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

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Answer 1 · 2017-06-04T17:32:32

The unit vector is $=<-3/sqrt13, 2/sqrt13,0>$

The vector perpendicular to 2 vectors is calculated with the determinant (cross product)

$| (veci,vecj,veck), (d,e,f), (g,h,i) |$

where $veca=〈d,e,f〉$ and $vecb=〈g,h,i〉$ are the 2 vectors

Here, we have $veca=〈2,3,-7〉$ and $vecb=〈-2,-3,2〉$

Therefore,

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

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

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

$=〈-15,10,0〉=vecc$

Verification by doing 2 dot products

$〈-15,10,0〉.〈2,3,-7〉=-15*2+10*3-7*0=0$

$〈-15,10,0〉.〈-2,-3,2〉=-15*-2+10*-3-0*2=0$

So,

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

The modulus of $vecc$ is $||vecc||=sqrt(15^5+10^2)=sqrt(325)$

The unit vector is

$hatc=vecc/||vecc||=1/325<-15,10,0>$

$=<-3/sqrt13, 2/sqrt13,0>$

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