}} \) \( \overrightarrow{\mathrm{RS}}=\hat{\ell}+\hat{\mathrm{j}}-\hat{\mathrm{k}} \)…
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