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

1 Answer
Dec 24, 2016

Answer:

The vector projection is #< 0,2,2 >#, the scalar projection is #2sqrt2#. See below.

Explanation:

Given #veca=< 0,1,3 ># and #vecb= < 0,4,4 >#, 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=< 0,1,3 >*< 0,4,4 >#

#=> (0*0)+(4*1)+(4*3)#

#=>0+4+12=16#

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((0)^2+(4)^2+(4)^2)#

#=>sqrt(0+16+16)=sqrt(32)#

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

#proj_(vecb)veca=(16)/sqrt(32)*(< 0,4,4 >)/sqrt(32)#

#=>(16 < 0,4,4 >)/32#

#=>(< 0,4,4 >)/2#

#=>< 0,2,2 >#

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 #16/sqrt(32)#, which further simplifies to #2sqrt2#. I've shown the simplification below.

#16/sqrt(32)#

#=>16/sqrt(16*2)#

#=>16/(4*sqrt2)#

#=>4/sqrt2#

#=>(4*sqrt2)/(sqrt2*sqrt2)#

#=>(4sqrt2)/2#

#=>2sqrt2#

Hope that helps!