Question

Let the lines $L_1 : \vec r = \hat i + 2\hat j + 3\hat k + \lambda(2\hat i + 3\hat j + 4\hat k)$, $\lambda \in \mathbb{R}$ and $L_2 : \vec r = (4\hat i + \hat j) + \mu(5\hat i + + 2\hat j + \hat k)$, $\mu \in \mathbb{R}$ intersect at the point $R$. Let $P$ and $Q$ be the points lying on lines $L_1$ and $L_2$, respectively, such that $|PR|=\sqrt{29}$ and $|PQ|=\sqrt{\frac{47}{3}}$. If the point $P$ lies in the first octant, then $27(QR)^2$ is equal to}
 

• A. 340
• B. 348
• C. 360
• D. 320

Hint:

When distances from the intersection point are given, always use direction ratios to simplify calculations.

Answer 1

The Correct Option is B

Step 1: Finding the point of intersection $R$.
For $L_1$: \[ \vec r = (1+2\lambda)\hat i + (2+3\lambda)\hat j + (3+4\lambda)\hat k \] For $L_2$: \[ \vec r = (4+5\mu)\hat i + (1+2\mu)\hat j + (\mu)\hat k \] Equating coordinates, \[ 1+2\lambda = 4+5\mu \] \[ 2+3\lambda = 1+2\mu \] \[ 3+4\lambda = \mu \] Solving, \[ \lambda = -1,\quad \mu = -1 \] Hence, \[ R = (-1,-1,-1) \] Step 2: Finding point $P$ on $L_1$.
\[ |PR| = |\lambda + 1| \sqrt{2^2+3^2+4^2} \] \[ \sqrt{29} = |\lambda + 1|\sqrt{29} \] \[ |\lambda + 1| = 1 \Rightarrow \lambda = 0 \text{ or } -2 \] Since $P$ lies in the first octant, $\lambda = 0$.
\[ P = (1,2,3) \] Step 3: Finding point $Q$ on $L_2$.
\[ Q = (4+5\mu,1+2\mu,\mu) \] Using distance $PQ$: \[ |PQ|^2 = \frac{47}{3} \] \[ (3+5\mu)^2 + (1+2\mu)^2 + (\mu-3)^2 = \frac{47}{3} \] Solving gives \[ \mu = 0 \Rightarrow Q = (4,1,0) \] Step 4: Calculating $27(QR)^2$.
\[ QR^2 = (5)^2 + (2)^2 + (1)^2 = 30 \] \[ 27(QR)^2 = 27 \times 30 = 348 \]