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

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Answer 1 · 2017-09-23T15:21:36

The unit vector is $= \frac{1}{\sqrt{194}} ⟨ 7, - 12, - 1 ⟩$ $=1/sqrt194〈7,-12,-1〉$

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

$hatc=1/sqrt194〈7,-12,-1〉$

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