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

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Answer 1 · 2017-02-15T14:06:47

The unit vector is $= < - \frac{10}{\sqrt{117}}, \frac{1}{\sqrt{117}}, \frac{4}{\sqrt{117}} >$ $=<-10/sqrt117,1/sqrt117,4/sqrt117>$

We calculate the vector that is perpendicular to the other 2 vectors by doing a cross product,

Let $\to a = < 1, - 2, 3 >$ $veca=<1,-2,3>$

$\to b = < 1, 2, 2 >$ $vecb=<1,2,2>$

$vecc=|(hati,hatj,hatk),(1,-2,3),(1,2,2)|$

$=hati|(-2,3),(2,2)|-hatj|(1,3),(1,2)|+hatk|(1,-2),(1,2)|$

$=hati(-10)-hatj(-1)+hatk(4)$

$=<-10,1,4>$

Verification

$veca.vecc=<1,-2,3>.<-10,1,4>=-10-2+12=0$

$vecb.vecc=<1,2,2>.<-10,1,4>=-10+2+8=0$

The modulus of $vecc=||vecc|=||<-10,1,4>||=sqrt(100+1+16)=sqrt117$

The unit vector $= vecc /(||vec||)$

$=1/sqrt117<-10,1,4>$

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