Question

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

Answer 1

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`