Unit vector which is normal to plane

Question

The question says to use the vector product to find a unit vector which is normal to the plane:

$$(3i+4j+2k)+ λ(2i+j+2k) + μ(i+2k)$$

My thoughts: vector product is $A\times B = |A||B|\sinθ$ and normal = 90°, therefore $A\times B = |A||B|$ as $sin 90^0 = 1$.

I am not sure how I could use this to find the unit vector normal to the plane.

Answer 1

The two vectors director of your plane are $$\vec{u}=(2,1,2) \; \text{ and } \vec{v}= (1,0,2)$$

the normal vector is then the vector product $$\vec{n}=\vec{u} \times \vec{v} =(2,-2,-1)$$

You can check that $$\vec{n}•\vec{u}=\vec{n}•\vec{v}=0$$

Answer 2

A normal vector to the plane is the cross-product of the two directing vectors of the planes. Its coordinates are the $2{\times}2$ minors of the matrix with columns the coordinates of the directing vectors: $$\begin{vmatrix}2&1\\1&0\\2&2\end{vmatrix}\rightsquigarrow \begin{pmatrix}\left|\begin{smallmatrix}1&0\\2&2\end{smallmatrix}\right|\\\left|\begin{smallmatrix}2&2\\2&1\end{smallmatrix}\right|\\\left|\begin{smallmatrix}2&1\\1&0\end{smallmatrix}\right|\end{pmatrix}=\begin{pmatrix}\phantom{-}2\\-2 \\-1\end{pmatrix}.$$ There remains to normalise this vector.