The magnitude of each vector is the Pythagorean sum of its components.
a. |v1| = √(2² + (-6)²) = √40 = 2√10
|v2| = √((-4)² + 7²) = √65
b. To make each vector into a unit vector, divide each component by the vector's magnitude.
u1 = v1/|v1| = (2/(2√10), -6/(2√10))
u1 = (√10/10, -3√10/10)
u2 = v2/|v2| = (-4/√65, 7/√65)
u2 = (-4√65/65, 7√65/65)
Answer:
5+5x-4y
Step-by-step explanation:
Answer:
The answer is 25.
Step-by-step explanation:
Answer is a idk the math teacher is not a teacher or teacher that she is going through so I am going
Answer: sin u = -5/13 and cos v = -15/17
Step-by-step explanation:
The nice thing about trig, a little information goes a long way. That’s because there is a lot of geometry and structure in the subject. If I have sin u = opp/hyp, then I know opp is the opposite side from u, and the hypotenuse is hyp, and the adjacent side must fit the Pythagorean equation opp^2 + adj^2 = hyp^2.
So for u: (-5)^2 + adj^2 = 13^2, so with what you gave us (Quad 3),
==> adj of u = -12 therefore cos u = -12/13
Same argument for v: adj = -15,
opp^2 + (-15)^2 = 17^2 ==> opp = -8 therefore sin v = -8/17
The cosine rule for cos (u + v) = (cos u)(cos v) - (sin u)(sin v) and now we substitute: cos (u + v) = (-12/13)(-15/17) - (-5/13)(-8/17)
I am too lazy to do the remaining arithmetic, but I think we have created a way to approach all of the similar problems.
