Question

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

Answer 1

The unit vector is `=1/sqrt194〈7,-12,-1〉`

Explanation:

The cross product of 2 vectors is calculated with the determinant

`| (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=〈3,2,-3〉` and `vecb=〈2,1,2〉`

Therefore,

`| (veci,vecj,veck), (3,2,-3), (2,1,2) |`

`=veci| (2,-3), (1,2) | -vecj| (3,-3), (2,2) | +veck| (3,2), (2,1) |`

`=veci(2*2+3*1)-vecj(3*2+3*2)+veck(3*1-2*2)`

`=〈7,-12,-1〉=vecc`

Verification by doing 2 dot products

`〈7,-12,-1〉.〈3,2,-3〉=7*3-12*2+1*3=0`

`〈7,-12,-1〉.〈2,1,2〉=7*2-12*1-1*2=0`

So,

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

The modulus of `vecc` is

`||vecc||=sqrt(7^2+(-12)^2+(-1)^2)=sqrt(49+144+1)=sqrt194`

Therefore,

The unit vector is

`hatc=1/sqrt194〈7,-12,-1〉`