Question

What is the projection of `(4 i + 4 j + 2 k)` onto `(i + j -7k)`?

Answer 1

The vector projection is `< -2/17,-2/17,14/17 >`, the scalar projection is `(-2sqrt(51))/17`. See below.

Explanation:

Given `veca=(4i+4j+2k)` and `vecb= (i+j-7k)`, 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, which can be written as `veca=< 4,4,2 >` and `vecb=< 1,1,-7 >`.

`veca*vecb=< 4,4,2 >*< 1,1,-7 >`

`=> (4*1)+(4*1)+(2*-7)`

`=>4+4-14=-6`

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+(-7)^2)`

`=>sqrt(1+1+49)=sqrt(51)`

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

`proj_(vecb)veca=(-6)/sqrt(51)*(< 1,1,-7 >)/sqrt(51)`

`=>(-6 < 1,1,-7 >)/51`

`=>-2/17< 1,1,-7 >`

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

`=>< -2/17,-2/17,+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 `-6/sqrt(51)`, which does not simplify any further, besides to rationalize the denominator if desired, giving `(-6sqrt(51))/51 => (-2sqrt(51))/17`

Hope that helps!