Question

What is the angle between `<-3,2,6 >` and `<6,1,-5>`?

Answer 1

I got `"2.56 rad"` or about `146.6^@`.

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Let `veca = << -3,2,6 >>` and `vecb = << 6,1,-5 >>`. This angle between `veca` and `vecb` is found by the following equation:

`\mathbf(veca cdot vecb = ||veca||cdot||vecb||costheta)`

where:

• `vecacdotvecb` is the dot product between `veca` and `vecb`.
• `||veca||` is the norm of `veca`.
• `theta` is the angle between `veca` and `vecb`.

The dot product in `RR^n` for some vectors `veca` and `vecb` is defined as follows:

`\mathbf(vecacdotvecb)`

`= << a_1,a_2, . . . , a_n >> cdot << b_1,b_2, . . . , b_n >>`

`= \mathbf(a_1b_1 + a_2b_2 + . . . + a_nb_n)`

And the norm in `RR^n` for some vector `vecu` is defined as:

`\mathbf(||veca||)`

`= sqrt(vecacdotveca`

`= sqrt(a_1a_1 + a_2a_2 + . . . + a_n a_n)`

`= \mathbf(sqrt(a_1^2 + a_2^2 + . . . + a_n^2))`

So, we evaluate the angle as follows:

`costheta = (veca cdot vecb)/(||veca||cdot||vecb||)`

`color(green)(theta = arccos((veca cdot vecb)/(||veca||cdot||vecb||)))`

First, we should find the dot product between `veca` and `vecb`.

`color(green)(vecacdotvecb)`

`= << -3,2,6 >> cdot << 6,1,-5 >>`

`= -3*6 + 2*1 + 6*-5`

`= -18 + 2 - 30`

`= color(green)(-46)`

Then, we should find each norm.

`color(green)(||veca||)`

`= sqrt(veca cdot veca)`

`= sqrt(<< -3,2,6 >>cdot<< -3,2,6 >>)`

`= sqrt(-3*-3 + 2*2 + 6*6)`

`= sqrt(9 + 4 + 36)`

`= color(green)(7)`

`color(green)(||vecb||)`

`= sqrt(vecb cdot vecb)`

`= sqrt(<< 6,1,-5 >>cdot<< 6,1,-5 >>)`

`= sqrt(6*6 + 1*1 + -5*-5)`

`= sqrt(36 + 1 + 25)`

`= color(green)(sqrt62)`

Finally, let's get the angle.

`color(blue)(theta) = arccos((veca cdot vecb)/(||veca||cdot||vecb||))`

`= arccos((-46)/(7sqrt62))`

`~~` `color(blue)("2.56 rad")` or about `color(blue)(146.6^@)`.

And you can see that from using Wolfram Alpha on this question as well.