Question

What is the projection of `<2,7,1 >` onto `<3,-5,7 >`?

Answer 1

The vector projection is `<-66/83,110/83,-154/83>` and the scalar projection is `-22/sqrt83`.

Explanation:

Given `veca= < 2, 7, 1>` and `vecb= < 3,-5,7 >,` we can find `proj_(vecb)veca`, the vector projection of `veca` onto `vecb` using the following formula:

`color(darkblue)(proj_(vecb)veca=((veca*vecb)/(abs(vecb)))vecb/abs(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 (a 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 given by:

`color(darkblue)(comp_(vecb)veca=(a*b)/(abs(b))`

Which may also written as `abs(proj_(vecb)veca)`.

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

`veca*vecb=< 2, 7, 1> * < 3,-5,7 >`

`=> (2*3)+(7*-5)+(1*7)`

`=>6-35+7=-22`

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

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

`abs(vecb)=sqrt((3)^2+(-5)^2+(7)^2)`

`=>sqrt(9+25+49)=color(darkblue)(sqrt(83))`

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

`proj_(vecb)veca=(-22)/sqrt(83)*(< 3,-5,7 >)/sqrt(83)`

`=>(-22 < 3,-5,7 >)/(83)`

`=><-66/83,110/83,-154/83>`

The scalar projection of `veca` onto `vecb` is just the first half of the formula, as described above. Therefore, the scalar projection is `-22/sqrt(83)`.

Hope that helps!