Video Transcript
Let the vector 𝐴 be equal to two, eight and the vector 𝐵 be equal to three, two. Find the angle between 𝐴 and 𝐵 giving your answer to two decimal places.
Let’s imagine vector 𝐴 and vector 𝐵 are beginning at the same point. Then it would look a little something like this. And the angle that were interested — let’s call that 𝜃 — is an acute angle. In fact, if we’re not told otherwise, we do indeed assume that 𝜃 is going to be an acute angle. And so we use something called the scalar product. This tells us that cos of 𝜃 is equal to the dot product of 𝐴 and 𝐵 over the product of the magnitudes of 𝐴 and 𝐵. Let’s begin by finding 𝐴 dot 𝐵. In this case, that’s the dot product of two, eight and three, two. That’s two times three plus eight times two which is equal to 22.
Next, we’ll find the magnitude of the vector 𝐴. That’s the square root of the sum of the squares of its components. So it’s the square root of two squared plus eight squared, which is the square root of 68. And similarly, the magnitude of 𝐵 is the square root of three squared plus two squared, which is the square root of 13. According to our scale of product formula, cos of 𝜃 equals 22 over the square root of 68 times the square root of 13. And due to the properties of the square root, we can say that the square root of 68 times the square to 13 is the square root of 68 times 13 which is 884. We solve this equation for 𝜃 by finding the inverse cos of both sides. So 𝜃 is the inverse cos of 22 over the square root of 884 which is 42.27368 and so on. Correct to two decimal places, that’s 42.27 degrees.
