Question

What is the projection of `<5,8,3 >` onto `<2,4,-2 >`?

Answer 1

The vector projection is `< 3,6,-3 >,` the scalar projection is `3sqrt6`.

Explanation:

Given `veca= < 5,8,3 >` and `vecb= < 2,4,-2 >,` we can find `proj_(vecb)veca`, the vector projection of `veca` onto `vecb` using the following formula:

`proj_(vecb)veca=((veca*vecb)/(|vecb|))vecb/|vecb|`

That is, the dot product of the two vectors divided by the magnitude of `vecb`, multiplied by `vecb` divided by its magnitude. The second quantity is a vector quantity, as we divide a vector by a scalar. Note that we divide `vecb` by its magnitude in order to obtain a unit vector (vector with magnitude of `1`). You might notice that the first quantity is scalar, as we know that when we take the dot product of two vectors, the resultant is a scalar.

Therefore, the scalar projection of `a` onto `b` is `comp_(vecb)veca=(a*b)/(|b|)`, also written `|proj_(vecb)veca|`.

We can start by taking the dot product of the two vectors.

`veca*vecb=< 5,8,3 > * < 2,4,-2 >`

`=> (5*2)+(8*4)+(3*-2)`

`=>10+32-6=36`

Then we can find the magnitude of `vecb` by taking the square root of the sum of the squares of each of the components.

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

`|vecb|=sqrt((2)^2+(4)^2+(2)^2)`

`=>sqrt(4+16+4)=sqrt(24)`

And now we have everything we need to find the vector projection of `veca` onto `vecb`.

`proj_(vecb)veca=(36)/sqrt(24)*(< 2,4,-2 >)/sqrt(24)`

`=>(36 < 2,4,-2 >)/24`

`=3/2< 2,4,-2 >`

`=>< 3,6,-3 >`

The scalar projection of `veca` onto `vecb` is just the first half of the formula, where `comp_(vecb)veca=(a*b)/(|b|)`. Therefore, the scalar projection is `36/sqrt(24)`. This could be simplified further if desired:

`=>36/(2sqrt6)`

`=>18/sqrt6`

`=>(18sqrt6)/6`

`=>3sqrt6`

Hope that helps!