Question

Two vectors and are given by and . If these two vectors are drawn starting at the same point, what is the angle between them

Answer 1

To find the angle between two vectors, we can use the dot product formula: A·B = |A||B|cosθ. Let's calculate the dot product first. A·B = (2)(3) + (4)(-5) + (6)(1) = -16. Next, let's find the magnitudes of each vector: |A| = sqrt(2^2 + 4^2 + 6^2) = sqrt(56). |B| = sqrt(3^2 + (-5)^2 + 1^2) = sqrt(35). Now we can find the angle: cosθ = A·B / (|A||B|) = -16 / (sqrt(56)sqrt(35)). θ = acos(cosθ) = acos(-16 / (sqrt(56)sqrt(35))).

Step-by-step explanation:

To find the angle between two vectors, we can use the dot product formula: A·B = |A||B|cosθ. Let's calculate the dot product first.

A·B = (2)(3) + (4)(-5) + (6)(1) = -16

Next, let's find the magnitudes of each vector: |A| = sqrt(2^2 + 4^2 + 6^2) = sqrt(56)

|B| = sqrt(3^2 + (-5)^2 + 1^2) = sqrt(35)

Now we can find the angle: cosθ = A·B / (|A||B|) = -16 / (sqrt(56)sqrt(35))

θ = acos(cosθ) = acos(-16 / (sqrt(56)sqrt(35)))

Answer 2

Answer: hello your question is incomplete below is the complete question

The Two vectors; A = 5i + 6j +7k and B = 3i -8j +2k.

answer;

angle = 102°

Step-by-step explanation:

multiplying the vectors

A.B = |A| * |B|* cosθ

hence : Cosθ = (Ai*Bi )+ (Aj*Bj) + ( Ak*Bk/ (√A^2 *√B^2 )

= 15 - 48 + 14 /(√25+26+29) * (√9+64+4)

= -0.206448454

θ = cos^-1 ( -0.206448454) = 101.9° ≈ 102°