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Let $A(3,-1,2)$, $B(2,1,5)$, and $C(1,-2,-2)$ be points in space. Let $\vec{AC} = \langle1, 2,3\rangle$ and $\vec{AB} = \langle -2,-1,-4\rangle$ be adjacent vectors. I want to find the distance from the point $A$ to the point $BC$.

The first thing I did was construct the vector $\vec{BC} = \langle -1,-3,-7\rangle$. I then found the area of the parallelogram $\vec{AB} \times \vec{AC}$, which is $5\sqrt{6}$. Intuitively, I believe the distance from $A$ to $\vec{BC}$ is

$$\frac{|\vec{AB}\times\vec{AC}|}{|BC|} = \frac{5\sqrt{6}}{\sqrt{59}}$$

but I do not know how to mathematize my intuition, and I'm not even sure if it is correct.

I also found related problems on the site, but many of the solutions introduce parameterizations of the line, which is fine, but I was wondering if there is a way to only use the base and height of the given parallelogram.

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  • $\begingroup$ The problem becomes simpler if you shift (=translate) $A,B,C$ into $0,B',C'$, i.e., in such a way that $A$ is translated to the origin. Then express the line under the parametric form $\vec{AM}=t\vec{AB} \ \iff \ M=A+t*(B-A)$ ; then look for the value of $t$ such that $\|\vec{OM}\|²$ minimum. $\endgroup$
    – Jean Marie
    Sep 12, 2017 at 22:01

1 Answer 1

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$$\vec{BC}(-1,-3,-7).$$ Thus, $$(x,y,z)=(1,-2,-2)+t(1,3,7)$$ is an equation of $BC$.

Let $K\in BC$ such that $AK\perp BC$.

We need to find the length of $AK$.

Indeed, $K(1+t,-2+3t,-2+7t)$, $\vec{AK}(-2+t,-1+3t,-4+7t)$ and since $$\vec{AK}\cdot\vec{BC}=0,$$ we obtain $$-2+t+3(-1+3t)+7(-4+7t)=0$$ or $$t=\frac{33}{59}.$$ Thus, $$\vec{AK}\left(-\frac{85}{59},\frac{40}{59},-\frac{5}{59}\right)$$ and $$AK=\frac{\sqrt{85^2+40^2+5^2}}{59}=\frac{5\sqrt{354}}{59}.$$

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