The shortest distance between the lines x - 1 2 = y + 8 - 7 = z - 4 5 and x - 1 2 = y - 2 1 = z - 6 - 3 is [2023]

Question

The shortest distance between the lines $\frac{x - 1}{2} = \frac{y + 8}{- 7} = \frac{z - 4}{5}$ and $\frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - 6}{- 3}$ is [2023]

Smart Achievers · Accepted Answer

(4)

$Since two lines L_{1} and L_{2} are given as$

$L_{1} \equiv \frac{x - 1}{2} = \frac{y + 8}{- 7} = \frac{z - 4}{5}, L_{2} \equiv \frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - 6}{- 3}$

$For line L_{1}, we have \vec{a} = \hat{i} - 8 \hat{j} + 4 \hat{k}$

$For line L_{2}, we have \vec{b} = \hat{i} + 2 \hat{j} + 6 \hat{k}$

$\vec{p} = 2 \hat{i} - 7 \hat{j} + 5 \hat{k}$ and $\vec{q} = 2 \hat{i} + \hat{j} - 3 \hat{k}$

$So, we have$ $\vec{p} \times \vec{q} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & - 7 & 5 \\ 2 & 1 & - 3 \end{matrix} |$

$= \hat{i} (21 - 5) - \hat{j} (- 6 - 10) + \hat{k} (2 + 14)$

$= 16 \hat{i} + 16 \hat{j} + 16 \hat{k} = 16 (\hat{i} + \hat{j} + \hat{k})$

$The shortest distance, d$

$=$ $| \frac{(\vec{a} - \vec{b}) \cdot (\vec{p} \times \vec{q})}{| \vec{p} \times \vec{q} |} |$

$=$ $| \frac{[\hat{i} (1 - 1) + \hat{j} (- 8 - 2) + \hat{k} (4 - 6)] \cdot 16 (\hat{i} + \hat{j} + \hat{k})}{\sqrt{16^{2} + 16^{2} + 16^{2}}} |$

$=$ $| \frac{(- 10 \hat{j} - 2 \hat{k}) \cdot (16 \hat{i} + 16 \hat{j} + 16 \hat{k})}{16 \sqrt{3}} |$

$=$ $| \frac{(- 10) (16) + (- 2) (16)}{16 \sqrt{3}} |$ $=$ $| \frac{192}{16 \sqrt{3}} |$ $=$ $| \frac{- 12}{\sqrt{3}} |$ $= 4 \sqrt{3}$

Solution

Q.
The shortest distance between the lines $\frac{x - 1}{2} = \frac{y + 8}{- 7} = \frac{z - 4}{5}$ and $\frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - 6}{- 3}$ is [2023]

1 $3 \sqrt{3}$

2 $2 \sqrt{3}$

3 $5 \sqrt{3}$

4 $4 \sqrt{3}$

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Solution

Q. The shortest distance between the lines x-12=y+8-7=z-45 and x-12=y-21=z-6-3 is [2023]

1 33

2 23

3 53