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.

Answer 2

We are given two lines in symmetric form and the shortest distance between them is \(3\sqrt{30}\). Let

\[ L_1:\ \frac{x-3}{3}=\frac{y-\alpha}{-1}=\frac{z-3}{1}=\lambda \quad\Rightarrow\quad \mathbf{r}_1=(3,\alpha,3),\ \mathbf{a}_1=\langle 3,-1,1\rangle, \] \[ L_2:\ \frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-\beta}{4}=\mu \quad\Rightarrow\quad \mathbf{r}_2=(-3,-7,\beta),\ \mathbf{a}_2=\langle -3,2,4\rangle. \]

Concept Used:

The shortest distance \(d\) between skew lines with position vectors \(\mathbf{r}_1,\mathbf{r}_2\) and direction vectors \(\mathbf{a}_1,\mathbf{a}_2\) is

\[ d=\frac{\left|(\mathbf{r}_2-\mathbf{r}_1)\cdot(\mathbf{a}_1\times\mathbf{a}_2)\right|}{\left|\mathbf{a}_1\times\mathbf{a}_2\right|}. \]

Step-by-Step Solution:

Step 1: Compute \(\mathbf{a}_1\times\mathbf{a}_2\) and its magnitude.

\[ \mathbf{a}_1\times\mathbf{a}_2= \begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ 3&-1&1\\ -3&2&4 \end{vmatrix} =\langle -6,\,-15,\,3\rangle,\qquad \left|\mathbf{a}_1\times\mathbf{a}_2\right|=\sqrt{36+225+9}=3\sqrt{30}. \]

Step 2: Form \(\mathbf{r}_2-\mathbf{r}_1\) and the scalar triple product.

\[ \mathbf{r}_2-\mathbf{r}_1=\langle -6,\,-7-\alpha,\,\beta-3\rangle, \] \[ (\mathbf{r}_2-\mathbf{r}_1)\cdot(\mathbf{a}_1\times\mathbf{a}_2) = (-6)(-6)+(-7-\alpha)(-15)+(\beta-3)(3) = 36+105+15\alpha+3\beta-9 =132+15\alpha+3\beta. \]

Step 3: Use the given shortest distance \(d=3\sqrt{30}\):

\[ \frac{\left|132+15\alpha+3\beta\right|}{3\sqrt{30}}=3\sqrt{30} \ \Rightarrow\ \left|132+15\alpha+3\beta\right|=(3\sqrt{30})^2=270. \] \[ \Rightarrow\ 15\alpha+3\beta=\pm 138 \ \Rightarrow\ 5\alpha+\beta=\pm 46. \]

Final Computation & Result

The positive value of \(5\alpha+\beta\) is 46.