What is the projection of #<7,1,1 ># onto #<-1,6,-8 >#?

1 Answer
Jan 3, 2017

Answer:

The vector projection is #< 9/101,-54/101,72/101 >#, the scalar projection is #(-9sqrt(101))/101#.

Explanation:

Given #veca= < 7,1,1 ># and #vecb= < -1,6,-8 >#, 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=< 7,1,1 > * < -1,6,-8 >#

#=> (7*-1)+(1*6)+(1*-8)#

#=>-7+6-8=-9#

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+(6)^2+(-8)^2)#

#=>sqrt(1+36+64)=sqrt(101)#

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

#proj_(vecb)veca=(-9)/sqrt(101)*(< -1,-6,-8 >)/sqrt(101)#

#=>(-9 < -1,6,-8 >)/101#

#=-9/101< -1,6,-8 >#

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

#=>< 9/101,-54/101,72/101 >#

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 #-9/sqrt(101)#, which does not simplify any further, besides to rationalize the denominator if desired, giving #(-9sqrt(101))/101#.

Hope that helps!