How do you find a unit vector perpendicular to the vectors 6i + 2j - k and -2i + 3j + 4k?

Question

2333 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-25T16:43:11

#(+-1/3)(1 , - 2. 2)#. These two are in opposite directions.

If

#a=a1_ i + a_2 j + a_3 k=(a1_ i, a_2, a_3)# and

# b = b_1 i + b_2 j + b_3 k=(b_1 i, b_2 , b_3)

then

#+-(a X b )/} a X b }#

#=(+- b_2a_2 b_3=-a_3, a_3 b_1-a_1 b_3, a_1 b_2-a_2 b_1)/}a X b }#

is a unit vector perpendicular to both #a and b#.

Here, # a = (6, 2, -1) and b =(-2, 3, 4).#

So, # a X b =(11, -22, 22) and } a X b }= sqrt(11^2+(-22)^2+22^2)=33.

And so, the answer is #+-(1/33)(11)(1, -2, 2)=(+-1/3)(1 , - 2. 2)#.

These two are in opposite directions,

like ,#+-k# perpendicular to both #i and j#.