Question

What is the projection of `<7,1,1 >` onto `<-1,6,-8 >`?

Answer 1

The vector projection is `< 9/101,-54/101,72/101 >`, the scalar projection is `(-9sqrt(101))/101`.

Explanation:

Given `veca= < 7,1,1 >` and `vecb= < -1,6,-8 >`, 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=< 7,1,1 > * < -1,6,-8 >`

`=> (7*-1)+(1*6)+(1*-8)`

`=>-7+6-8=-9`

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+(6)^2+(-8)^2)`

`=>sqrt(1+36+64)=sqrt(101)`

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

`proj_(vecb)veca=(-9)/sqrt(101)*(< -1,-6,-8 >)/sqrt(101)`

`=>(-9 < -1,6,-8 >)/101`

`=-9/101< -1,6,-8 >`

You can distribute the coefficient to each component of the vector and write as:

`=>< 9/101,-54/101,72/101 >`

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 `-9/sqrt(101)`, which does not simplify any further, besides to rationalize the denominator if desired, giving `(-9sqrt(101))/101`.

Hope that helps!