# Video: Determining the Components of a Vector

Given that the measure of the smaller angle between 𝐀 and 𝐁 is 150°, and |𝐁| = 54, determine the component of vector 𝐁 along 𝐀.

02:17

### Video Transcript

Given that the measure of the smaller angle between 𝐀 and 𝐁 is 150 degrees and the magnitude of 𝐁 equals 54, determine the component of vector 𝐁 along 𝐀.

Okay, in this exercise, we have these two vectors 𝐀 and 𝐁. And just to help us see how they relate to one another, let’s assume that they lie in the 𝑥𝑦-plane. We might then draw these two vectors like this. And we’re told that the smaller angle between these two vectors, that’s this one here, we can call it 𝜃, is 150 degrees. We’re asked to determine the component of vector 𝐁 along vector 𝐀.

Looking at our sketch though, we might wonder if the answer isn’t zero, because it looks as though none of vector 𝐁 lies along vector 𝐀. Here we have to be careful though because this phrase “along 𝐀” really means along the line that passes through vector 𝐀, in other words this dashed line, where the line is assumed to have a direction equal to the direction of vector 𝐀. In answering this question then, we’re going to be calculating this distance here. That’s the component of vector 𝐁 along 𝐀.

To get started solving for this, let’s recall that the scalar projection of one vector onto another is equal to the dot product of those vectors divided by the magnitude of the vector being projected onto. And this is also equal to the magnitude of the first vector, here we’ve called it 𝐕 one, multiplied by the cosine of the angle between the two vectors. It’s this form of the scalar projection equation that we can make use of in this particular exercise. After all, we know the magnitude of what we could call our first vector, the magnitude of 𝐁, and we also know the angle between our vectors.

What we want to calculate then is the magnitude of 𝐁 times the cos of 𝜃, or substituting in our given values 54 times the cos of 150 degrees. The cosine of this angle equals exactly negative the square root of three over two. So our scalar projection equals 54 times negative root three over two, or in simplified form negative 27 root three. This is the component of vector 𝐁 along vector 𝐀. And we can see from this result that, in general, a scalar projection can be negative. In this case, that came from the fact that the part of 𝐁 lying along the line through vector 𝐀 points in the opposite direction as vector 𝐀.