Question

If the shortest distance between the lines $\frac{x - \lambda}{2} = \frac{y - 4}{3} = \frac{z - 3}{4}$ and $\frac{x - 2}{4} = \frac{y - 4}{6} = \frac{z - 7}{8}$ is $\frac{13}{\sqrt{29}}$, then a value of $\lambda$ is:

• A. $-\frac{13}{25}$
• B. $\frac{13}{25}$
• C. 1
• D. -1

Answer 1

Answer: (C)

1

Answer 2

Let the parametric equations of the two lines be:

$\vec{r}_1 = (\lambda \hat{i} + 4 \hat{j} + 3 \hat{k}) + \alpha (2 \hat{i} + 3 \hat{j} + 4 \hat{k})$,

$\vec{r}_2 = (2 \hat{i} + 6 \hat{j} + 7 \hat{k}) + \beta (2 \hat{i} + 3 \hat{j} + 4 \hat{k})$.

Here, the direction vector for both lines is:

$\vec{b} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$,

and the position vectors of points on the lines are:

$\vec{a}_1 = \lambda \hat{i} + 4 \hat{j} + 3 \hat{k}$, $\vec{a}_2 = 2 \hat{i} + 6 \hat{j} + 7 \hat{k}$.

The formula for the shortest distance between two skew lines is:

$\text{Shortest distance} = \frac{|\vec{b} \times (\vec{a}_2 - \vec{a}_1) \cdot \vec{b}|}{|\vec{b}|} = \frac{13}{\sqrt{29}}$.

Substituting $\vec{a}_2 - \vec{a}_1$:

$\vec{a}_2 - \vec{a}_1 = (2 - \lambda) \hat{i} + 2 \hat{j} + 4 \hat{k}$.

The cross product $\vec{b} \times (\vec{a}_2 - \vec{a}_1)$ simplifies as follows:

$\vec{b} \times (\vec{a}_2 - \vec{a}_1) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\\ 2 & 3 & 4 \\\ 2 - \lambda & 2 & 4 \end{vmatrix}$.

Solving this determinant gives:

$\vec{b} \times (\vec{a}_2 - \vec{a}_1) = (-8 \hat{j} + 12 \hat{i} + 4 (2 - \lambda) \hat{j})$.

Taking the magnitude and using the formula for shortest distance:

$\text{Shortest distance} = \frac{|(-8 \hat{j} + 12 \hat{i})|}{13}$.

Finally solving the quadratic for $\lambda = 1$.