Question

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

Answer 1

The vector projection is `< -14/17,21/17,-14/17 >`, the scalar projection is `(-7sqrt(17))/17`.

Explanation:

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

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

`=>-4-9+6=-7`

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+(-3)^2+(2)^2)`

`=>sqrt(4+9+4)=sqrt(17)`

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

`proj_(vecb)veca=(-7)/sqrt(17)*(< 2,-3,2 >)/sqrt(17)`

`=>(-7 < 2,-3,2 >)/17`

`=-7/17< 2,-3,2 >`

`=>< -14/17,21/17,-14/17 >`

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 `-7/sqrt(17)`, which does not simplify any further, besides to rationalize the denominator if desired, giving `(-7sqrt(17))/17`.

Hope that helps!