Mathematics - Vector Question with Solution | TestHub

Given line $L_1:\frac{x+7}{-6}=\frac{y-6}{7}=\frac{z-0}{1}$

Let any point on it is given by ${\overrightarrow{a}}_1\left(-7,6,0\right)$ and $L_1$ is parallel to ${\overrightarrow{b}}_1\left(-6,7,1\right)$

Also given line $L_2:\frac{x-7}{-2}=\frac{y-2}{1}=\frac{z-6}{1}$

Let any point on it be ${\overrightarrow{a}}_2\left(7,2,6\right)$ and $L_2$ is parallel to ${\overrightarrow{b}}_2\left(-2,1,1\right)$

Now finding ${\overrightarrow{b}}_1\times {\overrightarrow{b}}_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -6& 7& 1\\ -2& 1& 1\end{array}\right|=3\hat{i}+2\hat{j}+4\hat{k}$

Now shortest distance between $L_1$ and $L_2$ will be

$=\left|\frac{\left({\overrightarrow{a}}_2-{\overrightarrow{a}}_1\right)·\left({\overrightarrow{b}}_1\times {\overrightarrow{b}}_2\right)}{\left|{\overrightarrow{b}}_1\times {\overrightarrow{b}}_2\right|}\right|=\left|\frac{\left(-14\hat{i}+4\hat{j}-6\hat{k}\right)·\left(3\hat{i}+2\hat{j}+4\hat{k}\right)}{\sqrt{9+4+16}}\right|$

$=2\sqrt{29}$