Question

What is the angle between `<-3,5,6>` and `<-1,9,-3>`?

Answer 1

`cos theta = (<-3,5,6> * <-1,9,-3>)/((sqrt((-3)^2+5^2+6^2)) sqrt((-1)^2+9^2+(-3)^2))`
`cos theta=((-3)(-1)+(5)(9)+(6)(-3))/((sqrt70)(sqrt91)) =30/sqrt6370 ->theta = cos^-1 (30/sqrt6370) = 67.92^@`

Explanation:

Find the dot product of the vectors then divide by the magnitude of each. Finally take the cosine inverse to find `theta`

Answer 2

Use the definition of a dot product to find that:

`theta=67.1^o` or `1.17` radians

Explanation:

The dot product is a very convenient tool that allows you to find the angle between any two vectors. It is given by:

`vec a*vec b=|\vec a|*|\vec b|*costheta`

Where `vec a*vec b` represents the dot product between `vec a` and `vec b`, given by:

`vec a*vec b=(a_1*b_1)+(a_2*b_2)+(a_3*b_3)+...`

Note: The dot product will always result in a scalar value.

`|vec a|` and `|vec b|` are the magnitudes of `vec a` and `vec b`, respectively. They are also scalar values, calculated by using:

`|vec a| = (a_1^2+a_2^2+a_3^2+...)^(1/2)`

With this information, the angle between the vectors <−3,5,6> and <−1,9,−3> can be found:

Step 1: Calculate the dot product
`vec a*vec b=(a_1*b_1)+(a_2*b_2)+(a_3*b_3)+...`
`vec a*vec b=(-3*-1)+(5*9)+(6*-3)`
`vec a*vec b=3+45+(-18)`
`vec a*vec b=30`

Step 2: Calculate the magnitude of each vector
`|vec a| = (a_1^2+a_2^2+a_3^2+...)^(1/2)`
`|vec a| = ((-3)^2+(5)^2+(6)^2)^(1/2)`
`|vec a| = (9+25+36)^(1/2)`
`|vec a| = (70)^(1/2)`
`|vec a| = 8.367`

Following the same procedure for `|vec b|` gives:
`|vec b|=9.220`

`|vec a|*|vec b|` then is: `8.367*9.220=77.143`

Step 3: Solve for `theta`:

`vec a*vec b=|\vec a|*|\vec b|*costheta`
`30=77.143*costheta`
`costheta=30/77.143`
`costheta=0.389`
`theta=cos^-1(0.389)`
`theta=67.1^o` or `1.17` radians