What is the unit vector that is orthogonal to the plane containing $(20 j + 31 k)$ $(20j +31k)$ and $(32 i - 38 j - 12 k)$ $(32i-38j-12k)$ ?

Question

1593 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-13T07:38:12

The unit vector is $= = \frac{1}{1507.8} < 938, 992, - 640 >$ $==1/1507.8<938,992,-640>$

The vector orthogonal to 2 vectros in a plane 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 $veca=〈0,20,31〉$ and $vecb=〈32,-38,-12〉$

Therefore,

$| (veci,vecj,veck), (0,20,31), (32,-38,-12) |$

$=veci| (20,31), (-38,-12) | -vecj| (0,31), (32,-12) | +veck| (0,20), (32,-38) |$

$=veci(20*-12+38*31)-vecj(0*-12-31*32)+veck(0*-38-32*20)$

$=〈938,992,-640〉=vecc$

Verification by doing 2 dot products

$〈938,992,-640〉.〈0,20,31〉=938*0+992*20-640*31=0$

$〈938,992,-640〉.〈32,-38,-12〉=938*32-992*38+640*12=0$

So,

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

The unit vector is

$hatc=vecc/||vecc||=(<938,992,-640>)/||<938,992,-640>||$

$=1/1507.8<938,992,-640>$

What is the unit vector that is orthogonal to the plane containing (20j +31k) (20j+31k) (20j +31k) and (32i-38j-12k) (32i−38j−12k) (32i-38j-12k) ?