Equations of Lines — Question 5

Question 5

Let $$P=(4,3,-1),\qquad L:\ \vec r=\left\langle 1,-1,2\right\rangle+t\left\langle 2,1,2\right\rangle.$$

Tasks

1. Find the point $H$ on $L$ closest to $P$.

2. Write an equation of the line through $P$ that intersects $L$ perpendicularly at $H$.

3. Find the distance from $P$ to $L$.

Question 5 – Solution

Strategy At the closest point, the displacement from the line to $P$ is perpendicular to the line’s direction vector.

Perpendicular foot Write $A=(1,-1,2)$ and $\vec d=\left\langle 2,1,2\right\rangle$. For $H=A+t\vec d$, $$(P-H)\cdot\vec d=0\quad\Longrightarrow\quad (P-A-t\vec d)\cdot\vec d=0.$$ Since $(P-A)\cdot\vec d=4$ and $\vec d\cdot\vec d=9$, $t=4/9$. Therefore $$\boxed{H=\left(\tfrac{17}{9},-\tfrac 59,\tfrac{26}{9}\right)}.$$

Perpendicular line and distance A direction from $P$ to $H$ is $\left\langle -19,-32,35\right\rangle$ after multiplying by $9$. Thus $$\boxed{\vec r=\left\langle 4,3,-1\right\rangle+u\left\langle -19,-32,35\right\rangle}.$$ Also, $$\boxed{d(P,L)=\|P-H\|=\frac{\sqrt{2610}}9=\frac{\sqrt{290}}3}.$$

Verification $\left\langle -19,-32,35\right\rangle\cdot\left\langle 2,1,2\right\rangle=-38-32+70=0$, and $H$ satisfies the original line equation.