What is the unit vector that is orthogonal to the plane containing $(29i-35j-17k)$ and $(32i-38j-12k)$ ?

Question

1347 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-18T07:45:59

The answer is $=1/299.7〈-226,-196,18〉$

The vector perpendiculatr 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 $veca=〈29,-35,-17〉$ and $vecb=〈32,-38,-12〉$

Therefore,

$| (veci,vecj,veck), (29,-35,-17), (32,-38,-12) |$

$=veci| (-35,-17), (-38,-12) | -vecj| (29,-17), (32,-12) | +veck| (29,-35), (32,-38) |$

$=veci(35*12-17*38)-vecj(-29*12+17*32)+veck(-29*38+35*32)$

$=〈-226,-196,18〉=vecc$

Verification by doing 2 dot products

$〈-226,-196,18〉.〈29,-35,-17〉=-226*29+196*35-17*18=0$

$〈-226,-196,18〉.〈32,-38,-12〉=-226*32+196*38-12*18=0$

So,

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

The unit vector is

$=1/sqrt(226^2+196^2+18^2)〈-226,-196,18〉$

$=1/299.7〈-226,-196,18〉$

What is the unit vector that is orthogonal to the plane containing (29i-35j-17k) (29i-35j-17k) and (32i-38j-12k) (32i-38j-12k) ?