The shortest distance between lines \( L_1:\displaystyle\frac{x+3}{-4}=\displaystyle\frac{6-y}{-3}=\displaystyle\frac{z}{2} \) and \( L_2:\displaystyle\frac{x+2}{-4}=\displaystyle\frac{y}{1}=\displaystyle\frac{z-7}{1} \) is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 9 units
In standard form, \( \frac{6-y}{-3}=\frac{y-6}{3} \), so
\( \displaystyle L_1:\ \frac{x+3}{-4}=\frac{y-6}{3}=\frac{z}{2},\quad \vec{a}_1=(-3,6,0),\ \vec{b}_1=(-4,3,2) \)
\( \displaystyle L_2:\ \frac{x+2}{-4}=\frac{y}{1}=\frac{z-7}{1},\quad \vec{a}_2=(-2,0,7),\ \vec{b}_2=(-4,1,1) \)
The direction ratios are not proportional, so the shortest distance formula applies:
\( \displaystyle d=\frac{|(\vec{a}_2-\vec{a}_1)\cdot(\vec{b}_1\times\vec{b}_2)|}{|\vec{b}_1\times\vec{b}_2|} \)
\( \displaystyle \vec{a}_2-\vec{a}_1=(1,-6,7) \)
\( \displaystyle \vec{b}_1\times\vec{b}_2=\begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k}\\-4 & 3 & 2\\-4 & 1 & 1\end{vmatrix} \)
\( \displaystyle =\hat{\imath}(3-2)-\hat{\jmath}(-4+8)+\hat{k}(-4+12) \)
\( \displaystyle =\hat{\imath}-4\hat{\jmath}+8\hat{k} \)
\( \displaystyle |\vec{b}_1\times\vec{b}_2|=\sqrt{1+16+64}=9 \)
\( \displaystyle (\vec{a}_2-\vec{a}_1)\cdot(\vec{b}_1\times\vec{b}_2)=1+24+56=81 \)
\( \displaystyle d=\frac{81}{9}=9\text{ units} \)
Tip: The term 6 − y hides a sign: write it as −(y − 6) before reading the point and direction. Taking −3 as the direction ratio of y changes the cross product.