What is the projection of #(3i + 2j - 6k)# onto # (3i – 4j + 4k)#?

1 Answer
Dec 26, 2016

Answer:

The vector projection is #< -69/41,92/41,-92/41 >#, the scalar projection is #(-23sqrt(41))/41#.

Explanation:

Given #veca=(3i+2j-6k)# and #vecb= (3i-4j+4k)#, 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=< 3,2,-6 ># and #vecb=< 3,-4,4 >#.

#veca*vecb=< 3,2,-6 >*< 3,-4,4 >#

#=> (3*3)+(2*-4)+(-6*4)#

#=>9-8-24=-23#

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((3)^2+(-4)^2+(4)^2)#

#=>sqrt(9+16+16)=sqrt(41)#

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

#proj_(vecb)veca=(-23)/sqrt(41)*(< 3,-4,4 >)/sqrt(41)#

#=>(-23 < 3,-4,4 >)/41#

#=>-23/41< 3,-4,4 >#

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

#=>< -69/41,92/41,-92/41 >#

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 #-23/sqrt(41)#, which does not simplify any further, besides to rationalize the denominator if desired, giving #(-23sqrt(41))/41#.

Hope that helps!