Equations of Lines — Question 2

Question 2 – Solution

Strategy Use the common symmetric ratio as a parameter. For the closest point, minimize the squared distance to avoid unnecessary square roots.

Representations Let the common ratio equal $t$. Then $$\boxed{\vec r=\left\langle 3,-2,1\right\rangle+t\left\langle 0,4,-3\right\rangle},\qquad x=3,\ y=-2+4t,\ z=1-3t.$$ The zero $x$-component of the direction means the entire line lies in the plane $x=3$ and is parallel to the $yz$-plane.

Closest point The squared distance from the origin is $$D^2(t)=9+(-2+4t)^2+(1-3t)^2=25t^2-22t+14.$$ Its minimum occurs at $t=22/50=11/25$. Hence $$\boxed{H=\left(3,-\tfrac 6{25},-\tfrac 8{25}\right)},\qquad \boxed{d(O,L)=\frac{\sqrt{229}}5}.$$

Verification The vector $H$ from the origin has dot product $0$ with $\left\langle 0,4,-3\right\rangle$, confirming that $OH$ is perpendicular to $L$.