What is the projection of # (-4i + 3k)# onto #(-2i -j + 2k)#?

1 Answer
Mar 4, 2017

Answer:

The vector projection is #<-28/9,-14/9,28/9>,# the scalar projection is #14/3#.

Explanation:

Given #veca= < -4, 0, 3># and #vecb= < -2,-1,2 >,# 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=< -4, 0, 3> * < -2,-1,2 >#

#=> (-4*-2)+(0*-1)+(3*2)#

#=>8+0+6=14#

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

#=>sqrt(4+1+4)=sqrt(9)=3#

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

#proj_(vecb)veca=(14)/3*(< -2,-1,2 >)/3#

#=>(14 < -2,-1,2 >)/9#

#=><-28/9,-14/9,28/9>#

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 #14/3#.

Hope that helps!