Question

The distance between two lines is$\frac{x+2}{2}=\frac{y-1}{3}=\frac{z+1}{1}$ and $\frac{x-1}{1}=\frac{y+1}{2}=\frac{z-2}{4}.$

• A. $\frac{47}{5\sqrt{6}}$
• B. $12\sqrt{3}$
• C. $\frac{1}{11}$
• D. $\frac{1}{\sqrt{6}}$

Answer 1

Answer: (A)

$\frac{47}{5\sqrt{6}}$

Given eq of lines

$\frac{x+2}{2}=\frac{y-2}{3}=\frac{z+1}{1}$-----(1)

$\frac{x-1}{1}=\frac{y+1}{2}=\frac{z-2}{4}$-----(2)

position vector of line (1)

$\overrightarrow{a}=-2\hat{i}+2\hat{j}-\hat{k}$

position vector of line (2)

$\overrightarrow{c}=\hat{i}-\hat{j}+2\hat{k}$

normal vector of line (1)

$\overrightarrow{b}=2\hat{i}+3\hat{j}+\hat{k}$

normal vector of line (2)

$\overrightarrow{d}=\hat{i}+2\hat{j}+4\hat{k}$

shortest between two skews line

$SD=\frac{|\overrightarrow{AC}\cdot (\overrightarrow{b}\times \overrightarrow{d}\left)\right|}{|\overrightarrow{b}\times \overrightarrow{d}|}$

$AC=\overrightarrow{c}-\overrightarrow{a}$

$AC=\hat{i}-\hat{j}+2\hat{k}-(-2\hat{i}+\hat{j}-\hat{k})$

$AC=3\hat{i}-2\hat{j}+3\hat{k}$

$\overrightarrow{b}\times \overrightarrow{d}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 3& 1\\ 1& 2& 4\end{array}\right|$

$\overrightarrow{b}\times \overrightarrow{d}=\hat{i}(12-2)-\hat{j}(8-1)+\hat{k}(4-3)$

$\overrightarrow{b}\times \overrightarrow{d}=10\hat{i}-7\hat{j}+\hat{k}$

$|\overrightarrow{b}\times \overrightarrow{d}|=\sqrt{{10}^2+(-7{)}^2+1^2}$

$|\overrightarrow{b}\times \overrightarrow{d}|=\sqrt{100+49+1}$

$|\overrightarrow{b}\times \overrightarrow{d}|=\sqrt{150}$

putting AC $\overrightarrow{b}\times \overrightarrow{d},|\overrightarrow{b}\times \overrightarrow{d}|$ in formula

$SD=\frac{\left|\right(3\hat{i}-2\hat{j}+3\hat{k})\cdot (10\hat{i}-7\hat{j}+\hat{k}\left)\right|}{\sqrt{150}}$

$SD=\frac{|30+14+3|}{\sqrt{150}}$

$SD=\frac{47}{\sqrt{150}}$

$SD=\frac{47}{5\sqrt{6}}$

Hence it is correct ans.