The shortest distance between lines \(\displaystyle\frac{-x - 3}{4} = \frac{y - 6}{3} = \frac{z}{2}\) and \(\displaystyle\frac{-x - 2}{4} = \frac{y}{1} = \frac{z - 7}{1}\) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 9 units
In standard form, with the coefficient of x equal to 1:
\(\displaystyle \frac{x + 3}{-4} = \frac{y - 6}{3} = \frac{z}{2}\)
\(\displaystyle \frac{x + 2}{-4} = \frac{y}{1} = \frac{z - 7}{1}\)
The first line passes through A(−3, 6, 0) with direction \(\vec{b}_1 = -4\hat{\imath} + 3\hat{\jmath} + 2\hat{k}\); the second through B(−2, 0, 7) with direction \(\vec{b}_2 = -4\hat{\imath} + \hat{\jmath} + \hat{k}\). The shortest distance between skew lines is
\(\displaystyle d = \frac{|\vec{AB}\cdot(\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}\)
\(\displaystyle \vec{b}_1 \times \vec{b}_2 = \begin{vmatrix}\hat{\imath}&\hat{\jmath}&\hat{k}\\-4&3&2\\-4&1&1\end{vmatrix}\)
\(= \hat{\imath}(3 - 2) - \hat{\jmath}(-4 + 8) + \hat{k}(-4 + 12)\)
\(= \hat{\imath} - 4\hat{\jmath} + 8\hat{k}\)
\(|\vec{b}_1 \times \vec{b}_2| = \sqrt{1 + 16 + 64} = 9\)
\(\vec{AB} = \hat{\imath} - 6\hat{\jmath} + 7\hat{k}\)
\(\vec{AB}\cdot(\vec{b}_1 \times \vec{b}_2) = 1 + 24 + 56 = 81\)
\(\displaystyle d = \frac{81}{9} = 9\) units