Find the distance from the point (1, 4, -1) to the line (0,0,0) + a*(3, 1, 2).

Step-by-step solution

1. 1

   Set up the cross-product formula. For a line through a point $Q$ with direction $\mathbf d$, the distance from $P$ is

   $d=\frac{\left\|\overrightarrow{QP}\times\mathbf d\right\|}{\|\mathbf d\|}$

   Here $Q$ is the origin, so $\overrightarrow{QP}=\mathbf p=(1,4,-1)$ and $\mathbf d=(3,1,2)$. Geometrically the numerator is the area of the parallelogram spanned by the two vectors, and dividing by the base $\|\mathbf d\|$ leaves the height — which is exactly the distance.

2. 2

   Compute the cross product.

   $\mathbf p\times\mathbf d=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\ 1&4&-1\\ 3&1&2\end{vmatrix}=\left(4\cdot 2-(-1)\cdot 1,\;(-1)\cdot 3-1\cdot 2,\;1\cdot 1-4\cdot 3\right)=(9,\,-5,\,-11)$

   The middle component uses the minus sign built into the determinant expansion, which is where sign errors concentrate.

3. 3

   Take the two lengths.

   $\left\|\mathbf p\times\mathbf d\right\|=\sqrt{81+25+121}=\sqrt{227},\qquad \|\mathbf d\|=\sqrt{9+1+4}=\sqrt{14}$

4. 4

   Divide to get the distance.

   $d=\frac{\sqrt{227}}{\sqrt{14}}=\sqrt{\frac{227}{14}}\approx\sqrt{16.2143}\approx 4.0267$

5. 5

   Cross-check with the projection method. The foot of the perpendicular is at parameter

   $a^{*}=\frac{\mathbf p\cdot\mathbf d}{\mathbf d\cdot\mathbf d}=\frac{3+4-2}{14}=\frac{5}{14}$

   giving the foot $\tfrac{5}{14}(3,1,2)=(1.0714,\,0.3571,\,0.7143)$. The distance from $P$ to that point is $\sqrt{(1-1.0714)^2+(4-0.3571)^2+(-1-0.7143)^2}=4.0267\;\checkmark$ — the same value by an independent route.