How do you find a Unit vector perpendicular to vector a=(4,-3,1) and vector b=(2,3,-1)?

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Answer 1 · 2017-02-21T17:18:22

$hatvecn=sqrt10/10(-vecj+3veck)$

$" A vector perpendicular to: " veca "& "vecb" can be found by finding the cross product" vecaxxvecb$

$"the corresponding unit vector is found by " hatvecu=vecu/|vecu|$

$veca=<4,-3,1>, vecb=<2,3,-1>$

$vecaxxvecb=|(veci,vecj,veck),(4,-3,1),(2,3,-1)|$

expanding by Row 1

$vecaxxvecb=veci|(-3,color(white)-1),(color(white)(-)3,-1)|-vecj|(4,1),(2,-1)|+veck|(4,-3),(2,3)|$

$vecaxxvecb=veci(3-3)+vecj(-4-2)+veck(12- -6)$

$vecaxxvecb=-6vecj+18veck$

since we are only wanting the direction we can simplify teh vector by removing any common factors.

so a perpendicular vector in this direction can be written

$vec n=-vecj+3veck=> |vecn|=sqrt(1^2+3^2)=sqrt10$

so a unit vector will be:

$hatvecn=1/sqrt10(-vecj+3veck)=sqrt10/10(-vecj+3veck)$