Question

If the shortest distance between the lines \[ \frac{x - \lambda}{3} = \frac{y - 2}{-1} = \frac{z - 1}{1} \] and \[ \frac{x + 2}{-3} = \frac{y + 5}{2} = \frac{z - 4}{4} \] is \[ \frac{44}{\sqrt{30}}, \] then the largest possible value of $|\lambda|$ is equal to ________.

Answer 1

Correct Answer: 43

Let
\[\vec{a}_1 = \lambda \hat{i} + 2 \hat{j} + \hat{k}\]
\[\vec{a}_2 = -2 \hat{i} - 5 \hat{j} + 4 \hat{k}\]
\[\vec{p} = 3 \hat{i} - \hat{j} + \hat{k}\]
\[\vec{q} = 3 \hat{i} + 2 \hat{j} + 4 \hat{k}\]
Now, the vector \( \vec{a}_1 - \vec{a}_2 \) is given by:
\[(\lambda + 2) \hat{i} + 7 \hat{j} - 3 \hat{k}\]
Calculate \( \vec{p} \times \vec{q} \):
\[\vec{p} \times \vec{q} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\3 & -1 & 1 \\3 & 2 & 4\end{vmatrix}= -6 \hat{i} - 15 \hat{j} + 3 \hat{k}\]
The shortest distance \( d \) is given by:
\[d = \frac{|(\vec{a}_1 - \vec{a}_2) \cdot (\vec{p} \times \vec{q})|}{|\vec{p} \times \vec{q}|}\]
Substitute the values:
\[d = \frac{| -6\lambda - 12 - 105 + 9|}{\sqrt{(-6)^2 + (-15)^2 + 3^2}} = \frac{|6\lambda + 126|}{3 \sqrt{30}}\]
Equating, we get:
\[\frac{44}{\sqrt{30}} = \frac{|6\lambda + 126|}{3 \sqrt{30}}\]
\[132 = |6\lambda + 126|\]
Solving for \( \lambda \):
\[\lambda = 1 \quad \text{or} \quad \lambda = -43\]
Thus, the largest possible value of \( |\lambda| \) is:
\[|\lambda| = 43\]