What is the projection of #<2,7,1 ># onto #<3,5,7 >#?
1 Answer
Answer:
The vector projection is
Explanation:
Given
#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
#color(darkblue)(comp_(vecb)veca=(a*b)/(abs(b))#
Which may also written as
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)#
#=>635+7=22#
Then we can find the magnitude of
#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
#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
Hope that helps!