Question

Whats the angle between the two?

`i-sqrt3k` and `2i+k-3j`

Answer 1

If we have two vectors `vec a=((x_0),(y_0),(z_0))` and `vec b((x_1),(y_1),(z_1))`, then the angle `theta` between them is related to as

`vec a*vec b=|vec a||vec b|cos(theta)`

or

`theta=arccos((vec a*vec b)/(|vec a||vec b|))`

In the problem, there are two vectors given to us: `vec a=((1),(0),(sqrt(3)))` and `vec b=((2),(-3),(1))`.

Then, `|vec a|=sqrt(1^2+0^2+sqrt(3)^2)=2` and `|vec b|=sqrt(2^2+(-3)^2+1^2)=sqrt(14)`.

Also, `vec a*vec b=1*2+0*(-3)+sqrt(3)*1=2+sqrt(3)`.

Therefore, the angle `theta` between them is

`theta=arccos((vec a*vec b)/(|vec a||vec b|))=arccos((2+sqrt(3))/(2*sqrt(14)))~~60.08^@`.