Consider vectors u=(2,3,5) and v=(1,-4,5). What is the angle θ between them

Answer 1

since z values are both the same, 5

just work with 2 dimensions to simplify (2,3) and (1,-4)

plot them to get a rough idea of the angle

1st point has tanA = 3/2, with the x axis

2nd point has tanB = -4/1 = -4 with the x axis

A = tan^-1(3/2) =56.31 degrees

B= tan^-1(-4) = - 75.96 degrees

add their reference angles to get: 56,31+ 75.96 = 132.27 degrees

or

cosT = cross product divided by product of magnitudes

= -10/sqr(13x17)

T = about cos^-1(-.673)=132.27 degrees

or in 3D

cos^-1(15/sqr(42x38))= 67.95

double it for 4th quadrant

135.9 degrees

close to 132.27

find the error you get 2 points extra credit

Answer 2

The angle θ is just going to be the arccos (〈u.v〉/ ||u||⋅||v|| ) .

where〈u.v〉is the inner product of the vectors u and v

and ||u||⋅||v|| are the product of the norms of u an v.

〈u.v〉= 2⋅1 + 3⋅ - 4 + 5⋅3 =5

||u|| = √4+9+25 = √38

||v|| =√1+16+9 = √26

θ = arccos (〈u.v〉/ ||u||⋅||v|| )

= arccos( 5/ (√38 * √26)

= arccos(5/31.432467291)

= arccos(0.15907119074)

= 80.847 degrees