What is the projection of <3,−6,2> onto <1,1,1>?

1 Answer
Morgan
Mar 5, 2017

The vector projection is <−13,−13,−13>, the scalar projection is −√33.

Explanation:

Given →a=<3,−6,2> and →b=<1,1,1>, we can find proj→b→a, the vector projection of →a onto →b using the following formula:

proj→b→a=⎛⎜ ⎜ ⎜⎝→a⋅→b∣∣∣→b∣∣∣⎞⎟ ⎟ ⎟⎠→b∣∣∣→b∣∣∣

That is, the dot product of the two vectors divided by the magnitude of →b, multiplied by →b divided by its magnitude. The second quantity is a vector quantity, as we divide a vector by a scalar. Note that we divide →b 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→b→a=a⋅b|b|, also written ∣∣proj→b→a∣∣.

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

→a⋅→b=<3,−6,2>⋅<1,1,1>

⇒(3⋅1)+(−6⋅1)+(2⋅1)

⇒3−6+2=−1

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

∣∣∣→b∣∣∣=√(bx)2+(by)2+(bz)2

∣∣∣→b∣∣∣=√(1)2+(1)2+(1)2

⇒√1+1+1=√3

And now we have everything we need to find the vector projection of →a onto →b.

proj→b→a=−1√3⋅<1,1,1>√3

⇒−<1,1,1>3

⇒<−13,−13,−13>

The scalar projection of →a onto →b is just the first half of the formula, where comp→b→a=a⋅b|b|. Therefore, the scalar projection is −1√3. You can rationalize the denominator to yield −√33 equivalently.

Hope that helps!

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