Question

What is the unit vector that is orthogonal to the plane containing `(3i + 2j - 3k)` and `(i -2j + 3k)`?

Answer 1

The answer is `=〈0,-3/sqrt13,-2/sqrt13〉`

Explanation:

We do a cross product to find the vector orthogonal to the plane

The vector is given by the determinant

`| (hati,hatj,hatk), (3,2,-3), (1,-2,3) |`

`=hati(6-6)-hatj(9--3)+hatk(-6-2)`

`=〈0,-12,-8〉`

Verification by doing the dot product

`〈0,-12,-8〉.〈3,2,-3〉=0-24+24=0`

`〈0,-12,-8〉.〈1,-2,3〉=0+24-24=0`

The vector is orthgonal to the other 2 vectors

The unit vector is obtained by dividing by the modulus

`∥〈0,-12,-8〉∥=sqrt(0+144+64)=sqrt208=4sqrt13`

Thre unit vector is `=1/(4sqrt13)〈0,-12,-8〉`

`=〈0,-3/sqrt13,-2/sqrt13〉`