Calculating the coordinates of a unit vector normal

Question

First of all I apologize if the question is elementary, I have not practiced math for a long time and have only just slowly picked up the fundamentals again.

(i) With reference to Figure Q4, calculate coordinates of the unit (magnitude is equal to 1) normal vector N.

(ii) What are the coordinates of the unit normal vector to the opposite side of the triangle?

I am not very sure about how to deal with this question, but my thought process was firstly to get the unit vector of the 2 points [-2,0,0]and[1,0,2], after that I am not particularly sure how to convert that to coordinates, and I am fairly certain I have not done the correct step.

May I have guidance on this matter, I am fairly certain I have to manipulate all 3 points available, but I do not have an idea on how to proceed.

Answer 1

Let $A=(1,0,2)$, $B=(0,2,0)$, and $C=(-2,0,0)$. Now, let $u=B-A=(-1,2,-2)$, and let $v=C-A=(-3,0,-2)$. Then, a vector normal to that triangle is$$u\times v=(-4,4,6).$$Its first coordinate is negative, which is what we want (take a look at the picture). So, the answer will be$$\frac{u\times v}{\lVert u\times v\rVert}=\frac1{\sqrt{17}}(-2,2,3).$$

Answer 2

One way to go is to take two pairs of differences, like $(-2,0,0)-(0,2,0)=(-2,-2,0)$, and $(-2,0,0)-(1,0,2)=(-3,0,-2)$. And then compute the cross product. Finally normalize the result.

So $(-2,-2,0)×(-3,0,-2)=\begin {vmatrix} i&j&k\\-2&-2&0\\-3&0&-2\end{vmatrix}=(4,-4,-6)$.

The norm of that vector is $\sqrt{4^2+(-4)^2+(-6)^2}=\sqrt{68}=2\sqrt{17}$.

So divide: $\dfrac1{2\sqrt{17}}(4,-4,-6)$.

Multiply by negative one to reverse it.