What is the unit vector that is orthogonal to the plane containing  <3, 2, 1>  and  <1, 1, 1> ?

Nov 10, 2016

The unit vector is =1/sqrt6〈1,-2,1〉

Explanation:

To calculate a vector ortogonal to 2 other vectors, we caculate the cross product.
The cross product is the determinant of ∣((veci,vecj,veck),(3,2,1),(1,1,1))∣

$= \vec{i} \left(2 - 1\right) - \vec{j} \left(3 - 1\right) + \vec{k} \left(3 - 2\right)$
So the vector ortogonal is vecv=〈1,-2,1〉
To verify, we do the dot products .
〈3,2,1〉.〈1,-2,1〉=3-4+1=0
〈1,1,1〉.〈1,-2,1〉=1-2+1=0
As the dot products are $= 0$, $\vec{v}$is ortogonal to the other 2 vectors.
To compute the unit vector, we divide by the modulus.

hatv=vecv/(∣vecv∣)
The modulus is $= \sqrt{1 + 4 + 1} = \sqrt{6}$

Therefore, hatv=1/sqrt6〈1,-2,1〉