What is the unit vector that is orthogonal to the plane containing $(2i + 3j – 7k)$ and $(3i – 4j + 4k)$ ?

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Answer 1 · 2017-04-10T11:59:21

The unit vector is $=〈-16/sqrt1386,-29/sqrt1386,-17/sqrt1386〉$

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 $veca=〈2,3,-7〉$ and $vecb=〈3,-4,4〉$

Therefore,

$| (veci,vecj,veck), (2,3,-7), (3,-4,4) |$

$=veci| (3,-7), (-4,4) | -vecj| (2,-7), (3,4) | +veck| (2,3), (3,-4) |$

$=veci(3*4-7*4)-vecj(2*4+7*3)+veck(-2*4-3*3)$

$=〈-16,-29,-17〉=vecc$

Verification by doing 2 dot products

$〈-16,-29,-17〉.〈2,3,-7〉=-16*2-29*3-7*17=0$

$〈-16,-29,-17〉.〈3,-4,4〉=-16*3+29*4-17*4=0$

So,

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

The unit vector is

$=vecc/||vecc||=1/sqrt(16^2+29^2+17^2)〈-16,-29,-17〉$

$=1/sqrt1386〈-16,-29,-17〉$

What is the unit vector that is orthogonal to the plane containing (2i + 3j – 7k) (2i + 3j – 7k) and (3i – 4j + 4k) (3i – 4j + 4k) ?