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

1 Answer
Jul 25, 2016

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

Explanation:

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.