Answer:
Option C.
Step-by-step explanation:
We know that for two vectors:
A = 〈a₁, a₂, a₃ 〉and B =〈b₁, b₂, b₃〉.
The dot product between A and B is:
A.B = IAI*IBI*cos(θ)
Where θ is the angle between the vectors.
And the dot product can be also written as:
A.B = a₁*b₁ + a₂*b₂ + a₃*b₃
With that, we can find the angle between our vectors.
In this case, the vectors are:
A = 〈6, –3, 1〉 and B = 〈8, 9, –11〉
The modules are:
IAI = √( 6^2 + (-3)^2 + 1^2) = 6.78
IBI = √( 8^2 + 9^2 + (-11)^2) = 16.31
And the dot product between A and B is:
A.B = 6*8 + (-3)*9 + 1*(-11) = 10
Then we have that:
10 = IAI*IBI*cos(θ) = ( 6.78)*(16.31)*cos(θ)
10/(6.78*16.31) = cos(θ)
Now remember the inverse cosine function, Acos(x), this function has the property:
Acos(cos(x)) = x
If we apply this to both sides, we get:
Acos(10/(6.78*16.31)) = Acos(cos(θ)) = θ = 84.8°
The correct option is C.
