A line parallel to $6\widehat{i}-3\widehat{j}+2\widehat{k}$ intersects the line $\frac{x}{1}=\frac{y}{2}=\frac{z}{2}$ at $A$ and the line $\frac{x+1}{1}=\frac{y+2}{2}=\frac{z+3}{2}$ at $B$, then the length of $AB$ is equal to

• A

  $\frac{3}{\sqrt{7}}$ units
• B

  $\frac{5}{\sqrt{439}}$ units
• Correct answer
  C

  $\frac{7}{\sqrt{85}}$ units
• D

  $\frac{9}{\sqrt{425}}$ units

Solution

Both the given lines are parallel to $\overrightarrow{b}=\widehat{i}+2\widehat{j}+2\widehat{k}$

Let the angle between the line and $6\widehat{i}-3\widehat{j}+2\widehat{k}$ is $\theta$
$\therefore \cos \theta =\frac{\left(\hat{i}+2\hat{j}+2\hat{k}\right)\cdot \left(6\hat{i}-3\hat{j}+2\hat{k}\right)}{3\times 7}=\frac{4}{21}$
Let the $1^{\mathrm{s}\mathrm{t}}$ line is passing through $\overrightarrow{a}=\overrightarrow{0}$ and the $2^{\mathrm{n}\mathrm{d}}$ line is passing through $\overrightarrow{c}=-\widehat{i}-2\widehat{j}-3\widehat{k}$
$AC=$ distance between the parallel lines $=\left|\frac{\left(\overrightarrow{a}-\overrightarrow{c}\right)\times \overrightarrow{b}}{\left|\overrightarrow{b}\right|}\right|$
$=\left|\frac{\left(\widehat{i}+2\widehat{j}+3\widehat{k}\right)\times \left(\widehat{i}+2\widehat{j}+2\widehat{k}\right)}{3}\right|=\frac{\sqrt{5}}{3}$
$\therefore AB=AC\sec \left(90-\theta \right)=AC\mathrm{cosec}\theta$
$=\frac{\sqrt{5}}{3}\times \frac{21}{\sqrt{425}}=\frac{7}{\sqrt{85}}$

Hence, the correct answer is option: 'C'.