Question

What is the projection of `<6,5,3 >` onto `<2,-1,8 >`?

Answer 1

The vector projection is `< 62/69,-31/69,248/69 >`, the scalar projection is `(31sqrt(69))/69`.

Explanation:

Given `veca= < 6,5,3 >` and `vecb= < 2,-1,8 >`, 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=< 6,5,3 > * < 2,-1,8 >`

`=> (6*2)+(5*-1)+(3*8)`

`=>12-5+24=31`

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+(8)^2)`

`=>sqrt(4+1+64)=sqrt(69)`

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

`proj_(vecb)veca=(31)/sqrt(69)*(< 2,-1,8 >)/sqrt(69)`

`=>(31 < 2,-1,8 >)/69`

`=31/69< 2,-1,8 >`

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

`=>< 62/69,-31/69,248/69 >`

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 `31/sqrt(69)`, which does not simplify any further, besides to rationalize the denominator if desired, giving `(31sqrt(69))/69`.

Hope that helps!