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

##### 1 Answer

The vector projection is

#### Explanation:

Given *vector* projection of

#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 aunit 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

scalarprojection 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.

Then we can find the magnitude of

And now we have everything we need to find the vector projection of

The scalar projection of

Hope that helps!