# What is the projection of #<0, 1, 3># onto #<0, 4, 4>#?

##### 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!