Answer:
Angle between vectors u and v: 26.56 degrees
Step-by-step explanation:
I just recently did this question, so here's my response and hopefully that helps you:
Vector u is <-3,-4> and vector v is <-2,-1>. To find the angle between the two vectors, you first need to find their dot product. This means that u*v = <-3,-4> * <-2,-1> = (-3)(-2) + (-4)(-1) = 6 + 4 = 10. Second, you find the product of each of the vectors' magnitudes. The magnitude for vector u is ||u|| = sqrt((-3)^2 + (-4)^2) = sqrt(9+16) = sqrt(25) = 5. The magnitude for vector v is ||v|| = sqrt((-2)^2 + (-1)^2) = sqrt(4+1) = sqrt(5). The product of the magnitudes would thus be ||u||||v|| = 5sqrt(5). Third, divide the dot product by the product of the magnitudes which would be u*v/||u||||v|| = 10/5sqrt(5) = 2sqrt(5)/5. Lastly, you take the arccosine of the result which would be theta=cos^-1(2sqrt(5)/5) = 26.56 degrees, which is the angle between vectors u and v.
The explanation for determining the angle between the two vectors and find the angle in degrees should be explained below.
Since
vector v is from (0,0) to ( -2,-1)
So,
[tex]\bar{v}=-2i-j[/tex]
Here, vector u is from (0,0) to (-3,-4)
So,
[tex]\bar{u}=-3i-j4[/tex]
Now here we used dot product concept
Now
[tex]\bar{u}.\bar{v}=|u|.|v|cos\theta\\ \bar{u}.\bar{v}=(-3i-4j).(-2i-j)=6+4=10\\ |u|=\sqrt{9+16}=5\\ |v|=\sqrt{4+1}=\sqrt{5}\\ cos\theta=10/(5*\sqrt5)=2/\sqrt5\\ \theta=cos^{-1}(2/\sqrt5)= 26.56^o[/tex]
Learn more about graph here: https://brainly.com/question/2111181
