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Question Video: Finding the Angle between Two Vectors Given Their Scalar Product Physics

Consider the two vectors 𝐩, with a magnitude of 7.7, and πͺ, with a magnitude of 15. The scalar product of the two vectors is 27. What is the angle between the vectors? Give your answer to one decimal place.

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Video Transcript

Consider the two vectors 𝐩, with a magnitude of 7.7, and πͺ, with a magnitude of 15. The scalar product of the two vectors is 27. What is the angle between the vectors? Give your answer to one decimal place.

Okay, so in this question, we’re told that we have two vectors. We’re given the magnitudes of each of them, and we’re told the value of their scalar product. The question is asking us to find the angle between the vectors. So we need a way of calculating this angle from the information given, that is, from the magnitudes of the two vectors and from the value of their scalar product. Luckily for us, the scalar product of two vectors can be defined in terms of the magnitude of the vectors and the angle between those vectors.

Consider two general vectors 𝐀 and 𝐁 with some angle πœƒ between them. The scalar product of 𝐀 and 𝐁 is defined as the magnitude of 𝐀 multiplied by the magnitude of 𝐁 multiplied by the cos of the angle πœƒ between them. But how does this help us? Well, in our situation, we know the value of the scalar product, and we know the magnitudes of the two vectors. In this formula, we want to find this angle. So let’s start by making cos of πœƒ the subject of this equation. We divide both sides of the equation by the magnitude of 𝐀 and by the magnitude of 𝐁. On the right-hand side of this equation, the magnitudes in the numerator and the denominator cancel each other out. Then, swapping the left- and right-hand sides of this equation, we can write that the cos of πœƒ is equal to the scalar product of 𝐀 and 𝐁 divided by the magnitude of 𝐀 multiplied by the magnitude of 𝐁.

To make πœƒ the subject, we then need to take the inverse cosine of both sides of the equation. Now, we have an expression for the angle between our two vectors given their magnitudes and the value of their scalar product. This is just what we said we needed in order to answer this question. In our case, we have vectors 𝐩 and πͺ. Let’s label the angle between them πœ‘ so as to distinguish it from the πœƒ that we used in our general case formula. Then, we can take our general case equation and write it for our specific case of 𝐩, πͺ, and πœ‘. We have that the angle between our vectors, that’s πœ‘, is equal to the inverse cos of the scalar product of 𝐩 and πͺ divided by the magnitude of 𝐩 multiplied by the magnitude of πͺ.

Let’s substitute in our values. We are told that the scalar product of 𝐩 and πͺ has a value of 27. We are told that the magnitude of 𝐩 is 7.7 and the magnitude of πͺ is 15. Substituting in these values gives us an expression that we can evaluate in order to find the angle πœ‘. If we do the calculation, we find that our value of the angle πœ‘ is 76.48109 and so on with further decimal places. Notice though that the question asks us to give our answer to one decimal place. So to this level of precision, our final result is that the angle πœ‘ between the two vectors 𝐩 and πͺ is equal to 76.5 degrees.

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