Find the angle between the vectors. (Round your answer to the nearest degree.) a = ‹-8, 6› b = ‹√11, 5›

Answer 1

(* A slight correction to Raymond's answer, which only works if you are trying to find the angle formed from the right triangle whose non-right-angle endpoints have coordinates given by the two vectors, which is not what the problem is asking. *)

We can approach this by using the dot product operation. If we have two vectors x = (x1, x2) and y = (y1, y2), then their dot product dot(x,y) satisfies the following:

dot(x,y) = x1*y1 + x2*y2 = ||x||*||y||*cosθ, where || || denotes the magnitude of a vector, and θ is the angle between the two vectors.

Solving for θ, we get:

θ = arccos( dot(x,y) / (||x||*||y||) ), where arccos( ) is the inverse cosine function.

Plugging in our vectors x = ( -8, 6 ) and y = ( sqrt(11), 5 ), we get:

θ = arccos( (-8*sqrt(11) + 6*5) / (sqrt(64+36)*sqrt(11+25)) )

θ = arccos( (-8sqrt(11) + 30) / (sqrt(100)*sqrt(36)) )

θ = arccos( (-8sqrt(11) + 30) / (10*6) )

θ = arccos( -8sqrt(11)/60 + 30/60 )

θ = arccos( -2sqrt(11)/15 + 1/2 )

Inputting this into a calculator gives θ = 86.69 degrees.

Answer 2

Let θ be the angle between a and b.

cosθ = (a • b) / [ (ll a ll )( ll b ll)] = (-8√11 + 30) / [(10)(6)] = (-8√11 + 30) / 60 = -2√11 / 15 + 1/2

cosθ = 0.05778

θ = Arccos(0.05778) ≈ 87°

Answer 3

it helps to plot the two points, connect them with a line and notice it is slightly downward sloping

construct a nearly right triangle with 3 sides, and find the angle opposite the side length of the line connecting the vectors' endpoints, Use law of cosines to find the angle which should be close to 90 degrees

length of sides are sqr(64+36) = 10

and sqr(11+25)=6

and the line segment connecting the two points = sqr(1+(sqr11+8)^2) = sqr(1+11+64+16sqr11) = sqr(76+16sqr11)

c^2 =a^2 +b^2-2abCosC

76+16sqr11 = 100+36 -2(60)CosT where T= the angle between the two vectors

16sqr11 -60= -120CosT

CosT = (60-16sqr11)/120= about .06

T = Cos^-1(60-16sqr11)/120)=about 87 degrees

or another method

180-arctan(6/8) -arctan(5/sqr11)

= 180-( 37+56) = 87 degrees