### Video Transcript

A woman started walking from home and walked six miles at 40 degrees north of east, then two miles at 15 degrees east of south, then five miles at 30 degrees south of west. If she walked straight home, how far would she have to walk and in what direction? Give the distance in miles correct to two decimal places and the direction in degrees correct to one decimal place.

It would be a good idea to represent this story using a diagram. She starts walking from home which we represent as a point. And from there, she walks six miles at 40 degrees north of east. This is a vector quantity. We’re given both the magnitude, six miles, and the direction, 40 degrees north of east. And as the direction is given with reference to the compass directions, it’s probably a good idea to mark a compass on our diagram. From home, this direction is east. And so, this direction is 40 degrees north of east. And letting the magnitude of this factor be six miles. After walking six miles at 40 degrees north of east, she is then at the end of the vector at the terminal point. But she doesn’t stop there, she continues to walk two miles at 15 degrees east of south. We mark in south. Then the displacements two miles at 15 degrees east of south is represented by the orange vector. She then travels five miles at 30 degrees south of west. So finally, she ends up at the terminal point of this blue vector here.

The question is how does she get home? In which direction does she have to walk and for how far? All the vectors in our diagram are placed nose to tail. This gives us a hint that we’re dealing with a vector addition problem. Her final displacement from home, when she’s starting to wonder how to get back home, is the sum of the purple, orange, and blue displacement vectors. Our task is to find the green vector which takes her back home and so makes her displacement from home just the zero vector. Let’s first focus on finding this sum.

To find this sum, we first write each vector in this sum in component form. Let’s give these vectors some names to make things easier for ourselves. We call them 𝑈, 𝑉, and 𝑊. So our sum is 𝑈 plus 𝑉 plus 𝑊. First, let’s find 𝑈 in component form. We know how to do this, given the magnitude of the vector and the measure 𝜃 of the angle, measured counterclockwise from the vector 𝐼. First of all, we’re going to have to decide in which directions the unit vectors 𝐼 and 𝐽 are going to point in. We choose to make 𝐼 point east and 𝐽 point north which is somehow conventional. We’re ready to substitute in then. The magnitude of 𝑈 is six. And the value of 𝜃, the direction measured counterclockwise from east, is 40 degrees. So here we have the components of 𝑈. Let’s clear some space and do the same for 𝑉.

We write down the formula in terms of the magnitude of 𝑉 and 𝜃, the measure of the angle measured counterclockwise from east. The magnitude of 𝑉 is two. But what is the value of 𝜃? We might be tempted to say that 𝜃 is 15 degrees because this is the value we see marked. 𝜃 must be measured counterclockwise from the vector 𝐼. And as 𝐼 points east, this means counterclockwise from east. We just happened to be lucky with 𝑈 that the angle marked was already measured counterclockwise from east, as required. Unfortunately, for 𝑉, we’re going to have to do some more work. Let’s mark in the direction east from the initial point and mark in the angle we want which we call 𝜑. Looking at the angles around this initial point, we should be able to see that 𝜑 plus 15 degrees must be 90 degrees because the angle between east and south is 90 degrees. And so 𝜑 is 75 degrees. But 𝜃 is measured counterclockwise from east. And we see that we have 75 degrees clockwise from east. So 𝜃 is negative 75 degrees. And we substitute this value in. It’s time again to clear some space and find the components of 𝑊.

Again, the magnitude of 𝑊 is the easy bit. We’re told that this is five. The difficult bit is making sure the we measure 𝜃 in the right way. Again, we mark in the direction east from the initial point. We see that 𝜑 is measured clockwise from east because it is supplementary with the 30-degree angle marked. And so 𝜃, measured counterclockwise, is negative 150 degrees. Alternatively, we could have gone the other way around the circle adding 180 degrees to 30 degrees. And hence, getting a value of 210 degrees. Both approaches are valid. And both give the same components of 𝑊.

Now we have the components of 𝑈, 𝑉, and 𝑊. We can find their sum just by adding the components. We add up the 𝑥-components of the vectors using a calculator in degree mode to find that the 𝑥-component of the final vector is 0.783777 dot dot dot. Similarly, we add up the 𝑦-components of the vectors to find out the 𝑦-component of her final displacement is negative 0.575125 dot dot dot. To get back home from her final position, she has to go in the opposite direction. So we want the green vector that we’re looking for to be the additive inverse of this final displacement. The displacement she has to walk to get home is therefore negative 0.78377 dot dot dot, 0.575125 dot dot dot. Each component is the opposite of the component in the final displacement. You can check that the sum of 𝑈, 𝑉, 𝑊, and this green vector is zero. And so if she follows this path, she really does get back to where she started.

We’re almost done now. We found the displacement that she has to walk to get home. But we need to give this displacement in terms of the distance in miles and the direction in degrees she needs to walk. In other words, we need to take our vector, which we have in component form, and write it in magnitude-direction form. We’ve made some room to work this out. The distance we’re looking for is the magnitude of this displacement vector. And we know how to find the magnitude of this vector. It’s the square root of the sum of squares of the components.

Substituting the components into this formula, of course using the full six decimal places of accuracy rather than just the two I’ve written here, we get 0.97, correct to the two decimal places required. And of course, all the quantities we’re given in miles. And so this is a distance in miles.

Now we need to find the direction. The easiest way to do this is to use the components of the vector to draw an accurate diagram of it. If we put the initial point here, then the 𝑥-component tells us that the terminal point is negative 0.783777 dot dot dot units to the right, or equivalently, 0.783777 units to the left of the initial point. The 𝑦-component tells us that the terminal point is 0.575125 dot dot dot units above the initial point. And so we can mark the terminal point and draw in the vector. Remember, we’re looking for the direction of the vector. We can see that this direction is 𝜃 north of west, where 𝜃 is marked.

Our only question now is, what is the value of 𝜃? Well, luckily, we have a right triangle. And in this right triangle, tan 𝜃 is the length of the opposite side over the length of the adjacent side. And we can use the inverse tangent function on our calculator, making sure we’re in degree mode first, to find the value of 𝜃. We get that 𝜃 is 36.3 degrees, to the required accuracy of one decimal place. So the direction is approximately 36.3 degrees north of west.

And putting it all together, we get our final answer. That’s to get home, she has to walk 0.97 miles, 36.3 degrees north of west.