Question

What is the angle between vectors A=2i+3j+k and B=i+2j-4k?

Answer 1

`theta = 76.5^"o"`

Explanation:

We're asked to find the angle between two vectors, given their unit vector notations.

To do this, we can use the equation

`vecA · vecB = ABcostheta`

rearranging to solve for angle, `theta`:

`costheta = (vecA · vecB)/(AB)`

`theta = arccos((vecA · vecB)/(AB))`

where

• `vecA * vecB` is the dot product of the two vectors, which is

`vecA * vecB = A_xB_x + A_yB_y + A_zB_z`

`= (2)(1) + (3)(2) + (1)(-4) = color(red)(4`

• `A` and `B` are the magnitudes of vectors `vecA` and `vecB`, which are

`A = sqrt(2^2 + 3^2 + 1^2) = color(green)(sqrt14`

`B = sqrt(1^2 + 2^2+ (-4)^2) = color(purple)(sqrt21`

Therefore, we have

`theta = arccos(color(red)(4)/(color(green)(sqrt14)color(purple)(sqrt21))) = arccos(4/(7sqrt6)) = color(blue)(76.5^"o"`