Question Video: Finding the Angle between Two Vectors Given Their Vector Product Physics • First Year of Secondary School

Video Transcript

The diagram shows two vectors 𝐀 and 𝐁. The magnitude of the vector product of 𝐀 and 𝐁 is 6.8. Calculate the angle between the two vectors, 𝜃. Give your answer to two significant figures.

Alright, so this is a question about vector products. We are given a diagram showing two vectors, 𝐀 and 𝐁. This diagram tells us the length or magnitude of each of the two vectors. In the question, they also tell us the magnitude of the vector product of the two vectors. We are asked to calculate the angle 𝜃 between the two vectors 𝐀 and 𝐁.

So we know the magnitude of the vector product of 𝐀 and 𝐁. And we know the magnitudes of each of the vectors 𝐀 and 𝐁 individually. We need to use this information in order to calculate the angle 𝜃 between them. Luckily, there exists an equation which relates all of these quantities. We’ll consider two general vectors, which we’ll label lowercase 𝐚 and lowercase 𝐛. We’re using the lowercase letters to distinguish the general case from our specific vectors from the question.

Let’s assume that 𝐚 and 𝐛 have some angle 𝜙 between them. Again we’re using a different Greek character for the angle to make it clear that we’re talking about a general case rather than our specific angle 𝜃 from the question. The magnitude of the vector product of two general vectors 𝐚 and 𝐛 can be written as the magnitude of 𝐚 multiplied by the magnitude of 𝐛 multiplied by the sin of the angle 𝜙 between 𝐚 and 𝐛.

So how does this help us? Well, in our case, we know the magnitude of the vector product and we know the magnitudes of the two individual vectors. And so the only thing that we don’t know in this equation is the value of the angle between the vectors. In order to answer the question, we need to rearrange this equation to make the angle the subject. So let’s take our general equation and start by dividing both sides of the equation by the magnitude of vector 𝐚 multiplied by the magnitude of vector 𝐛.

On the right-hand side of this equation, these magnitudes cancel with the corresponding magnitudes in the numerator. So we have that the magnitude of the vector product of 𝐚 and 𝐛 divided by the magnitude of 𝐚 times the magnitude of 𝐛 is equal to the sin of the angle between 𝐚 and 𝐛. Since we’re aiming to make the angle 𝜙 the subject, it makes sense to swap the left- and right-hand sides of this equation. Doing this, we can say that the sin of the angle 𝜙 is equal to the magnitude of the vector product of 𝐚 and 𝐛 divided by the magnitude of 𝐚 times the magnitude of 𝐛.

Finally, to make the angle itself the subject, we need to take the inverse sin of both sides of the equation. Then on the left-hand side, the inverse sin of the sin of 𝜙 just gives us 𝜙. And we have that the angle 𝜙 between two general vectors 𝐚 and 𝐛 is given by the inverse sin of the magnitude of the vector product of 𝐚 and 𝐛 divided by the magnitude of 𝐚 times the magnitude of 𝐛.

Now that we have this expression, we can apply it to the vectors capital 𝐀 and capital 𝐁 that we are given in the question. We need to work out the angle 𝜃 between these two vectors. We know that the magnitude of the vector product of 𝐀 and 𝐁 is 6.8. We know that the magnitude of 𝐀 is five, and we know that the magnitude of 𝐁 is four.

Evaluating this expression inside the inverse sin gives us 0.34. Then taking the inverse sin of 0.34, we get a value of 19.87687 and so on with more decimal places. Looking back at the question, we see that we are told to give our answer to two significant figures. And so our result rounds up to give us our answer to the question that, to two significant figures, the angle 𝜃 between vectors 𝐀 and 𝐁 is equal to 20 degrees.