Thread: Vector differentiation problems

1) For each of the following pairs of vectors, solve the triangle formed by using the origin and their two heads as the three vertices of the triangle. Calculate angle measures to two decimal places.

a) a = ( -1,2) and b=(2,3)
b) a= (-1,2,3) and b= (2,2,1)


2) Given A(1,-2,3), B(-2,3,-2), c(4,4,5) AND d(-1,0,0)

find angle a<(AB,AC)

1 a)

We have three points.

a = (-1,2)
b = (2,3)
c = (0,0)

These form a triangle were we want to find the angles. One way to solve the problem could be to use the law of cosines, but since you're talking about vector I assume you're supposed to use the following relationship (which is derived using the law of cosines)
u*v = |u||v|cos(θ) (were * denotes the dot product)

Now we could treat a and b as vectors from the origin to the point a, respectively. Therefore we have
a*b = |a||b|cos(θ) => -1*2+2*3 = sqrt((-1)^2+2^2)*sqrt(2^2+3^2)*cos(θ)

Try now and come back if you still don't get the idea.

Hi fkf, thanks for the help here's what i did"

i used the formula (0 = theta) cos0= a*b / | a| | b | to get my first angle of 60.26 using that i use the law of cosines to find a*b and got 4 (rounding up)

i proceeded to then find the angle between b and a*b and got cos0 = 6.93/ sqroot 13 x 4 ( or square root 16) to get 61.29 degrees after inputting in my calculator.

THEN i subtrading 180- 60.26-61.29 = 58.45 to get my last angle between a and (a*b)


please correct me if i have done anything wrong,
thanks!

any ideas on #2 ? im not sure how to approach this

I used AB*AC / |AB| |AC| and got 78.2 degrees as my angle, am i doing it right?

For each of the following pairs of vectors, solve the triangle formed by using the origin and their two heads as the three vertices of the triangle. Calculate angle measures to two decimal places. supreme hat red bull caps cheap caps