Determine the unit vector through a given plane

Question

The plane $B$ in $\mathbb{R}^3$ is passing through points $(1,0,0), (0,1,0), (0,0,-1)$. Determine the unit vectors which are perpendicular/orthogonal to the plane $B$.

This is one of problem given by my teacher before. Only this I can't solve. Recalling that two vectors are said to be orthogonal if their dot products is zero. Yes, this is easy to apply to most problem I encounter, but how can I apply to this one? Do you have some nice idea?

Answer 1

Denote $p_i$ the given points for $i=1,2,3$ so $u_1=p_1-p_2=(1,-1,0)$ and $u_2=p_1-p_3=(1,0,1)$ are two vectors which span the plane. A vector $v=(x,y,z)$ orthogonal to the plane iff it's orthogonal to $u_i$ so $$v\cdot u_1=0\quad \text{and}\quad v\cdot u_2=0$$ hence we get $x-y=x+z=0\iff v\in\operatorname{span}(1,1,-1)$. Since $v$ is unit vector then

$$v=\pm\frac1{\sqrt3}(1,1,-1)$$

Answer 2

HINT

Two ways

• obtain the plane equation solving the sistem of equations $ax+by+cz=d$ obtained by the condition that the plane passes through the given points
• more quick and clever, use cross product between any pair of vectors as for example $u=P_1-P_2$ and $v=P_1-P_3\implies n=u\times v$

Finally normalize $n\implies \hat n=\frac{n}{|n|}$.

Answer 3

The normal vector to the plane passing through the points P(1,0,0), Q(0,1,0), R(0,0,-1) is the cross product of vectors PQ and PR.

Thus we find $$\det \begin{bmatrix} i&j&k\\-1&1&0\\-1&0&-1\end{bmatrix}$$

$$ N = <-1,-1,1> $$

To find the unit vectors, we normalize $N$ to get

$$ U= \frac {\pm 1}{\sqrt 3}<-1,-1,1>$$