Three Dimensional Geometry Question 3
Question 3 - 24 January - Shift 1
The shortest distance between the lines
$\frac{x-2}{3}=\frac{y+1}{2}=\frac{z-6}{2}$ and $\frac{x-6}{3}=\frac{1-y}{2}=\frac{z+8}{0}$ is equal to _______
Show Answer
Answer: 14
Solution:
Formula: Skew Lines
Shortest distance between the lines
$ \begin{aligned} & =\frac{\left|\begin{array}{ccc} 4 & 2 & -14 \\ 3 & 2 & 2 \\ 3 & -2 & 0 \end{array}\right|}{\left | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 2 \\ 3 & -2 & 0 \end{array} \right\rvert,} \\ & =\frac{16+12+168}{|4 \hat{i}+6 \hat{j}-12 \hat{k}|}=\frac{196}{14}=14 \\ & \end{aligned} $
