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.
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
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.
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.