Question

What is the projection of `<0, 1, 3>` onto `<0, 4, 4>`?

Answer 1

The vector projection is `< 0,2,2 >`, the scalar projection is `2sqrt2`. See below.

Explanation:

Given `veca=< 0,1,3 >` and `vecb= < 0,4,4 >`, 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=< 0,1,3 >*< 0,4,4 >`

`=> (0*0)+(4*1)+(4*3)`

`=>0+4+12=16`

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((0)^2+(4)^2+(4)^2)`

`=>sqrt(0+16+16)=sqrt(32)`

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

`proj_(vecb)veca=(16)/sqrt(32)*(< 0,4,4 >)/sqrt(32)`

`=>(16 < 0,4,4 >)/32`

`=>(< 0,4,4 >)/2`

`=>< 0,2,2 >`

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 `16/sqrt(32)`, which further simplifies to `2sqrt2`. I've shown the simplification below.

`16/sqrt(32)`

`=>16/sqrt(16*2)`

`=>16/(4*sqrt2)`

`=>4/sqrt2`

`=>(4*sqrt2)/(sqrt2*sqrt2)`

`=>(4sqrt2)/2`

`=>2sqrt2`

Hope that helps!