Question

What is the projection of `<0, 1, 3>` onto `<3, 2, 1>`?

Answer 1

The vector projection, `mathbf p`, of `mathbfa` on `mathbfb` is:

`mathbf p = ( mathbf a cdot mathbf(hatb) )\ mathbf (hatb) = (abs(mathbf a) abs( mathbf(hatb)) cos theta) \ mathbf (hatb)`

`implies mathbf p = (abs(mathbf a) cos theta) \ mathbf (hatb`

Now:

`mathbf a cdot mathbf b = abs(mathbf a) abs( mathbf b) cos theta`

`implies cos theta = (mathbf a cdot mathbf b)/( abs(mathbf a) abs( mathbf b))`

`implies mathbf p = abs(mathbf a) (mathbf a cdot mathbf b)/( abs(mathbf a) abs( mathbf b)) mathbf (hatb`

`= (mathbf a cdot mathbf b)/( abs( mathbf b)) ( mathbf b)/( abs( mathbf b))`

`= (mathbf a cdot mathbf b)/( abs( mathbf b)^2) mathbf b`

`= (langle 0,1,3 rangle cdot langle3,2,1 rangle)/( 3^2 + 2^2 + 1^2) langle3,2,1 rangle`

`= (5)/(14) langle3,2,1 rangle`