Question

What is the angle between `<-1,8,2 >` and `< 0,-3,1>`?

Answer 1

Approximately
`2.563color(white)(l)"rad"`

or equivalently,
`147^sf("o")`.

Explanation:

Let `theta` be the angle between the two vectors . By the geometric definition of scalar products,

`"Angle between vector " bb(a)" and "bb(b)=arccos ((bb(a)*bb(b))/(abs(bb(a))*abs(bb(b))))`

Where `bb(a)*bb(b)` the dot product of `bb(a)` and `bb(b)`, and `abs(bb(a))*abs(bb(b))` the product of their magnitude.

Let `bb(a)=<-1,8,2>` and `bb(b)=<0,-3,1>`,

`theta=arccos((<-1,8,2>*<0,-3,1>)/(sqrt(1+8^2+2^2)*sqrt(3^2+1^2)))`
`color(white)(theta)=arccos((-24+2)/(sqrt(69)*sqrt(10)))`
`color(white)(theta)=arccos(-22/sqrt(690))`
`color(white)(theta)~~2.563color(white)(l)"rad"=147^sf("o")`