Three Dimensional Geometry Question 11
Question 11 - 25 January - Shift 2
The shortest distance between the lines $x+1=2 y=-$ $12 z$ and $x=y+2=6 z-6$ is
(1) 2
(2) 3
(3) $\frac{5}{2}$
(4) $\frac{3}{2}$
Show Answer
Answer: (1)
Solution:
Formula: Skew Lines
$\frac{x+1}{1}=\frac{y}{\frac{1}{2}}=\frac{z}{\frac{-1}{12}}$ and $\frac{x}{1}=\frac{y+2}{1}=\frac{z-1}{\frac{1}{6}}$
$\Rightarrow$ Shortest distance $=\frac{(\vec{b}-\vec{a}) \cdot(\vec{p} \times \vec{q})}{|\vec{p} \times \vec{q}|}$
S.D. $=(-\hat{i}+2 \hat{j}-\hat{k}) \cdot \frac{(\vec{p} \times \vec{q})}{|\vec{p} \times \vec{q}|}$
S.D. $=\frac{(-\hat{i}+2 \hat{j}-\hat{k}) \cdot(2 \hat{i}-3 \hat{j}+6 \hat{k})}{\sqrt{2^{2}+3^{2}+6^{2}}}=|\frac{-14}{7}|=2$
