The shortest distance between the line $x = y = z$ and the line of intersection of $2x + y + z - 1 = 0$ and $3x + y + 2z - 2 = 0$ is
Step-by-Step Solution
Key Concept: The direction vector of the line of intersection is the cross product of the normal vectors of the two planes, and a point on the line is found by setting one coordinate and solving.
The direction vector of the line is parallel to the cross product of the normal vectors of the two planes. Computing $\vec{n_1} \times \vec{n_2} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 2 & 1 & 1 \\ 3 & 1 & 2 \end{vmatrix} = \vec{i}(2-1) - \vec{j}(4-3) + \vec{k}(2-3) = \vec{i} - \vec{j} - \vec{k}$. For a point on the line, setting $z = 0$ and solving the system $2x + y = 1$ and $3x + y = 2$ gives $x = 1, y = -1$, so the point is $(1, -1, 0)$. The equation of the line is $\frac{x-1}{1} = \frac{y+1}{-1} = \frac{z}{-1}$. The shortest distance formula between two skew lines gives $\frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}$.
Correct Answer: 1