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

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Answer 1 · 2017-03-25T12:41:20

The answer is $=<4/sqrt90,5/sqrt90,-7/sqrt90>$

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 $veca=〈-3,1,-1〉$ and $vecb=〈1,2,2〉$

Therefore,

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

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

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

$=〈4,5,-7〉=vecc$

Verification by doing 2 dot products

$〈4,5,-7〉.〈-3,1,-1〉=-12+5+7=0$

$〈4,5,-7〉.〈1,2,2〉=4+10-14=0$

So,

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

The unit vector is

$=1/sqrt(16+25+49)*<4,5,-7>$

$=<4/sqrt90,5/sqrt90,-7/sqrt90>$

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