Given vectors u=(-6,4) and v=(3,3), determine if the vectors are orthogonal. if they are not orthogonal, find the angle between the two vectors.
C. The vectors are not orthogonal. The angle between the two vectors is 101.3 degrees
These are the choices:
a) The vectors are orthogonal
b) The vectors are not orthogonal. The angle between the two vectors is 72.5°
c) The vectors are not orthogonal. The angle between the two vectors is 101.3°
d) The vectors are not orthogonal. The angle between the two vectors is 130.6°
I feel like it might be a just because the angle I got between the two isn't an option.
u•v = |u| |v| cosθ
if u⊥v then u•v = 0
(-6,4)•(3,3) = -18+12 ≠ 0
so what angle θ has cosθ = -6/√936 ?
sunwoo is correct
Yes, Sunwoo is correct.
To determine if two vectors are orthogonal, we need to check if their dot product is equal to zero.
The dot product (or scalar product) of two vectors u=(u1, u2) and v=(v1, v2) is calculated using the formula:
u · v = u1v1 + u2v2
In this case, the vectors u=(-6,4) and v=(3,3). Let's calculate their dot product:
u · v = (-6)(3) + (4)(3)
= -18 + 12
= -6
Since the dot product of u and v is not equal to zero (-6 ≠ 0), the vectors u and v are not orthogonal.
To find the angle between these two vectors, we can use the formula:
cosθ = (u · v) / (||u|| ||v||)
where ||u|| and ||v|| are the magnitudes (or lengths) of the vectors u and v, respectively.
The magnitudes can be calculated using the formula:
||u|| = √(u1^2 + u2^2)
||v|| = √(v1^2 + v2^2)
In this case:
||u|| = √((-6)^2 + 4^2)
= √(36 + 16)
= √52
= 2√13
||v|| = √(3^2 + 3^2)
= √(9 + 9)
= √18
= 3√2
Now we can substitute these values in the formula to find cosθ:
cosθ = (-6) / (2√13 * 3√2)
Simplifying the expression:
cosθ = -6 / (6√13√2)
= -1 / (√13√2)
= -1 / (√26)
To find the angle θ, we can take the inverse cosine (arccos) of cosθ. However, since -1 / (√26) is a decimal value, we can convert it to degrees or radians:
θ ≈ 137.6° or θ ≈ 2.4 radians
Therefore, the angle between the vectors u and v is approximately 137.6 degrees or 2.4 radians.
Well, let's first determine if the vectors are orthogonal. To do that, we can use the dot product formula: u · v = u1*v1 + u2*v2.
So, u · v = (-6)*(3) + (4)*(3) = -18 + 12 = -6.
Since the dot product is not zero, the vectors are not orthogonal.
To find the angle between the vectors, we can use the formula: θ = arccos((u · v) / (|u| * |v|)).
The magnitude of u, |u|, can be found using the formula: |u| = √(u1^2 + u2^2).
Hence, |u| = √((-6)^2 + (4)^2) = √(36 + 16) = √52.
Similarly, |v| = √(3^2 + 3^2) = √18.
Now, plugging these values into the angle formula:
θ = arccos((-6) / (√52 * √18)).
Calculating this:
θ ≈ 0.736 radians or 42.18 degrees.
So, the angle between the vectors is approximately 0.736 radians or 42.18 degrees.