Let the line be . Let . Find the distance from to the line .

Solid Geometry Advanced Solution – Problem 11: Find the distance from to the line

Question

Let the line $l$ be $\mathbf r=(0,1,1)+t(1,1,0)$ . Let $P(1,0,0)$ . Find the distance from $P$ to the line $l$ .

Step-by-step solution

(1) Take a point on the line: $Q(0,1,1)$ (when $t=0$ ). The direction vector is $\mathbf u=(1,1,0)$ .

(2) The distance from a point to a line in space is $d=\frac{|\overrightarrow{QP}\times\mathbf u|}{|\mathbf u|}.$ Compute $\overrightarrow{QP}=P-Q=(1,-1,-1)$ .

(3) Cross product: $\overrightarrow{QP}\times\mathbf u=(1,-1,-1)\times(1,1,0)=(1,-1,2).$ So $|\overrightarrow{QP}\times\mathbf u|=\sqrt{1+1+4}=\sqrt6$ , and $|\mathbf u|=\sqrt2$ .

Hence $d=\frac{\sqrt6}{\sqrt2}=\sqrt3.$

Final answer

The distance from $P$ to line $l$ is $\sqrt3$ .

Marking scheme

Step 1 — Setup

Checkpoint: pick a point $Q$ on the line and identify the direction vector $\mathbf u$ (2 pts)

Step 2 — Key Calculation

Checkpoint: compute $\overrightarrow{QP}\times\mathbf u$ and its norm correctly (3 pts)

Step 3 — Final Answer

Checkpoint: apply $d=\frac{|\overrightarrow{QP}\times\mathbf u|}{|\mathbf u|}$ to obtain $\sqrt3$ (2 pts)

Zero credit if: uses a planar distance formula.

Deductions: -1 pt for a cross product arithmetic slip that is later corrected.

Solid Geometry Advanced – Problem 11: Find the distance from to the line