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

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:
$|\lambda| = 43$