A woman leaves home and walks three miles west and then two miles southwest. How far from home is she? In what direction must she walk to directly head home? Give the distance, in miles, correct to two decimal places and the direction, in degrees, to one decimal place.
In this question, we’re given some information about a woman walking from her home. First, we’re told that she walks three miles west, and then after that she walks two miles southwest. The first part of this question wants us to determine how far from her home is she after she walks in this manner. We need to give our answer to this part of the question in miles, and we need to give our answer correct to two decimal places. The second part of this question wants us to determine the direction she needs to walk to head directly home. We need to give the answer to this part of the question in degrees, and we need to give our answer correct to one decimal place.
We have several different methods of answering this question. The first thing we’re going to want to do is sketch a picture of her walk. To do this, we’re going to need to start by picking a direction to be north. We’ll choose up to be north, giving us the following compass directions. Let’s sketch her journey piece by piece. First, she starts at her home, and then we’re told she walks three miles to the west. After this, we’re told that she walks two miles southwest. We then need to use this information to determine how far she is from her home and what direction she needs to walk home. There’s a few different methods of doing this.
For example, from our sketch, we could do this by using trigonometry. The distance she is from home is the unknown side length in this triangle. We can find the opposite angle to this triangle by using what we know about compass directions. The angle between the eastern direction and the southwestern direction will be 135 degrees, because it’s 45 degrees plus 90 degrees. This is enough information to use the law of cosines to find the distance she needs to travel home. Remember, her current position is one of the vertices of our triangle. And the direction she needs to travel home can be represented as an angle above the eastern direction.
There’s a lot of different ways of finding this angle. For example, we could use trigonometry. However, we could also do this by forming the following parallelogram. We know the angle opposite to 135 degrees will also be 135 degrees. We can then use this to find the other angles in our parallelogram. They’ll both be equal to 45 degrees. Then we can find one of the other angles in our triangle, for example, by using the law of sines, since we know all three lengths and one of the angles in our triangle. But this is only one way of answering this question. We’re going to go for a method involving vectors.
The start of this method is exactly the same. We sketch out the information we’re given in the question. However, this time, instead of representing these as lengths in a triangle, we’re going to represent these as vectors since they have both magnitude and direction. We can give these vectors names. We’ll call the vector representing the first part of her journey 𝐮, the vector representing the second part of her journey 𝐯, and the vector representing the direct journey home 𝐰. In our diagram, we can clearly see the directions of these vectors. However, we should write down their magnitude. The magnitude of vector 𝐮 will be the distance walked in the first part of her journey, three miles, and the magnitude of 𝐯 will be the distance walked in the second part of her journey, two miles.
And in our sketch, we can see something interesting about vector 𝐰. Vectors 𝐮, 𝐯, and 𝐰 represent a triangle. Transversing these three vectors, we end up at the same point we started. In other words, vector 𝐮 plus vector 𝐯 plus vector 𝐰 is the zero vector. Remember, the question wants us to find the distance traveled to directly walk home and the direction needed to travel. In other words, we want to find an expression for vector 𝐰. There’s several different methods we could use to do this. For example, we could do this directly from our sketch, which would involve using trigonometry. However, this would be incredibly similar to the method we already showed. Instead, we’re going to represent vectors 𝐮 and 𝐯 in terms of their components.
Let’s start with vector 𝐮. Vector 𝐮 represents traveling three miles west. We’ll represent moving left as a negative change and moving east as a positive change. And we’ll represent moving north as a positive change and south as a negative change. So our vector 𝐮 will be the vector negative three, zero. The first part of her journey is moving three miles west. And we don’t move north or south at all. Let’s now try and do the same for vector 𝐯. This is more difficult, however, because vector 𝐯 is traveling southwest. And remember, when we’re traveling southwest, we’re moving an equal amount west as we are south. So the first and second component of vector 𝐯 will be equal. And because we’re moving south and west, both of these will be negative.
𝐯 is the vector negative 𝑎, negative 𝑎, the sum positive value of 𝑎. But this is not the only piece of information we know about vector 𝐯. We know the magnitude of vector 𝐯 is two miles. So we can use this to find the value of 𝑎. Let’s find the magnitude of the vector negative 𝑎, negative 𝑎. And to do this, it might be helpful to recall to find the magnitude of a vector, we need to find the square root of the sum of the squares of its components. So the magnitude of the vector 𝑏, 𝑐 is equal to the square root of 𝑏 squared plus 𝑐 squared. So we have two miles is the magnitude of vector 𝐯 is equal to the square root of negative 𝑎 squared plus negative 𝑎 squared.
Of course, we can simplify this. Negative 𝑎 all squared is just equal to 𝑎 squared. So we can add these together to get that the magnitude of vector 𝐯 is the square root of two 𝑎 squared. And we can simplify this further by distributing the square root over our two factors. This gives us the square root of two multiplied by the square root of 𝑎 squared. And we can simplify this. The square root of 𝑎 squared is equal to 𝑎. And remember, our value of 𝑎 has to be positive. If our value of 𝑎 was negative, then our vector 𝐯 would be pointing northeast. This gives us 𝑎 multiplied by root two, and this is all equal to the value of two miles.
So we get two miles is equal to the square root of two. We can solve for 𝑎 by dividing through by root two. We get that 𝑎 is equal to two divided by the square root of two miles. And there’s a few different ways of simplifying this. For example, we can multiply both the numerator and the denominator by the square root of two. Alternatively, we can notice that we can rewrite our numerator as the square root of two all squared. Then we can cancel the square root of two in our denominator with one of the factors of root two. In either method, we show that our value of 𝑎 is equal to the square root of two miles. So we can substitute this value of 𝑎 into our vector 𝐯 and then clear our working.
𝐯 is the vector negative root two, negative root two. Now we can combine our component-wise definitions of vector 𝐮 and vector 𝐯 with our equation 𝐮 plus 𝐯 plus 𝐰 is the zero vector. Using the component-wise definitions of vector 𝐮 and vector 𝐯 and writing the zero vector as the vector zero, zero, we get the vector negative three, zero plus the vector negative root two, negative root two plus 𝐰 is equal to the vector zero, zero. We can rearrange this equation for 𝐰 by subtracting our two vectors from the equation, giving us that vector 𝐰 is the vector zero, zero minus the vector negative three, zero minus the vector negative root two, negative root two.
So all we need to do now is evaluate this expression for 𝐰. There’s a few different ways of doing this. We’re going to distribute the negative over both of our vectors, and we’re also going to remove the vector zero, zero from this equation since it doesn’t affect the value. Remember, to multiply a vector by a scalar, we just need to multiply all of the components by our scalar. In this case, both of our scalars are negative one. So we just need to flip the sign of all of our components. We get that our vector 𝐰 is the vector three, zero plus the vector root two, root two.
Finally, to add these two vectors together, we just need to add the corresponding components together. And by doing this, we get our vector 𝐰 is the vector three plus root two, root two. We can then use this component-wise definition of the vector 𝐰 to answer our question. First, though, let’s clear some space.
The first part of this question wants us to determine how far from her home is the woman. Another way of saying this is, how far does she have to walk to get back home? This is the magnitude of vector 𝐰. And we know the components of vector 𝐰, so we can calculate its magnitude. The magnitude of vector 𝐰 is the square root of three plus root two all squared plus root two all squared. And of course, in this case, this represents a distance in miles, so we can give this the unit of miles.
We can then calculate this. We get 4.635 and this expansion continues miles. And we remember the question wants us to give our answer in miles to two decimal places. So we need to round this to the nearest two decimal places. We do this by looking at the third decimal place. We can see this is five, which is greater than or equal to five. So we need to round up, giving us to two decimal places 4.64 miles.
The second part of this question wants us to determine the direction she needs to walk home. And we need to give our answer in degrees to one decimal place. There’s actually a lot of different ways of doing this. We’re just going to sketch our vector 𝐰. We know that the vector 𝐰 has horizontal component three plus root two and vertical component root two. So we can represent it graphically like shown. Well, it’s important to remember that right on our diagram represents east and up on our diagram represents north. So to find the direction she needs to walk home, we can find the direction by finding the following angle on our diagram. This angle in degrees will be the angle north of east that she needs to walk home.
And this is a right-angled triangle. And we want to find the angle where we know the opposite length and the adjacent length. So, by using trigonometry, if we call this angle 𝜃, the tangent of 𝜃 should be equal to the length of the opposite side divided by the length of the adjacent side, root two divided by three plus root two. So we can solve for 𝜃 by using trigonometry. 𝜃 will be equal to the inverse tangent of root two divided by three plus root two. And it’s important to remember 𝜃 is the angle in a right-angled triangle. It’s going to be an acute angle. And we can calculate this making sure our calculator is set to degrees mode. We get 17.76 and this expansion continues degrees.
Well, the question wants us to give our answer to one decimal place. So we look at the second decimal place to determine whether we need to round up or round down. The second decimal place is six, which is greater than or equal to five. So we need to round up. So this gives us 17.8 degrees. But remember, this is just the measure of our angle. We need to represent this as a direction. So we’ll say that it’s 17.8 degrees north of the direction east. And this is our final answer.
In this question, we were given a real-world problem involving a woman’s walk from home. We were able to convert this into a problem involving vectors. And we were able to use this to determine the total distance she would need to walk directly home and the direction. We were able to show she would need to move, to two decimal places, 4.64 miles to get home and in a direction of 17.8 degrees north of east to one decimal place.