Question

What is the projection of `(3i + 2j - 6k)` onto `(3i – 4j + 4k)`?

Answer 1

The vector projection is `< -69/41,92/41,-92/41 >`, the scalar projection is `(-23sqrt(41))/41`.

Explanation:

Given `veca=(3i+2j-6k)` and `vecb= (3i-4j+4k)`, 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=< 3,2,-6 >` and `vecb=< 3,-4,4 >`.

`veca*vecb=< 3,2,-6 >*< 3,-4,4 >`

`=> (3*3)+(2*-4)+(-6*4)`

`=>9-8-24=-23`

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((3)^2+(-4)^2+(4)^2)`

`=>sqrt(9+16+16)=sqrt(41)`

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

`proj_(vecb)veca=(-23)/sqrt(41)*(< 3,-4,4 >)/sqrt(41)`

`=>(-23 < 3,-4,4 >)/41`

`=>-23/41< 3,-4,4 >`

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

`=>< -69/41,92/41,-92/41 >`

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 `-23/sqrt(41)`, which does not simplify any further, besides to rationalize the denominator if desired, giving `(-23sqrt(41))/41`.

Hope that helps!