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

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Answer 1 · 2018-01-08T14:42:30

The unit vector is $=1/sqrt(549)(-12i-18j-9k)$

A vector perpendicular to 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 $veca=〈3,-1,-2〉$ and $vecb=〈3,-4,4〉$

Therefore,

$| (veci,vecj,veck), (3,-1,-2), (3,-4,4) |$

$=veci| (-1,-2), (-4,4) | -vecj| (3,-2), (3,4) | +veck| (3,-1), (3,-4) |$

$=veci(-1*4-(-2)*-4)-vecj(3*4-3*-2)+veck(-4*3-3*-1)$

$=〈-12,-18,-9〉=vecc$

Verification by doing 2 dot products

$〈3,-1,-2〉.〈-12,-18,-9〉=-3*12+1*18+2*9=0$

$〈3,-4,4〉.〈-12,-18,-9〉=-3*12+4*18-4*9=0$

So,

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

The unit vector $hatc$ in the direction of $vecc$ is

$hatc=(vecc)/sqrt((-12)^2+(-18)^2+(-9)^2)=vecc/sqrt(549)$

$=1/sqrt(549)(-12i-18j-9k)$

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