Question

Let P(α, β, γ) be the image of the point 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 PQR is λ and λ² = 14K, then K is equal to:

(1) 36

(2) 72

(3) 18

(4) 81

Answer 1

Correct option is (4) 81

\(\mathrm{RQ}=\sqrt{1+64+4}=\sqrt{69}\)

\(\overrightarrow{\mathrm{RQ}}=\hat{\ell}-8 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}\)

\(\overrightarrow{\mathrm{RS}}=\hat{\imath}+\hat{\mathrm{j}}-\hat{\mathrm{k}}\)

\(\cos \theta=\frac{\overrightarrow{\mathrm{RQ}} \cdot \overrightarrow{\mathrm{RS}}}{|\overrightarrow{\mathrm{RQ}}||\overrightarrow{\mathrm{RS}}|}=\left|\frac{1-8-2}{\sqrt{69} \sqrt{3}}\right|=\frac{9}{3 \sqrt{23}}\)

\(\cos \theta=\frac{3}{\sqrt{23}}=\frac{\mathrm{RS}}{\mathrm{RQ}}=\frac{\mathrm{RS}}{\sqrt{69}}\)

\(\mathrm{RS}=3 \sqrt{3}\)

\(\sin \theta=\frac{\sqrt{14}}{\sqrt{23}}=\frac{\mathrm{QS}}{\sqrt{69}}\)

\(\mathrm{QS}=\sqrt{42}\)

area \(=\frac{1}{2} \cdot 2 \mathrm{QS} \cdot \mathrm{RS}=\sqrt{42} \cdot 3 \sqrt{3}\)

\(\lambda=9 \sqrt{14}\)

\(\lambda^{2}=81.14=14 \mathrm{k}\)

\(\mathrm{k}=81\)