Find the vector passing through a given point which is orthogonal to a given triangle in space

Question

I'm given this problem where I have 3 points in space $A(3, -1, 2)$, $B(-2,1,2)$ and $C(2, 0, 5)$.

I need to find the vector passing through point $A$ that is perpendicular to the triangle made by $ABC$.

I'm fairly certain that to find the vector perpendicular to the triangle $ABC$ I'd just need to find the cross product of $AB$ and $AC$. However I'm stuck on how to put this into a vector that passes through $A$.

Answer 1

You would have that $AB \times AC$ is perpendicular to the triangle. Then, the line passing through AB with that direction vector is $OA + t (AB \times AC)$ where $O$ is the origin.

Answer 2

You are right about the cross product of $AB=(-5,2,0), AC=(-1,1,3)$. The cross product $\vec{d}$ is the direction vector of the line. The equation of the line should be

$$\vec{r}=\vec{A}+t\vec{d} \implies \begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}3\\-1\\2\end{pmatrix}+t\vec{d}$$

Answer 3

Hints:

== The given triangle lays on the plane $\;A+t\vec{AB}+s\vec{AC}\;,\;\;t,s\in\Bbb R$

== The (a) normal vector to a plane $\;a+t\vec b+s\vec c\;$ is given by the vector (cross) product

$$\vec b\times\vec c:=\begin{vmatrix}e_1&e_2&e_3\\b_1&b_2&b_3\\c_1&c_2&c_3\end{vmatrix}=\left(b_2c_3-b_3c_2\,,\,\,b_3c_1-b_1c_3\,,\,\,b_1c_2-b_2c_1\right)$$