Question Video: Finding the Angle between Two Given Vectors in a Three-Dimensional Plane Mathematics

Find the angle πœƒ between the vectors 𝐕 = βˆ’π’ + 2𝐣 + 𝐀 and 𝐖= βˆ’3𝐒 + 6𝐣 + 3𝐀.

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Video Transcript

Find the angle πœƒ between the vectors 𝐕 negative 𝐒 plus two 𝐣 plus 𝐀 and 𝐖 negative three 𝐒 plus six 𝐣 plus three 𝐀.

In this question, we’re given two vectors 𝐕 and 𝐖 in terms of the unit directional vectors 𝐒, 𝐣, and 𝐀. We need to use this to determine the angle πœƒ between these two vectors. To answer this question, let’s start by recalling how we find the angle between two vectors. We recall if πœƒ is the angle between two vectors 𝐀 and 𝐁, then the cos of πœƒ will be equal to the dot product of vectors 𝐀 and 𝐁 divided by the magnitude of vector 𝐀 multiplied by the magnitude of vector 𝐁.

The same is also true in reverse. If πœƒ satisfies this equation, then πœƒ can be said to be an angle between vectors 𝐀 and 𝐁. However, usually when we’re talking about the angle between two vectors, we mean the smallest nonnegative angle between these two vectors. We can find this by taking the inverse cosine of both sides of this equation because, for example, in degrees, the inverse cosine function has a range between zero and 180 degrees. Therefore, we can find the angle πœƒ between the vectors 𝐕 and 𝐖 by finding the dot product between 𝐕 and 𝐖 and the magnitudes of vector 𝐕 and vector 𝐖.

Let’s start by finding the dot product between vector 𝐕 and vector 𝐖. That’s the dot product between the vector negative 𝐒 plus two 𝐣 plus 𝐀 and the vector negative three 𝐒 plus six 𝐣 plus three 𝐀. Remember, to find the dot product between two vectors, we need to find the sum of the products of the corresponding components of these two vectors. We could write these vectors in component form; however, it’s not necessary. The components of the vectors will be the coefficients of the unit directional vectors 𝐒, 𝐣, and 𝐀. Therefore, in the dot product between vectors 𝐕 and 𝐖, the sum of the products of the corresponding components of each vector is negative one times negative three plus two times six plus one multiplied by three. And if we calculate this expression, we see it’s equal to 18.

We now need to determine the magnitude of vector 𝐕 and the magnitude of vector 𝐖. To do this, we recall the magnitude of a vector is equal to the square root of the sums of the squares of its components. In other words, the magnitude of the vector π‘Žπ’ plus 𝑏𝐣 plus 𝑐𝐀 is equal to the square root of π‘Ž squared plus 𝑏 squared plus 𝑐 squared. We can use this to find the magnitude of vector 𝐕 and the magnitude of vector 𝐖. The magnitude of vector 𝐕 will be equal to the magnitude of negative 𝐒 plus two 𝐣 plus 𝐀. We need to find the square root of the sum of the squares of its components. Remember, the components of this vector will be the coefficients of the unit directional vectors 𝐒, 𝐣, and 𝐀. The magnitude of vector 𝐕 will be the square root of negative one all squared plus two squared plus one squared, which we can calculate is equal to the square root of six.

We can do the same to find the magnitude of 𝐖. The square root of the sum of the squares of the components of 𝐖 will be the square root of negative three all squared plus six squared plus three squared, which we can calculate is equal to the square root of 54. We can simplify this even further by noticing that 54 is equal to three squared multiplied by six. We can then use this to simplify root 54 to be the square root of three squared multiplied by root six, which is, of course, three times root six.

We’re now ready to use our formula to find the value of πœƒ, the angle between the two vectors 𝐕 and 𝐖. First, we know the cos of πœƒ will be equal to the dot product between vectors 𝐕 and 𝐖 divided by the magnitude of vector 𝐕 times the magnitude of vector 𝐖. Next, we can substitute the values we found for the dot product between 𝐕 and 𝐖, the magnitude of vector 𝐕 and the magnitude of vector 𝐖, into this equation. This gives us the cos of πœƒ is equal to 18 divided by root six multiplied by three root six. We can simplify the right-hand side of this equation. In our denominator, we have root six times three root six. Root six times root six is equal to six, and three times six is equal to 18. So, the right- hand side of this equation simplifies to give us 18 divided by 18, which is, of course, just equal to one. So, our equation simplifies to give us the cos of πœƒ is equal to one.

Finally, we can solve for our value of πœƒ by taking the inverse cosine of both sides of this equation. Doing this gives us that πœƒ is equal to the inverse cos of one, which we know is equal to zero. In this case, we’ll use degrees. So, πœƒ is equal to zero degrees.

Now, we could stop here. We’ve shown that the angle πœƒ between our two vectors 𝐕 and 𝐖 is equal to zero degrees. However, it’s worth pointing something out. Because the angle between these two vectors is equal to zero degrees, we can conclude that the two vectors must have the same direction. In other words, these two vectors must be parallel, and we can show this. If we were to multiply all of the components of vector 𝐕 by three, we would get back to 𝐖. In other words, three 𝐕 is equal to 𝐖. It’s also worth pointing out if the scalar value had instead been negative, then we know that 𝐕 and 𝐖 would point in opposite directions. However, in this case, the angle between the two vectors would have been 180 degrees.

Therefore, we were able to show the angle between two vectors 𝐕 negative 𝐒 plus two 𝐣 plus 𝐀 and 𝐖 negative three 𝐒 plus six 𝐣 plus three 𝐀 is equal to zero degrees.

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