Question

What is the projection of `< 2 , -6, 0>` onto `< 5, -3, 8>`?

Answer 1

Let `veca = << 2,-6,0 >>` and `vecb = << 5,-3,8 >>`.

The projection of `veca` onto `vecb` is written as:

`\mathbf("proj"_(vecb)veca = (vecacdotvecb)/(||vecb||cdot||vecb||)vecb)`

So first, we have to find the dot product between `veca` and `vecb`.

`color(green)(vecacdotvecb)`

`= << 2,-6,0 >> cdot << 5,-3,8 >>`

`= 2*5 + -6*-3 + 0*8`

`= 10 + 18 = color(green)(28)`

And the norm of `vecb` is:

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

`= sqrt(vecbcdotvecb)`

`= sqrt(<< 5,-3,8 >>cdot<< 5,-3,8 >>)`

`= sqrt(5*5+ -3*-3+8*8)`

`= sqrt(25+9+64)`

`= sqrt(98) = color(green)(7sqrt2)`

Thus, the projection of `veca` onto `vecb` is:

`color(blue)("proj"_(vecb)veca)`

`= 28/(7sqrt2*7sqrt2)<< 5,-3,8 >>`

`= 28/98<< 5,-3,8 >>`

`= 2/7<< 5,-3,8 >>`

`= color(blue)(<< 10/7,-6/7,16/7 >>)` <-- let this be `vecc`.

At this point, to check that this worked, you should perform the following to determine whether the angle between the vectors is indeed `0^@`:

`vecb cdot vecc = ||vecb||cdot||vecc||costheta`

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

`= arccos((5*10/7 + -3*-6/7 + 8*16/7)/(sqrt(5^2 + (-3)^2 + 8^2)*sqrt((10/7)^2 + (-6/7)^2 + (16/7)^2)))`

`= arccos(((50 + 18 + 128)/7)/(7sqrt2*sqrt((100+36+256)/49)))`

`= arccos(((196)/7)/(7sqrt2*sqrt8))`

`= arccos((28)/(28))`

`= color(blue)(0^@)`