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

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Answer 1 · 2017-02-20T10:12:27

The unit vector is $=<5/sqrt42,4/sqrt42,1/sqrt42>$

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

Let $veca=<-1,1,1>$

$vecb=<1,-2,3>$

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

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

$=hati(5)-hatj(-4)+hatk(1)$

$=<5,4,1>$

Verification

$veca.vecc=<-1,1,1>.<5,4,1>=-5+4+1=0$

$vecb.vecc=<1,-2,3>.<5,4,1>=5-8+3=0$

The modulus of $vecc=||vecc||=||<5,4,1>||=sqrt(25+16+1)=sqrt42$

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

$=1/sqrt42<5,4,1>$

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