Question

Let the line $L_1$ be parallel to the vector $-3\hat{i }+ 2\hat{j} + 4\hat{k}$ and pass through the point $(2, 6, 7)$, and the line $L_2$ be parallel to the vector $2\hat{i} + \hat{j} + 3\hat{k}$ and pass through the point $(4, 3, 5)$. If the line $L_3$ is parallel to the vector $-3\hat{i} + 5\hat{j} + 16\hat{k}$ and intersects the lines $L_1$ and $L_2$ at the points $C$ and $D$, respectively, then $|\vec{CD}|^2$ is equal to :

• A. 290
• B. 89
• C. 312
• D. 171

Hint:

When a line intersects two other lines, its direction vector must be proportional to the vector connecting arbitrary points on those two lines.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We define points $C$ and $D$ on lines $L_1$ and $L_2$ using parametric coordinates and use the condition that $\vec{CD}$ is parallel to a given direction.

Step 2: Detailed Explanation:
Point $C$ on $L_1$: $\vec{c} = (2 - 3\lambda, 6 + 2\lambda, 7 + 4\lambda)$.
Point $D$ on $L_2$: $\vec{d} = (4 + 2\mu, 3 + \mu, 5 + 3\mu)$.
Vector $\vec{CD} = \vec{d} - \vec{c} = (2 + 2\mu + 3\lambda, -3 + \mu - 2\lambda, -2 + 3\mu - 4\lambda)$.
Since $\vec{CD}$ is parallel to $\vec{v} = (-3, 5, 16)$:
\[ \frac{2 + 2\mu + 3\lambda}{-3} = \frac{-3 + \mu - 2\lambda}{5} = \frac{-2 + 3\mu - 4\lambda}{16} = k \]
Solving first two: $10 + 10\mu + 15\lambda = 9 - 3\mu + 6\lambda \implies 13\mu + 9\lambda = -1$.
Solving last two: $-48 + 16\mu - 32\lambda = -10 + 15\mu - 20\lambda \implies \mu - 12\lambda = 38$.
Solving these equations gives $\lambda = -3$ and $\mu = 2$.
Now, $k = \frac{2 + 2(2) + 3(-3)}{-3} = \frac{-3}{-3} = 1$.
So $\vec{CD} = 1 \cdot (-3, 5, 16)$.
\[ |\vec{CD}|^2 = (-3)^2 + 5^2 + 16^2 = 9 + 25 + 256 = 290. \]

Step 3: Final Answer:
$|\vec{CD}|^2 = 290$.