Let the foot of perpendicular from P(5, 1, -3) on the line L1: x-1=y-2=z and L2: x-1=y=z-1 is Q and R, respectively.

Question

Let the foot of perpendicular from P(5, 1, -3) on the line \(L_{1}: x-1=y-2=z\) and \(L_{2}: x-2=y=z-1\) is Q and R, respectively. The area of triangle P and R is equal to

(1) \(\frac{7}{2}\)

(2) \(\frac{7}{\sqrt{2}}\)

(3) \(\frac{7 \sqrt{3}}{2}\)

(4) 7

Answer 1

Correct option is: (3) \(\frac{7 \sqrt{3}}{2}\) 

P(5, 1, -3)

\(L_{1}: x-1=y-2=z=\lambda\)

\(L_{2}: x-2=y=z-1=\mu\)

Any point of \(L_{1}\) is \(Q(\lambda+1, \lambda+2, \lambda)\)

Any point of \(L_{2}\ \text{is}\ R(\mu+2, \mu, \mu+1)\)

Now \(P Q<\lambda-4, \lambda+1, \lambda+3\rangle \cdot\langle 1,1,1\rangle=0\)

\(\lambda-4+\lambda+1+\lambda+3=0\)

\(3 \lambda=0\)

\(\Rightarrow \lambda=0\)

\(\therefore\ Q(1,2,0)\)