Question

Consider two lines $\frac{x+3}{-4}=\frac{y-6}{3}=\frac{z}{2}$and$\frac{x-2}{-4}=\frac{y+1}{1}=\frac{z-6}{1}$. Which of the following are correct?

• A. The given lines are coplanar.
• B. The given lines are non-coplanar.
• C. The shortest distance between the given lines is 9.
• D. $(\hat{i}-4\hat{j}+8\hat{k})$ is a vector perpendicular to both given lines.

Answer 1

Answer: (B), (C), (D)

The correct options are
B The shortest distance between the given lines is 9.
C The given lines are non-coplanar.
D $(\hat{i}-4\hat{j}+8\hat{k})$ is a vector perpendicular to both given lines.

Lines $L_1=\frac{x+3}{-4}=\frac{y-6}{3}=\frac{z}{2}and{L}_2=\frac{x-2}{-4}=\frac{y+1}{1}=\frac{z-6}{1}$

For coplanarity, $\left|\begin{array}{ccc}5& -7& 6\\ -4& 3& 2\\ -4& 1& 1\end{array}\right|$

$\ne 0$

Hence, the lines are non-coplanar.

Let the shortest distance be $d$

$d=\left|\frac{(\overrightarrow{a_1}-\overrightarrow{a_2}).(\overrightarrow{b_1}\times \overrightarrow{b_2})}{|\overrightarrow{b_1}\times \overrightarrow{b_2}|}\right|$

$\overrightarrow{a_1}=-3\hat{i}+6\hat{j};\overrightarrow{a_2}=2\hat{i}-\hat{j}+6\hat{k}$

$\overrightarrow{b_1}=-4\hat{i}+3\hat{j}+2\hat{k};\overrightarrow{b_2}=-4\hat{i}+\hat{j}+\hat{k}$

$d=\frac{\left|\right(-5\hat{i}+7\hat{j}-6\hat{k}).(\hat{i}-4\hat{j}+8\hat{k}\left)\right|}{|\hat{i}-4\hat{j}+8\hat{k}|}=9$

Unit vector perpendicular to both lines is $\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -4& 3& 2\\ -4& 1& 1\end{array}\right|$

$=\hat{i}-4\hat{j}+8\hat{k}$

Hence, options 'B', 'C' and 'D' are correct.