What is the projection of # (4 i + 4 j + 2 k)# onto #(i + j -7k)#?

1 Answer
Dec 26, 2016

Answer:

The vector projection is #< -2/17,-2/17,14/17 >#, the scalar projection is #(-2sqrt(51))/17#. See below.

Explanation:

Given #veca=(4i+4j+2k)# and #vecb= (i+j-7k)#, 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, which can be written as #veca=< 4,4,2 ># and #vecb=< 1,1,-7 >#.

#veca*vecb=< 4,4,2 >*< 1,1,-7 >#

#=> (4*1)+(4*1)+(2*-7)#

#=>4+4-14=-6#

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((1)^2+(1)^2+(-7)^2)#

#=>sqrt(1+1+49)=sqrt(51)#

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

#proj_(vecb)veca=(-6)/sqrt(51)*(< 1,1,-7 >)/sqrt(51)#

#=>(-6 < 1,1,-7 >)/51#

#=>-2/17< 1,1,-7 >#

You can distribute the coefficient to each component of the vector and write as:

#=>< -2/17,-2/17,+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 #-6/sqrt(51)#, which does not simplify any further, besides to rationalize the denominator if desired, giving #(-6sqrt(51))/51 => (-2sqrt(51))/17#

Hope that helps!