The shortest distance between lines $\frac{-x-3}{4}=\frac{y-6}{3}=\frac{z}{2}$ and $\frac{-x-2}{4}=\frac{y}{1}=\frac{z-7}{1}$ is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 9 units
Given lines:
\(\frac{-x - 3}{4} = \frac{y - 6}{3} = \frac{z}{2}\)
\(\frac{-x - 2}{4} = \frac{y}{1} = \frac{z - 7}{1}\)
Parametric form:
Line 1:
\[ \begin{cases} \frac{-x - 3}{4} = t \Rightarrow x = -4t - 3 \\ \frac{y - 6}{3} = t \Rightarrow y = 3t + 6 \\ \frac{z}{2} = t \Rightarrow z = 2t \end{cases} \]
Line 2:
\[ \begin{cases} \frac{-x - 2}{4} = s \Rightarrow x = -4s - 2 \\ \frac{y}{1} = s \Rightarrow y = s \\ \frac{z - 7}{1} = s \Rightarrow z = s + 7 \end{cases} \]
Direction vectors:
\(\vec{d_1} = (-4, 3, 2)\)
\(\vec{d_2} = (-4, 1, 1)\)
Points on lines at parameter 0:
\(\vec{A} = (-3, 6, 0)\)
\(\vec{B} = (-2, 0, 7)\)
Vector between points:
\(\vec{AB} = \vec{B} - \vec{A} = (1, -6, 7)\)
Cross product:
\[ \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -4 & 3 & 2 \\ -4 & 1 & 1 \\ \end{vmatrix} = (1, -4, 8) \]
Magnitude of cross product:
\(|\vec{d_1} \times \vec{d_2}| = \sqrt{1^2 + (-4)^2 + 8^2} = 9\)
Dot product:
\(\vec{AB} \cdot (\vec{d_1} \times \vec{d_2}) = 81\)
Shortest distance:
\[ \frac{|\vec{AB} \cdot (\vec{d_1} \times \vec{d_2})|}{|\vec{d_1} \times \vec{d_2}|} = \frac{81}{9} = 9 \]