The line is parallel to the vector and passes through the point (7, 6, 2) and the line is parallel to the vector and passes through the point (5,3, 4). The shortest distance between the lines and is : [2025]

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Solution

Q.
The line $L_{1}$ is parallel to the vector $\vec{a} = - 3 \hat{i} + 2 \hat{j} + 4 \hat{k}$ and passes through the point (7, 6, 2) and the line $L_{2}$ is parallel to the vector $\vec{b} = 2 \hat{i} + \hat{j} + 3 \hat{k}$ and passes through the point (5,3, 4). The shortest distance between the lines $L_{1}$ and $L_{2}$ is : [2025]

1 $\frac{23}{\sqrt{38}}$

2 $\frac{21}{\sqrt{57}}$

3 $\frac{23}{\sqrt{57}}$

4 $\frac{21}{\sqrt{38}}$

Ans.
(1)

We have, $L_{1} : 7 \hat{i} + 6 \hat{j} + 2 \hat{k} + λ (- 3 \hat{i} + 2 \hat{j} + 4 \hat{k})$

$L_{2} : 5 \hat{i} + 3 \hat{j} + 4 \hat{k} + μ (2 \hat{i} + \hat{j} + 3 \hat{k})$

Here, ${\vec{a}}_{1} = 7 \hat{i} + 6 \hat{j} + 2 \hat{k}, {\vec{a}}_{2} = 5 \hat{i} + 3 \hat{j} + 4 \hat{k}$

${\vec{b}}_{1} = - 3 \hat{i} + 2 \hat{j} + 4 \hat{k} and {\vec{b}}_{2} = 2 \hat{i} + \hat{j} + 3 \hat{k}$

Now, ${\vec{b}}_{1} \times {\vec{b}}_{2} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ - 3 & 2 & 4 \\ 2 & 1 & 3 \end{matrix} |$

$= \hat{i} (6 - 4) - \hat{j} (- 9 - 8) + \hat{k} (- 3 - 4)$

$= 2 \hat{i} + 17 \hat{j} - 7 \hat{k}$

Also, ${\vec{a}}_{2} - {\vec{a}}_{1} = - 2 \hat{i} - 3 \hat{j} + 2 \hat{k}$

Shortest distance between $L_{1}$ and $L_{2}$

$= \frac{| ({\vec{a}}_{2} - {\vec{a}}_{1}) \cdot ({\vec{b}}_{1} \times {\vec{b}}_{2}) |}{| {\vec{b}}_{1} \times {\vec{b}}_{2} |} = \frac{| - 4 - 51 - 14 |}{\sqrt{4 + 289 + 49}}$

$= \frac{69}{\sqrt{342}} = \frac{69}{3 \sqrt{38}} = \frac{23}{\sqrt{38}}$

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