Question

What is the unit vector that is orthogonal to the plane containing `(2i + 3j – 7k)` and `(3i – 4j + 4k)`?

Answer 1

The unit vector is `=〈-16/sqrt1386,-29/sqrt1386,-17/sqrt1386〉`

Explanation:

The vector perpendicular to 2 vectors is calculated with the determinant (cross product)

`| (veci,vecj,veck), (d,e,f), (g,h,i) |`

where `〈d,e,f〉` and `〈g,h,i〉` are the 2 vectors

Here, we have `veca=〈2,3,-7〉` and `vecb=〈3,-4,4〉`

Therefore,

`| (veci,vecj,veck), (2,3,-7), (3,-4,4) |`

`=veci| (3,-7), (-4,4) | -vecj| (2,-7), (3,4) | +veck| (2,3), (3,-4) |`

`=veci(3*4-7*4)-vecj(2*4+7*3)+veck(-2*4-3*3)`

`=〈-16,-29,-17〉=vecc`

Verification by doing 2 dot products

`〈-16,-29,-17〉.〈2,3,-7〉=-16*2-29*3-7*17=0`

`〈-16,-29,-17〉.〈3,-4,4〉=-16*3+29*4-17*4=0`

So,

`vecc` is perpendicular to `veca` and `vecb`

The unit vector is

`=vecc/||vecc||=1/sqrt(16^2+29^2+17^2)〈-16,-29,-17〉`

`=1/sqrt1386〈-16,-29,-17〉`