Find the shortest distance between the lines given by $\mathbf{r} = (8 + 3\lambda)\hat{i} - (9 + 16\lambda)\hat{j} + (10 + 7\lambda)\hat{k}$ and $\mathbf{r} = 15\hat{i} + 29\hat{j} + 5\hat{k} + \mu(3\hat{i} + 8\hat{j} - 5\hat{k})$.

$14$ units

The correct answer is Option (2) → 2 ##

We have, $\mathbf{r} = (8 + 3\lambda)\hat{i} - (9 + 16\lambda)\hat{j} + (10 + 7\lambda)\hat{k}$

$= 8\hat{i} - 9\hat{j} + 10\hat{k} + 3\lambda\hat{i} - 16\lambda\hat{j} + 7\lambda\hat{k}$

$= 8\hat{i} - 9\hat{j} + 10\hat{k} + \lambda(3\hat{i} - 16\hat{j} + 7\hat{k})$

On comparing with $\mathbf{r} = \mathbf{a}_1 + \lambda \mathbf{b}_1$, we get

$\mathbf{a}_1 = 8\hat{i} - 9\hat{j} + 10\hat{k} \text{ and } \mathbf{b}_1 = 3\hat{i} - 16\hat{j} + 7\hat{k} \dots (i)$

Also, $\mathbf{r} = 15\hat{i} + 29\hat{j} + 5\hat{k} + \mu(3\hat{i} + 8\hat{j} - 5\hat{k})$

On comparing with $\mathbf{r} = \mathbf{a}_2 + \mu \mathbf{b}_2$, we get

$\mathbf{a}_2 = 15\hat{i} + 29\hat{j} + 5\hat{k} \text{ and } \mathbf{b}_2 = 3\hat{i} + 8\hat{j} - 5\hat{k} \dots (ii)$

Now, shortest distance between two lines is given by $\frac{|(\mathbf{b_1} \times \mathbf{b_2}) \cdot (\mathbf{a_2} - \mathbf{a_1})|}{|\mathbf{b_1} \times \mathbf{b_2}|}$

$∴\mathbf{b_1} \times \mathbf{b_2} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\3 & -16 & 7 \\3 & 8 & -5\end{vmatrix}$

$= \hat{i}(-80 - 56) - \hat{j}(-15 - 21) + \hat{k}(24 + 48)$

$= -136\hat{i} + 36\hat{j} + 72\hat{k} $

$= 24\hat{i} + 36\hat{j} + 72\hat{k}$

Now, $|\mathbf{b_1} \times \mathbf{b_2}| = \sqrt{(24)^2 + (36)^2 + (72)^2}$

$= \sqrt{576 + 1296 + 5184} = \sqrt{7056} = 84$

and $(\mathbf{a_2} - \mathbf{a_1}) = (15-8)\hat{i} + (29+9)\hat{j} + (5-10)\hat{k}$

$= 7\hat{i} + 38\hat{j} - 5\hat{k}$

$∴\text{Shortest distance}= \frac{|(24\hat{i} + 36\hat{j} + 72\hat{k}) \cdot (7\hat{i} + 38\hat{j} - 5\hat{k})|}{84}$

$= \frac{|168 + 1368 - 360|}{84} = \frac{1176}{84} = 14 \text{ units}$