Question

Let the shortest distance between the lines $\frac{x-3}{3} = \frac{y-\alpha}{-1} = \frac{z-3}{1}$ and $\frac{x+3}{-3} = \frac{y+7}{2} = \frac{z-\beta}{4}$ be $3\sqrt{30}$. Then the positive value of $5\alpha + \beta$ is

• A. 42
• B. 46
• C. 48
• D. 40

Hint:

The shortest distance between two skew lines can be found using the vector cross product and dot product.

Answer 1

The Correct Option is B

1. Identify the points and direction vectors: 
- Line 1: $\frac{x-3}{3} = \frac{y-\alpha}{-1} = \frac{z-3}{1}$ 
- Point $A(3, \alpha, 3)$ 
- Direction vector $\vec{p} = 3\hat{i} - \hat{j} + \hat{k}$ 
- Line 2: $\frac{x+3}{-3} = \frac{y+7}{2} = \frac{z-\beta}{4}$ 
- Point $B(-3, -7, \beta)$ 
- Direction vector $\vec{q} = -3\hat{i} + 2\hat{j} + 4\hat{k}$ 
2. Calculate $\vec{p} \times \vec{q}$: \[ \vec{p} \times \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} 3 & -1 & 1 \\ -3 & 2 & 4 \end{vmatrix} = 6\hat{i} + 15\hat{j} - 9\hat{k} \] 
3. Calculate $\vec{BA}$: \[ \vec{BA} = (3 + 3)\hat{i} + (\alpha + 7)\hat{j} + (3 - \beta)\hat{k} = 6\hat{i} + (\alpha + 7)\hat{j} + (3 - \beta)\hat{k} \] 
4. Use the distance formula: \[ \frac{|\vec{BA} \cdot (\vec{p} \times \vec{q})|}{|\vec{p} \times \vec{q}|} = 3\sqrt{30} \] \[ \frac{|6 \cdot 6 + 15(\alpha + 7) - 9(3 - \beta)|}{\sqrt{6^2 + 15^2 + (-9)^2}} = 3\sqrt{30} \] \[ 36 + 15(\alpha + 7) - 9(3 - \beta) = 270 \] \[ 15\alpha + 3\beta = 138 \] \[ 5\alpha + \beta = 46 \] Therefore, the correct answer is (2) 46.