How do you find a unit vector that is orthogonal to both u= - 6i + 4j + k and v= 3i + j + 5k?

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Answer 1 · 2017-05-17T05:34:14

The unit vector is $=1/1774〈19,33,-18〉$

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 $veca=〈-6,4,1〉$ and $vecb=〈3,1,5〉$

Therefore,

$| (veci,vecj,veck), (-6,4,1), (3,1,5) |$

$=veci| (4,1), (1,5) | -vecj| (-6,1), (3,5) | +veck| (-6,4), (3,1) |$

$=veci(4*5-1*1)-vecj(-6*5-1*3)+veck(-6*1-4*3)$

$=〈19,33,-18〉=vecc$

Verification by doing 2 dot products

$〈19,33,-18〉.〈-6,4,1〉=-6*19+4*33-1*18=0$

$〈19,33,-18〉.〈3,1,5〉=19*3+1*33-5*18=0$

So,

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

The unit vector is

$hatc=vecc/(||vecc||)=1/sqrt(19^2+33^2+18^2)*〈19,33,-18〉$

$=1/1774〈19,33,-18〉$