Question

The length of the shortest distance between the lines $\vec{r}=3\hat{i}+5\hat{j}+7\hat{k}+\lambda(\hat{i}-2\hat{j}+\hat{k})$ and $\vec{r} = -\hat{i}-\hat{j}-\hat{k}+\mu (7\hat{i}-6\hat{j}+\hat{k})$ is

• A. 83 units
• B. $\sqrt{6}$ units
• C. $\sqrt{3}$ units
• D. $2\sqrt{29}$ units

Answer 1

The Correct Option is D

Shortest distance PQ = $\bigg| \frac{(\vec{b_1} \times \vec{b_2} ) \ . \ (\vec{a_2} - \vec{a_1})}{|(\vec{b_1} \times \vec{b_2})|} \bigg|$
Now, $\vec{a_2} \ - \ \vec{a_1} \ = \ -\hat{i}-\hat{j}-\hat{k}-3\hat{i}-5\hat{j}-7\hat{k}$
$\Rightarrow \ \ \ \vec{a_2} - \vec{a_1} \ = \ -4\hat{i}-6\hat{j}-8\hat{k}$

And $\vec{b_1} \ \times \ \vec{b_2} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
1 & -2 & 1 \\
7 & -6 & 1 \end{vmatrix}$
$\Rightarrow \ \ \vec{b_1} \times \vec{b_2} = \hat{i} (-2+6)-\hat{j}(1-7)+\hat{k}(-6+14)$
$\Rightarrow \ \ \vec{b_1} \times \vec{b_2} \ = \ 4\hat{i}+6\hat{j}+8\hat{k}$
$\therefore $Shortest distance
$PQ= \bigg| \frac{(4\hat{i} + 6\hat{j} +8 \hat{k}) \ . \ (-4\hat{i}-6\hat{j}-8\hat{k})}{\sqrt{16+36+64}}\bigg|$
$\Rightarrow \ \ PQ \ = \ \bigg| \frac{-16-36-64}{\sqrt{116}} \bigg|$
$=|\frac{-116}{\sqrt{116}}| \ =\sqrt{116} \ = 2\sqrt{29}$
$\therefore \ \ PQ=2\sqrt{29} \ units$