Answer to Question #166198 in Physics for gibson mkandawire

Question

Find the angle between the vectors 𝑨⃗⃗ =(𝟑𝒊̂+𝟐𝒋̂+𝒌̂) and 𝑨⃗⃗ =(𝟓𝒊̂−𝟐𝒋̂−𝟑𝒌̂)

Answer 1

The angle between two vectors can be found as follows:

$$A\cdot B=|A|\cdot|B|\cdot cos\theta,$$ $$cos\theta=\dfrac{A\cdot B}{|A|\cdot|B|},$$ $$cos\theta=\dfrac{A_x\cdot B_x+A_y\cdot B_y+A_z\cdot B_z}{\sqrt{A_x^2+A_y^2+A_z^2}\cdot\sqrt{B_x^2+B_y^2+B_z^2}},$$ $$\theta=cos^{-1}(\dfrac{A_x\cdot B_x+A_y\cdot B_y+A_z\cdot B_z}{\sqrt{A_x^2+A_y^2+A_z^2}\cdot\sqrt{B_x^2+B_y^2+B_z^2}}),$$ $$\theta=cos^{-1}(\dfrac{3\cdot5+2\cdot(-2)+1\cdot(-3)}{\sqrt{(3)^2+(2)^2+(1)^2}\cdot\sqrt{(5)^2+(-2)^2+(-3)^2}}),$$ $$\theta=69.7^{\circ}.$$