Question

Let the values of $ p $, for which the shortest distance between the lines $ \frac{x + 1}{3} = \frac{y}{4} = \frac{z}{5} $ and $ \vec{r} = (p \hat{i} + 2 \hat{j} + \hat{k}) + \lambda (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) $ is $ \frac{1}{\sqrt{6}} $, be $ a, b $, where $ a<b $. Then the length of the latus rectum of the ellipse $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $ is:

• A. 9
• B. \( \frac{3}{2} \)
• C. \( \frac{2}{3} \)
• D. 18

Hint:

To find the shortest distance between two skew lines, use the cross product of the direction vectors and the formula for the distance.

Answer 1

The Correct Option is C

The shortest distance between two skew lines is given by the formula: \[ d = \frac{| (\vec{a} - \vec{b}) \cdot (\vec{p} \times \vec{q}) |}{|\vec{p} \times \vec{q}|} \] where \( \vec{a} = -\hat{i} + 0 \hat{j} + 0 \hat{k} \), \( \vec{b} = \pi \hat{i} + 2 \hat{j} + \hat{k} \), \( \vec{p} = 3 \hat{i} + 4 \hat{j} + 5 \hat{k} \), and \( \vec{q} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k} \). 
We compute \( \vec{a} - \vec{b} \) as: \[ \vec{a} - \vec{b} = (-1 - \pi) \hat{i} - 2 \hat{j} - \hat{k} \] Now, we compute the cross product \( \vec{p} \times \vec{q} \): \[ \vec{p} \times \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} 3 & 4 & 5 2 & 3 & 4 \end{vmatrix} \] \[ = \hat{i} \begin{vmatrix} 4 & 5 3 & 4 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & 5 \\ 2 & 4 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & 4 \\ 2 & 3 \end{vmatrix} \] \[ = \hat{i} (16 - 15) - \hat{j} (12 - 10) + \hat{k} (9 - 8) \] \[ = \hat{i} - 2 \hat{j} + \hat{k} \] Now, we calculate the magnitude of the cross product: \[ |\vec{p} \times \vec{q}| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{6} \] Using the formula for the shortest distance: \[ d = \frac{| (-1 - \pi) \hat{i} - 2 \hat{j} - \hat{k} \cdot \hat{i} - 2 \hat{j} + \hat{k} |}{\sqrt{6}} = \frac{1}{\sqrt{6}} \] This yields the condition for the distance, and solving for the length of the latus rectum of the ellipse: \[ \text{L.R.} = \frac{2a^2}{b} \] 
Thus, the correct answer is \( \frac{2}{3} \).