Question

What is the projection of `<3, -6, 2>` onto `<1, 1, 1>`?

Answer 1

The vector projection is `<-1/3,-1/3,-1/3>`, the scalar projection is `-sqrt3/3`.

Explanation:

Given `veca= < 3, -6, 2>` and `vecb= < 1,1,1 >,` 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=< 3, -6, 2> * < 1,1,1 >`

`=> (3*1)+(-6*1)+(2*1)`

`=>3-6+2=-1`

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((1)^2+(1)^2+(1)^2)`

`=>sqrt(1+1+1)=sqrt(3)`

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

`proj_(vecb)veca=(-1)/sqrt3*(< 1,1,1 >)/sqrt3`

`=>(- < 1,1,1 >)/3`

`=><-1/3,-1/3,-1/3>`

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 `-1/sqrt3`. You can rationalize the denominator to yield `-sqrt3/3` equivalently.

Hope that helps!