given factors 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.

Answer 1

To determine if the vectors are orthogonal, we need to find the dot product of the two vectors.

The dot product of two vectors is given by:

u • v = (-6)(3) + (4)(3)
u • v = -18 + 12
u • v = -6

If the dot product of two vectors is 0, then the vectors are orthogonal. Since the dot product of u and v is not 0, these vectors are not orthogonal.

To find the angle between the two vectors, we can use the formula:

θ = arccos((u • v) / (||u|| * ||v||))

Where ||u|| and ||v|| are the magnitudes of the vectors.

First, calculate the magnitudes of the vectors:

||u|| = sqrt((-6)^2 + 4^2)
||u|| = sqrt(36 + 16)
||u|| = sqrt(52)
||u|| = 2(sqrt(13))

||v|| = sqrt(3^2 + 3^2)
||v|| = sqrt(9 + 9)
||v|| = sqrt(18)
||v|| = 3(sqrt(2))

Now, plug in the values to calculate the angle:

θ = arccos(-6 / (2(sqrt(13)) * 3(sqrt(2))))
θ = arccos(-6 / (6(sqrt(26))))
θ = arccos(-1 / sqrt(26))
θ = arccos(-1)
θ = cos^(-1)(-1)
θ = π

Therefore, the angle between the two vectors is π radians.

Answer 2

you got this far ok: θ = arccos(-1 / sqrt(26))
and then go off the rails, with the next step: θ = arccos(-1)
arccos(-1 / sqrt(26)) = 1.76819 or about 0.5628π
makes sense, since -1/√26 is small and negative, so θ will be just a bit greater than π/2

Answer 3

Apologies for the error in my previous response. Thank you for pointing it out.

Using the correct calculation:

θ = arccos(-1 / sqrt(26))

θ ≈ arccos(-0.377964)

θ ≈ 1.7682 radians or about 0.5628π radians

Therefore, the angle between the vectors u and v is approximately 1.7682 radians or about 0.5628π radians. Thank you for correcting the mistake.