What is the projection of #<5,8,3 ># onto #<2,4,-2 >#?

1 Answer
Feb 28, 2017

Answer:

The vector projection is #< 3,6,-3 >,# the scalar projection is #3sqrt6#.

Explanation:

Given #veca= < 5,8,3 ># and #vecb= < 2,4,-2 >,# 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=< 5,8,3 > * < 2,4,-2 >#

#=> (5*2)+(8*4)+(3*-2)#

#=>10+32-6=36#

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

#=>sqrt(4+16+4)=sqrt(24)#

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

#proj_(vecb)veca=(36)/sqrt(24)*(< 2,4,-2 >)/sqrt(24)#

#=>(36 < 2,4,-2 >)/24#

#=3/2< 2,4,-2 >#

#=>< 3,6,-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 #36/sqrt(24)#. This could be simplified further if desired:

#=>36/(2sqrt6)#

#=>18/sqrt6#

#=>(18sqrt6)/6#

#=>3sqrt6#

Hope that helps!