Three Dimensional Geometry Question 16
Question 16 - 29 January - Shift 2
Shortest distance between the lines
$\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5}$ and $\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}$ is
(1) $2 \sqrt{3}$
(2) $4 \sqrt{3}$
(3) $3 \sqrt{3}$
(4) $5 \sqrt{3}$
Show Answer
Answer: (2)
Solution:
Formula: Skew Lines
$\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5} \quad \vec{a}=\hat{i}-8 \hat{j}+4 \hat{k}$
$\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3} \quad \vec{b}=\hat{i}+2 \hat{j}+6 \hat{k}$
$\vec{p}=2 \hat{i}-7 \hat{j}+5 \hat{k}, \vec{q}=2 \hat{i}+\hat{j}-3 \hat{k}$
$\overrightarrow{{}p} \times \overrightarrow{{}q}= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -7 & 5 \\ 2 & 1 & -3\end{vmatrix} $
$=\hat{i}(16)-\hat{j}(-16)+\hat{k}(16)$
$=16(\hat{i}+\hat{j}+\hat{k})$
$d=|\frac{(a-b) \cdot(\overrightarrow{{}p} \times \overrightarrow{{}q})}{|\overrightarrow{{}p} \times \overrightarrow{{}q}|}|=|\frac{(-10 \hat{j}-2 \hat{k}) \cdot 16(\hat{i}+\hat{j}+\hat{k})}{16 \sqrt{3}}|$
$=|\frac{-12}{\sqrt{3}}|=4 \sqrt{3}$
