What is the unit vector that is orthogonal to the plane containing # (3i + 2j - 6k) # and # (3i – 4j + 4k) #?

1 Answer
Feb 17, 2016

Answer:

#u_n = (-16i-30j-18k)/38.5#
Notice in the picture I actually drew the unit vector in the opposite direction, i.e.: #u_n = (16i+30j+18k)/38.5#
It does matter it depend what you're are rotating to what as you apply the Right Hand Rule...

Explanation:

As you can see you vectors - let's call them
#v_(red)=3i +2j -6k# and #v_(blue) = 3i -4j +4k#
This two vector constitute a plane see the figure.
The vector formed by their x-product => #v_n=v_(red)xxv_(blue)#
is an orthogonal vector. The unit vector is obtained by normalizing the #u_n = v_n/|v_n|#

Now let's sub and calculate our orthonormal vector #u_n#
#v_n = [(i,j,k ), (3,2,-6 ), (3,-4,4)]#
#v_n = i[(2, -6), (-4, 4 )] -j [(3, -6 ), (3, 4 )] +k[(3,2), (3, -4 )]#
#v_n = ((2*4 ) - (-4*-6))i - ((3*4 ) - (3*-6))j + (( 3*-4) - (3*2))k #
#v_n = (8-24)i-(12+18)j+ (-12-6 ) = -16i-30j-18k#
#|v_n| =sqrt(16^2 +30^2+18^2) = sqrt(256+900+324 )~~ 38.5 #
#u_n = (-16i-30j-18k)/38.5#

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