Mathematics - Vector Question with Solution | TestHub

Given,

Equation of the lines

$x+1=2 y=-12z$ and $x=y+2=6z-6$

$\Rightarrow \frac{x+1}{1}=\frac{y}{\frac{1}{2}}=\frac{z}{\frac{-1}{12}}$ and $\frac{x}{1}=\frac{y+2}{1}=\frac{z-1}{\frac{1}{6}}$

We know that, shortest distance between the lines is given by $\mathrm{S}.\mathrm{D}=\frac{\left(\overrightarrow{b}-\overrightarrow{a}\right)·\left(\overrightarrow{p}\times \overrightarrow{q}\right)}{\left|\overrightarrow{p}\times \overrightarrow{q}\right|}$

Here $\overrightarrow{p}=\hat{i}+\frac{\hat{j}}{2}+\frac{-\hat{k}}{12} \& \overrightarrow{q}=\hat{i}+\hat{j}+\frac{\hat{k}}{6}$ and $\overrightarrow{a}=\hat{i} \& \overrightarrow{b}=2\hat{j}-\hat{k}$

 $\Rightarrow \mathrm{S}.\mathrm{D}=\left(-\hat{i}+2\hat{j}-\hat{k}\right)·\frac{\left(\overrightarrow{p}\times \overrightarrow{q}\right)}{\left|\overrightarrow{p}\times \overrightarrow{q}\right|}$

Now solving $\overrightarrow{p}\times \overrightarrow{q}\equiv \left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& \frac{1}{2}& \frac{-1}{12}\\ 1& 1& \frac{1}{6}\end{array}\right|$

$\Rightarrow \overrightarrow{p}\times \overrightarrow{q}\equiv \frac{1}{6}\hat{i}-\frac{1}{4}\hat{j}+\frac{1}{2}\hat{k}$

$\Rightarrow \overrightarrow{p}\times \overrightarrow{q}\equiv 2\hat{i}-3\hat{j}+6\hat{k}$

Hence, $\mathrm{S}.\mathrm{D}=\frac{\left(-\hat{i}+2\hat{j}-\hat{k}\right)·\left(2\hat{i}-3\hat{j}+6\hat{k}\right)}{\sqrt{2^2+3^2+6^2}}=\left|\frac{-14}{7}\right|=2$