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

Given #veca= < 3, -6, 2># and #vecb= < 1,1,1 >,# 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=< 3, -6, 2> * < 1,1,1 >#

#=> (3*1)+(-6*1)+(2*1)#

#=>3-6+2=-1#

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((1)^2+(1)^2+(1)^2)#

#=>sqrt(1+1+1)=sqrt(3)#

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

#proj_(vecb)veca=(-1)/sqrt3*(< 1,1,1 >)/sqrt3#

#=>(- < 1,1,1 >)/3#

#=><-1/3,-1/3,-1/3>#

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 #-1/sqrt3#. You can rationalize the denominator to yield #-sqrt3/3# equivalently.

Hope that helps!