Question

What is the projection of `(-4i + 3k)` onto `(-2i -j + 2k)`?

Answer 1

The vector projection is `<-28/9,-14/9,28/9>,` the scalar projection is `14/3`.

Explanation:

Given `veca= < -4, 0, 3>` and `vecb= < -2,-1,2 >,` 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=< -4, 0, 3> * < -2,-1,2 >`

`=> (-4*-2)+(0*-1)+(3*2)`

`=>8+0+6=14`

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

`=>sqrt(4+1+4)=sqrt(9)=3`

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

`proj_(vecb)veca=(14)/3*(< -2,-1,2 >)/3`

`=>(14 < -2,-1,2 >)/9`

`=><-28/9,-14/9,28/9>`

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 `14/3`.

Hope that helps!