Lines in 3D Math Example 4

Solution

1. 1
   Direction vector: $\vec{d} = \langle 2, 1, -1 \rangle$. Points on the lines: $P_1 = (1, 0, -1)$ and $P_2 = (3, 1, 0)$. $\vec{P_1P_2} = \langle 2, 1, 1 \rangle$.
2. 2
   Distance $= \frac{|\vec{P_1P_2} \times \vec{d}|}{|\vec{d}|}$. Cross product: $\vec{P_1P_2} \times \vec{d} = \langle(1)(−1)−(1)(1), (1)(2)−(2)(−1), (2)(1)−(1)(2)\rangle = \langle -2, 4, 0 \rangle$. $|\vec{P_1P_2} \times \vec{d}| = \sqrt{4+16+0} = \sqrt{20} = 2\sqrt{5}$. $|\vec{d}| = \sqrt{4+1+1} = \sqrt{6}$. Distance $= \frac{2\sqrt{5}}{\sqrt{6}} = \frac{2\sqrt{30}}{6} = \frac{\sqrt{30}}{3}$.