# Question Video: Finding a Sum of Vectors on the Edges of a Polygon Graphically Mathematics • 12th Grade

Fill in the blank: In the following figure, π¨π© + π©π + ππ + ππ = οΌΏ.

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### Video Transcript

Fill in the blank. In the following figure, the vector from π΄ to π΅ added to the vector from π΅ to πΆ added to the vector from πΆ to π· added to the vector from π· to πΈ is equal to what.

In this question, weβre given a diagram containing an irregular pentagon π΄π΅πΆπ·πΈ. We need to determine the sum of four of the vectors representing the sides of this pentagon. And there are in fact two different ways we can go about answering this question, and weβll go through both of these. Letβs start by looking at the vectors weβre asked to sum, the vector from π΄ to π΅ added to the vector from π΅ to πΆ added to the vector from πΆ to π· added to the vector from π· to πΈ.

If we add these vectors onto our diagram, we can see that these are the vectors starting at vertex π΄ going around the sides of this pentagon all the way up to vertex πΈ. And this gives us several different methods we can use to answer this question. One way is to add the vector from πΈ to π΄ onto this diagram. Adding this vector onto our diagram and remembering when we add vectors together, weβre adding the displacements of the vectors together, we can notice something interesting. The sum of all of these vectors is going to be the zero vector. One way of seeing this is to start at vertex π΄ and consider the displacement when we add each of these vectors in turn.

Starting at vertex π΄ and traveling along the vector from π΄ to π΅ will end us at the point π΅. We can continue this process. We can travel along the vector from π΅ to πΆ to end up at the point πΆ, then travel along the vector from πΆ to π· to end up at the point π·. And if we keep following this process, we will end up back at the point π΄. Combining the displacements and directions of all five of these vectors ends us back at the point we started, a displacement of zero in both the horizontal and vertical directions. Therefore, the sum of these five vectors is equal to the zero vector. We get the following vector equation. And we can rearrange this equation to find an expression for the blank in the given equation. We subtract the vector from πΈ to π΄ from both sides of the equation.

This gives us that the vector from π΄ to π΅ added to the vector from π΅ to πΆ added to the vector from πΆ to π· added to the vector from π· to πΈ must be equal to negative the vector from πΈ to π΄. And of course, we could fill in the blank with this expression. However, itβs worth noting we can simplify this vector expression on the right-hand side of the equation. We can recall that multiplying a vector by negative one is the same as leaving its magnitude unchanged and switching its direction. In other words, negative the vector from πΈ to π΄ will be just equal to the vector from π΄ to πΈ. Therefore, this is one way of showing that the sum of these four vectors in the diagram will be the vector from π΄ to πΈ. And itβs worth pointing out this process will work for more complicated polygons as well.

There is another method worth going through to answer this question. And to go through this method, letβs start by recalling the triangle rule for vector addition. This tells us we can add two vectors together given graphically or given in terms of their vertices if the terminal point of one vector is coincident with the initial point of the other vector. In other words, the vector from π to π added to the vector from π to π will be equal to the vector from π to π. This then gives us two different methods we can use to evaluate the given addition. We can do it from the figure or we can also do this just from the given expression.

To use this method in the given figure, we can add each of the vectors separately. First, we can add the vector from π΄ to π΅ to the vector from π΅ to πΆ. We can see that this starts at the initial point π΄ and terminates at the point πΆ, so itβs equivalent to the vector from π΄ to πΆ. This adds the first two vectors in this given expression. We can then add on the next vector, the vector from πΆ to π·. We can do this once again by using the triangle rule, since the vector from π΄ to πΆ terminates at the point πΆ and the vector from πΆ to π· has initial point at vertex πΆ. This then gives us the vector from π΄ to π·. In other words, weβve shown that the sum of the first three vectors in this given expression simplifies to give us the vector from π΄ to π·.

And of course, we can apply this process one more time to add on the vector from π· to πΈ. When we apply this process one final time, we can see that our initial point will be vertex π΄ and the terminal point of our vector will be vertex πΈ. In other words, all of these vectors sum to give us the vector from π΄ to πΈ, which agrees with our other method of answering this question. And as discussed previously, we can also do this directly from the given vector expression. First, remember we can add the vectors in any order. So letβs start by adding the first two vectors by using the triangle rule for vector addition. The terminal point of the first vector is point π΅, and the initial point of the second vector is point π΅. So this will be the vector from π΄ to πΆ.

We can follow the same process for the vector from π΄ to πΆ and the vector from πΆ to π·. We can add these two vectors together using the triangle rule to get the vector from π΄ to π·. And we need to add this to the vector from π· to πΈ. We do this one final time by using the triangle rule. And when we do this, we once again get the vector from π΄ to πΈ. Therefore, we were able to show several different methods of determining the vector from π΄ to π΅ added to the vector from π΅ to πΆ added to the vector from πΆ to π· added to the vector from π· to πΈ is equal to the vector from π΄ to πΈ.