What is the projection of #<6,5,3 ># onto #<2,-1,8 >#?

1 Answer
Dec 30, 2016

Answer:

The vector projection is #< 62/69,-31/69,248/69 >#, the scalar projection is #(31sqrt(69))/69#.

Explanation:

Given #veca= < 6,5,3 ># and #vecb= < 2,-1,8 >#, 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=< 6,5,3 > * < 2,-1,8 >#

#=> (6*2)+(5*-1)+(3*8)#

#=>12-5+24=31#

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((2)^2+(-1)^2+(8)^2)#

#=>sqrt(4+1+64)=sqrt(69)#

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

#proj_(vecb)veca=(31)/sqrt(69)*(< 2,-1,8 >)/sqrt(69)#

#=>(31 < 2,-1,8 >)/69#

#=31/69< 2,-1,8 >#

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

#=>< 62/69,-31/69,248/69 >#

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 #31/sqrt(69)#, which does not simplify any further, besides to rationalize the denominator if desired, giving #(31sqrt(69))/69#.

Hope that helps!