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

1 Answer
Jan 15, 2017

Answer:

The vector projection is #< -14/17,21/17,-14/17 >#, the scalar projection is #(-7sqrt(17))/17#.

Explanation:

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

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

#=>-4-9+6=-7#

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+(-3)^2+(2)^2)#

#=>sqrt(4+9+4)=sqrt(17)#

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

#proj_(vecb)veca=(-7)/sqrt(17)*(< 2,-3,2 >)/sqrt(17)#

#=>(-7 < 2,-3,2 >)/17#

#=-7/17< 2,-3,2 >#

#=>< -14/17,21/17,-14/17 >#

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 #-7/sqrt(17)#, which does not simplify any further, besides to rationalize the denominator if desired, giving #(-7sqrt(17))/17#.

Hope that helps!