The shortest distance between the lines \( \displaystyle\frac{x+1}{7}=\displaystyle\frac{-y-1}{6}=\displaystyle\frac{z+1}{1} \) and \( \displaystyle\frac{x-3}{1}=\displaystyle\frac{-y+5}{2}=\displaystyle\frac{z-7}{1} \) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 2\sqrt{29} \)
The y-terms are not in standard form; rewritten, \( \frac{-y-1}{6}=\frac{y+1}{-6} \) and \( \frac{-y+5}{2}=\frac{y-5}{-2} \).
Line 1 passes through (−1, −1, −1) with direction (7, −6, 1).
Line 2 passes through (3, 5, 7) with direction (1, −2, 1).
\( \displaystyle \vec{b}_1\times\vec{b}_2=\begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k}\\7 & -6 & 1\\1 & -2 & 1\end{vmatrix} \)
\( \displaystyle =-4\hat{\imath}-6\hat{\jmath}-8\hat{k} \)
\( \displaystyle |\vec{b}_1\times\vec{b}_2|=\sqrt{16+36+64} \)
\( \displaystyle =\sqrt{116}=2\sqrt{29} \)
\( \displaystyle \vec{a}_2-\vec{a}_1=4\hat{\imath}+6\hat{\jmath}+8\hat{k} \)
By the shortest distance formula for skew lines,
\( \displaystyle d=\frac{|(\vec{a}_2-\vec{a}_1)\cdot(\vec{b}_1\times\vec{b}_2)|}{|\vec{b}_1\times\vec{b}_2|} \)
\( \displaystyle =\frac{|-16-36-64|}{2\sqrt{29}} \)
\( \displaystyle =\frac{116}{2\sqrt{29}}=2\sqrt{29} \)
NCERT (2026-27 edition):
Shortest Distance between Two Lines
