How do you find a unit vector perpendicular to both (2,1,1) and the x-axis?

1 Answer
Nov 23, 2016

Please see the explanation for steps leading to: hatC = sqrt(2)/2hatj - sqrt(2)/2hatk

Explanation:

Given: barA = 2hati + hatj+ hatk

Let barB = "a vector along the x axis" =hati

The vector barC =barA xx barB will be perpendicular to both but is will not be a unit vector.

barC =barA xx barB = | (hati, hatj, hatk, hati, hatj), (2,1,1,2,1), (1,0,0,1,0) |=

{1(0) - 1(0)}hati + {1(1) - 2(0)}hatj + {2(0) - (1)(1)}hatk

barC = hatj - hatk

Compute the magnitude of barC

|barC| = sqrt(1^2 + 1^2) = sqrt(2)

The unit vector perpendicular to both is:

hatC = barC/|barC|

hatC = (hatj - hatk)/sqrt(2)

hatC = sqrt(2)/2hatj - sqrt(2)/2hatk