Line through two points and decomposition of a vector

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Let $s$ be the line through $A(1,0,-1)$ and $B(4,-1,1)$.

(a) Write the parametric equations of $s$.

(b) Compute the distance of $Q(4,2,1)$ from $s$.

(c) Find the component of $\overrightarrow{AQ}$ parallel to $\overrightarrow{AB}$ and the component perpendicular to it.

Solution

The direction vector is $\overrightarrow{AB}=(3,-1,2)$.

(a) $s:\ x=1+3t,\ y=-t,\ z=-1+2t$.

(b)-(c) We have $\overrightarrow{AQ}=(3,2,2)$. The scalar projection of $\overrightarrow{AQ}$ onto $\overrightarrow{AB}$ is $\dfrac{\overrightarrow{AQ}\cdot\overrightarrow{AB}}{|\overrightarrow{AB}|^2}=\dfrac{11}{14}$, so the parallel component is $\dfrac{11}{14}(3,-1,2)\approx(2.357,-0.786,1.571)$ and the perpendicular component is $\overrightarrow{AQ}-\text{parallel}\approx(0.643,2.786,0.429)$. The required distance is the magnitude of the perpendicular component: $d(Q,s)=|\text{perp}|\approx2.891$. $\boxed{d(Q,s)\approx2.891;\ \text{parallel}\approx(2.357,-0.786,1.571);\ \text{perp}\approx(0.643,2.786,0.429)}$