Question

Find the measure of the angle between the vectors 〈6, –3, 1〉 and 〈8, 9, –11〉

Answer 1

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.