What is the unit vector that is normal to the plane containing <2i+7j-2k> and <8i-2j+3k>?

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Answer 1 · 2017-05-13T17:57:19

The unit vector is $= \frac{1}{\sqrt{4373}} \cdot ⟨ 17, - 22, - 60 ⟩$ $=1/sqrt(4373)*〈17,-22,-60〉$

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 $veca=〈2,7,-2〉$ and $vecb=〈8,-2,3〉$

Therefore,

$| (veci,vecj,veck), (2,7,-2), (8,-2,3) |$

$=veci| (7,-2), (-2,3) | -vecj| (2,-2), (8,3) | +veck| (2,7), (8,-2) |$

$=veci(7*3-2*2)-vecj(2*3+8*2)+veck(2*-2-7*8)$

$=〈17,-22,-60〉=vecc$

Verification by doing 2 dot products

$〈17,-22,-60〉.〈2,7,-2〉=17*2-22*7-60*-2=0$

$〈17,-22,-60〉.〈8,-2,3〉=17*8-22*-2-60*3=0$

So,

$vecc$ is perpendicular to $veca$ and $vecb$

The unit vector is

$hatc=vecc/||c||=1/sqrt(4373)*〈17,-22,-60〉$