Question

What is the angle between `<-1,8,6 >` and `< -6,3,0>`?

Answer 1

`theta~~63.58^o`

Explanation:

The equation for the angle between two vectors is given by:

`cos(theta)=(veca*vecb)/(|veca|*|vecb|)`

Where `veca=< -1,8,6 >` and `vecb= < -6,3,0 >`

First we find the dot product of the two vectotrs:

`veca*vecb= < -1,8,6 > *< -6,3,0 >`

`=>(-1*-6)+(8*3)+(6*0)`

`=>6+24=30`

Next, we find the product of the magnitude of each vector:

`|veca|=sqrt((a_x)^2+(a_y)^2+(a_z)^2)`

`=>sqrt((-1)^2+(8)^2+(6)^2)`

`=>sqrt(1+64+36)`

`=>sqrt(101)`

`|vecb|=sqrt((b_x)^2+(b_y)^2+(b_z)^2)`

`=>sqrt((-6)^2+(3)^2+(0)^2)`

`=>sqrt(36+9+0)`

`=>sqrt(45)`

And `sqrt(45)*sqrt(101)=sqrt(4545)`

Revisiting our original equation, we can solve for `theta` by taking the inverse cosine:

`theta=cos^-1((veca*vecb)/(|veca|*|vecb|))`

Substituting in our values found above:

`theta=cos^-1((veca*vecb)/(|veca|*|vecb|))`

`=>theta=cos^-1((30)/(sqrt(4545)))`

This comes out to `~~63.58^o` or `~~1.11rad`

Explanation of equation:

We can derive the given equation using a geometric interpretation of the vectors.

Given vectors `veca` and `vecb`:

By the law of cosines, given:

`c^2=a^2+b^2-2abcos(theta)`

Applying this to our vector-made triangle:

`|veca-vecb|^2=|veca|^2+|vecb|^2-2|veca|*|vecb|cos(theta)`

Simplifying:

`=>(veca-vecb)*(veca-vecb)=(veca*veca)+(vecb*vecb)-2|veca|*vec|b|cos(theta)`

`=>cancel(veca^2)-2(veca*vecb)cancel(+vecb^2)=cancel(veca^2)+cancel(vecb^2)-2|veca|*vec|b|cos(theta)`

`=>-2(veca*vecb)=-2|veca|*vec|b|cos(theta)`

`=>(veca*vecb)=|veca|*vec|b|cos(theta)`

Solve for `cos(theta)`:

`=>cos(theta)=(veca*vecb)/(|veca|*vec|b|)`