Question

16. If a = 3i+j+2k and 5 = 2î – 2j+4k →
(a) find the magnitude of axb
(b) find a unit vector perpendicular to both ā and 5
(c)find the cosine and sine of the angle between the vector ā and
and
B,​

Answer 1

Answer:

Explanation:

θ=76.5°

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

To do this, we can use the equation

→A⋅→B=ABcosθ

rearranging to solve for angle, θ:

cosθ=→A⋅→BAB

θ=arccos⎛⎝→A⋅→BAB⎞⎠

where

→A⋅→B is the dot product of the two vectors, which is

→A⋅→B=AxBx+AyBy+AzBz

=(2)(1)+(3)(2)+(1)(−4)=4

A and B are the magnitudes of vectors →A and →B, which are

A=√22+32+12=√14

B=√12+22+(−4)2=√21

Therefore, we have