Point-line distance

Statement

Calculate the distance of the point $P(3;-1;2)$ from the line $\vec S=(1;0;-1)+t(2;1;3)$.

Solution

Base point $S_0=(1;0;-1)$, direction $\vec v=(2;1;3)$ with $|\vec v|^2=14$. The vector from the point of the line to $P$ is $\vec a = P-S_0 = (2;-1;3).$ Projection: $\vec a\cdot\vec v = 4-1+9 = 12$, so the perpendicular component has magnitude given by $|\vec a_\perp|^2 = |\vec a|^2 - \frac{(\vec a\cdot\vec v)^2}{|\vec v|^2} = 14 - \frac{144}{14} = \frac{26}{7}.$ $d = \sqrt{\frac{26}{7}} = \frac{\sqrt{182}}{7} \approx \boxed{1{,}93}.$

Links

Topics: Analytic geometry in space
Concepts: Parallel component · Point-line distance · Line · Line in space
Skills: Analytic geometry · Using formulae
Exercise types: Geometric problem