Consider the lines L1: x - 1 = y - 2 = z and L2: x - 2 = y = z - 1.

Question

Consider the lines L1: x - 1 = y - 2 = z and L2: x - 2 = y = z - 1. Let the feet of the perpendiculars from the point P(5, 1, -3) on the lines L1 and L2 be Q and R respectively. If the area of the triangle PQR is A, then 4A2 is equal to :

Accepted Answer

$$\begin{aligned}
& \mathrm{L}_1: \frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}-2}{1}=\frac{\mathrm{z}-0}{2} \
& \text { Let } \mathrm{Q}(\lambda+1, \lambda+2, \lambda) \
& \overrightarrow{\mathrm{PQ}}=(\lambda-4, \lambda-1, \lambda+3) \
& \overrightarrow{\mathrm{PQ}} \cdot \overrightarrow{\mathrm{~m}}=0
\end{aligned}$$

$$\begin{aligned}
&\begin{aligned}
& \Rightarrow \lambda-4+\lambda+1, \lambda+3=0 \
& \Rightarrow 3 \lambda=0 \
& \Rightarrow \lambda=0 \
& \Rightarrow \mathrm{Q}(1,2,0) \
& \mathrm{L}_2: \frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}-0}{1}=\frac{\mathrm{z}-1}{2}
\end{aligned}\
\&\text { Let } \mathrm{R}(\mu+2, \mu, \mu+1) \overrightarrow{\mathrm{PR}}=(\mu-3, \mu-1, \mu+4)\
\&\overrightarrow{\mathrm{PR}} \cdot \overrightarrow{\mathrm{n}}=0\
\&\mu-3+\mu-1+\mu+4=0\
\&\neq \mu=0\
\&\mathrm{R}(2,0,1)
\end{aligned}$$

$$\begin{aligned}
& \text { Area of } \triangle \mathrm{PQR}(\mathrm{~A})=\frac{1}{2}|\overrightarrow{\mathrm{PQ}} \times \overrightarrow{\mathrm{PR}}| \
& \mathrm{A}=\frac{1}{2}|(-4 \hat{\mathrm{i}}+\hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \times(-3 \hat{\mathrm{i}}+\hat{\mathrm{j}}+4 \hat{\mathrm{k}})| \
& \mathrm{A}=\frac{1}{2}|7(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})| \
& \left|\[\begin{gathered} \hat{\mathrm{i}} \quad  \hat{\mathrm{j}} \quad  \hat{\mathrm{k}} \ -4 \quad  1 \quad  3 \ -3 \quad  -1 \quad  4 \end{gathered}\end{aligned}$$\right| \
\& =7 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}+7 \hat{\mathrm{k}} \
\& 4 \mathrm{~A}\^2=49 \times 3=147
\]

Solutions · 1