### Instruction: Let $A$ and $B$ be two opposite corners of a unit cube, and let $C$ be another vertex of the cube other than $A$ and $B.$  Find the distance from $C$ to line $AB.$.### Response: I need to find the shortest distance from $C$ to line $AB.$ I know that this distance is the same as the length of the perpendicular segment from $C$ to line $AB.$
I can use vectors to represent the points $A, B,$ and $C.$ Since the cube is unit, I can assume that $A = (0, 0, 0), B = (1, 1, 1),$ and $C = (1, 0, 0).$
To find the vector $\vec{AB},$ I can subtract the coordinates of $A$ from the coordinates of $B.$ This gives me $\vec{AB} = (1, 1, 1) - (0, 0, 0) = (1, 1, 1).$
To find the vector $\vec{AC},$ I can subtract the coordinates of $A$ from the coordinates of $C.$ This gives me $\vec{AC} = (1, 0, 0) - (0, 0, 0) = (1, 0, 0).$
Now I need to find the projection of $\vec{AC}$ onto $\vec{AB}.$ This is the part of $\vec{AC}$ that is parallel to $\vec{AB}.$ The formula for the projection of $\vec{u}$ onto $\vec{v}$ is $\frac{\vec{u} \cdot \vec{v}}{|\vec{v}|^2} \vec{v},$ where $\vec{u} \cdot \vec{v}$ is the dot product of $\vec{u}$ and $\vec{v},$ and $|\vec{v}|$ is the magnitude of $\vec{v}.$
To find the dot product of $\vec{AC}$ and $\vec{AB},$ I can multiply the corresponding coordinates and add them up. This gives me $\vec{AC} \cdot \vec{AB} = (1, 0, 0) \cdot (1, 1, 1) = 1.$
To find the magnitude of $\vec{AB},$ I can use the Pythagorean theorem on each coordinate. This gives me $|\vec{AB}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}.$
Therefore, the projection of $\vec{AC}$ onto $\vec{AB}$ is $\frac{\vec{AC} \cdot \vec{AB}}{|\vec{AB}|^2} \vec{AB} = \frac{1}{3} \vec{AB} = \frac{1}{3} (1, 1, 1) = \left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right).$
This means that the part of $\vec{AC}$ that is perpendicular to $\vec{AB}$ is $\vec{AC}$ minus the projection of $\vec{AC}$ onto $\vec{AB}.$ This gives me $\vec{AC} - \frac{1}{3} \vec{AB} = (1, 0, 0) - \left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right) = \left(\frac{2}{3}, -\frac{1}{3}, -\frac{1}{3}\right).$