Question

If `veca = <8 ,1 ,-5 >`, `vecb = <6 ,-2 ,-8 >` and `vecc=veca-vecb`, what is the angle between `veca` and `vecc`?

Answer 1

I get about `84.84^@` or `"1.48 rad"`.

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The angle between `veca` and `vecc` is ultimately determined from the following relationship:

`\mathbf(vecacdotvecc = || veca ||cdot|| vecc || costheta)`

Therefore, we will need to find `vecc` by subtracting `veca` and `vecb`.

`color(green)(veca - vecb)`

`= << 8,1,-5 >> - << 6,-2,-8 >>`

`= << 8-6, 1-(-2), -5-(-8) >>`

`= color(green)(<< 2,3,3 >> = vecc)`

Next, the angle between `veca` and `vecc` is gotten like so:

`costheta = (vecacdotvecc)/(|| veca ||cdot|| vecc ||)`

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

Afterwards, we do not yet know `veca cdot vecc`, so we'll need to evaluate the dot product between `veca` and `vecc`, which is just a sum of product terms consisting of corresponding vector coordinates.

`color(green)(veca cdot vecc)`

`= << 8,1,-5 >> cdot << 2,3,3 >>`

`= 8*2 + 1*3 + -5*3`

`= 16 + 3 - 15 = color(green)(4)`

And lastly, we'll need to know the product of the norms of `veca` and `vecc`. The norm of some arbitrary vector `vecv` is

`\mathbf(|| vecv || = sqrt(vecvcdotvecv)),`

so we'll need to dot `veca` and `vecc` with themselves and take the square root to get `|| veca ||` and `|| vecc ||`, respectively.

`color(green)(|| veca ||) = sqrt(veca cdot veca)`

`= sqrt(<< 8,1,-5 >>cdot<< 8,1,-5 >>)`

`= sqrt(8*8 + 1*1 + (-5)*(-5))`

`= sqrt(64 + 1 + 25)`

`= color(green)(sqrt(90))`

And now for `vecc`:

`color(green)(|| vecc ||) = sqrt(vecc cdot vecc)`

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

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

`= sqrt(4+9+9)`

`= color(green)(sqrt(22))`

Finally, we can combine all this to get the angle between `veca` and `vecc` as:

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

`= arccos(4/(sqrt(90)*sqrt(22)))`

`= arccos(4/(3sqrt(10)*sqrt(22)))`

`= arccos(4/(3sqrt(220)))`

`= arccos(cancel(4)^(2)/(3*cancel(2)sqrt(55)))`

`= arccos(2/(3*sqrt(55)))`

`~~` `color(blue)("1.48 rad")` or about `color(blue)(84.84^@)`.

Indeed, that should be the answer, as we see the result here to be:

`arccos(2/(3sqrt55))`.