The two vectors 𝐯 and 𝐮 are shown in the diagram. The following parallelogram is drawn based on the two vectors 𝐯 and 𝐮. Using the properties of parallelograms, find 𝜃.
So, the first thing we need to do if we’re dealing with a parallelogram is mark on our parallel sides. And that, you can see from the diagram we have two pairs of parallel sides. So, the first relationship that we can use is this one here. So, we can see that our 50 degrees and our 𝜃 have either side of them parallel lines. So therefore, we can use the relationship known as the supplementary angles relationship.
This tells us that supplementary angles sum to 180 degrees. So therefore, we can say that 𝜃 is gonna be equal to 180 minus 50. That’s because, as we said, they are supplementary angles. And as you can see, I’ve written that down next to the 180 minus 50. That’s because whenever you’re doing this type of question, you need to give reasoning when you’re trying to find your answers. So therefore, we can say the value of 𝜃 is gonna be 130 degrees. And I’ve also shown this now on our diagram. Well now, there is in fact a second part to the question. And now that we’ve answered the first part, let’s move on to the second part.
So now, the second part of the question says that given that 𝐯 equals 13 and 𝐮 equals 11, so that’s our vectors, use the cosine rule to find the magnitude of 𝐯 plus 𝐮. Give your answer to two decimal places.
So the first thing we need to do is remember what the cosine rule tells us. And it tells us that if we have a triangle, 𝑎𝑏𝑐, then 𝑎 squared equals 𝑏 squared plus 𝑐 squared minus two cos 𝐴, where 𝐴 is the angle opposite the side 𝑎. So now that we have the cosine rule, let’s apply it to our situation that we’ve got in the second part of our question.
Well, we can apply the cosine rule to our scenario because we’ve got a triangle, and we’ve got our angle 𝐴, because that’s our 130 degrees. And then we’ve got 𝑎, the side, because this is the side opposite to 130 degrees, which is 𝐯 plus 𝐮. And therefore, as we said that 𝐯 plus 𝐮 is equal to 𝑎, we can say that 𝑎 squared is gonna be equal to 11 squared plus 13 squared, because that was our 𝐮 and our 𝐯, minus two multiplied by 11 multiplied by 13 cos of 130. And we got these values because our 𝑏 and 𝑐, or 𝐮 and 𝐯, which were 11 and 13 have also shown that on our diagram, because I’ve put 𝐮 on the right-hand side as well, because this is gonna be vector 𝐮 as well as the left-hand side. So, what we’re gonna get is that 𝑎 squared is equal to 290 minus 286 cos 130.
So now what we need to do, we need to take the square root of both sides of the equation. And that’s because we’re trying to find 𝑎 because 𝑎 is our 𝐯 plus 𝐮. And if you square root 𝑎 squared, you just get 𝑎. And whatever you do to one side of the equation, you have to do to the other. And when we do that, we get 𝑎 is equal to 21.767, et cetera.
So, have we finished here? Well, no, because we want the answer rounded to two decimal places. And to round our answer to two decimal places, I’ve put a line after the second decimal place. Now, look to the digit to the right of it cause this is gonna be our deciding digit. And if this is five or above, it means we’re gonna round the six up to a seven. Therefore after rounding, and given that 𝐯 equals 13 and 𝐮 equals 11, we’ve used the cosine rule to find the magnitude of 𝐯 plus 𝐮. And, the answer is 𝐯 plus 𝐮 is equal to 21.77, and that’s to two decimal places.