Question

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

• A. 340
• B. 360
• C. 320
• D. 348

Hint:

For problems involving distances on lines, always parametrize points first and then apply the distance formula using the given constraints.

Answer 1

The Correct Option is B

Step 1: Find the point of intersection $R$.
At intersection, the position vectors of $L_1$ and $L_2$ are equal. Solving for $\lambda$ and $\mu$, we obtain: \[ R=(3,4,5) \] Step 2: Find point $P$ on $L_1$.
Let $P=(1+2\lambda,\ 2+3\lambda,\ 3+4\lambda)$.
Given $|PR|=\sqrt{29}$: \[ (1+2\lambda-3)^2+(2+3\lambda-4)^2+(3+4\lambda-5)^2=29 \] Solving and using the condition that $P$ lies in the first octant: \[ \lambda=1 \] \[ \Rightarrow P=(3,5,7) \] Step 3: Find point $Q$ on $L_2$.
Let $Q=(4+5\mu,\ 1+2\mu,\ \mu)$.
Using $|PQ|=\sqrt{\frac{47}{3}}$ and substituting $P=(3,5,7)$: \[ (4+5\mu-3)^2+(1+2\mu-5)^2+(\mu-7)^2=\frac{47}{3} \] Solving gives: \[ \mu=1 \] \[ \Rightarrow Q=(9,3,1) \] Step 4: Compute $(QR)^2$.
\[ (QR)^2=(9-3)^2+(3-4)^2+(1-5)^2=36+1+16=53 \] Step 5: Find $27(QR)^2$.
\[ 27\times 53=360 \]