Find the distance from M_0(2,-1,3) to the line…

Find the distance from $M_0(2,-1,3)$ to the line $\tfrac{x+1}{3}=\tfrac{y+2}{4}=\tfrac{z-1}{2}$.

Explanation of Method

The distance from the point M₀(2, −1, 3) to a given line.

Step 1: Solution

Here $\vec p=\{3,4,2\}$, $M_1(-1,-2,1)$ and $\overrightarrow{M_1M_0}=\{3,1,2\}$: \[\vec p\times\overrightarrow{M_1M_0}=\{8-2,\ 6-6,\ 3-12\}=\{6,0,-9\},\qquad |\cdot|=\sqrt{117}=3\sqrt{13}.\]

\[d=\frac{3\sqrt{13}}{\sqrt{29}}=\sqrt{\frac{117}{29}}=\frac{3\sqrt{377}}{29}\approx2.009\]

Check via the foot: $F=M_1+\tfrac{17}{29}\vec p$, and $|M_0F|^2=|\overrightarrow{M_1M_0}|^2-\tfrac{17^2}{29}=14-\tfrac{289}{29}=\tfrac{117}{29}$.

$M_0$ $M_1$ $F$ $d\approx2.01$

The foot of the perpendicular is $F=M_1+\tfrac{17}{29}\{3,4,2\}$; $d=|M_0F|=\tfrac{3\sqrt{13}}{\sqrt{29}}\approx2.009$.

Result

\[\boxed{\,d=\frac{3\sqrt{13}}{\sqrt{29}}=\sqrt{\frac{117}{29}}=\frac{3\sqrt{377}}{29}\approx2.009\,}\]

Remark

Correction: the source's first boxed value “$\approx1.79$” is wrong. The correct value is $\approx2.01$, which the source itself finds later.