What is the projection of #<2,7,1 ># onto #<3,-5,7 >#?

1 Answer
Aug 18, 2017

Answer:

The vector projection is #<-66/83,110/83,-154/83># and the scalar projection is #-22/sqrt83#.

Explanation:

Given #veca= < 2, 7, 1># and #vecb= < 3,-5,7 >,# we can find #proj_(vecb)veca#, the vector projection of #veca# onto #vecb# using the following formula:

#color(darkblue)(proj_(vecb)veca=((veca*vecb)/(abs(vecb)))vecb/abs(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 (a 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 given by:

#color(darkblue)(comp_(vecb)veca=(a*b)/(abs(b))#

Which may also written as #abs(proj_(vecb)veca)#.

We can start by taking the dot product of the two vectors.

#veca*vecb=< 2, 7, 1> * < 3,-5,7 >#

#=> (2*3)+(7*-5)+(1*7)#

#=>6-35+7=-22#

Then we can find the magnitude of #vecb# by taking the square root of the sum of the squares of each of the components.

#abs(vecb)=sqrt((b_x)^2+(b_y)^2+(b_z)^2)#

#abs(vecb)=sqrt((3)^2+(-5)^2+(7)^2)#

#=>sqrt(9+25+49)=color(darkblue)(sqrt(83))#

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

#proj_(vecb)veca=(-22)/sqrt(83)*(< 3,-5,7 >)/sqrt(83)#

#=>(-22 < 3,-5,7 >)/(83)#

#=><-66/83,110/83,-154/83>#

The scalar projection of #veca# onto #vecb# is just the first half of the formula, as described above. Therefore, the scalar projection is #-22/sqrt(83)#.

Hope that helps!