### Video Transcript

Which of the vectors π, π, π, π, and π shown in the diagram is equal to π¨ plus π©?

The diagram weβre given is this set of Cartesian axes with several vectors all represented as arrows. The vector π¨ is represented by this arrow. The vector π© is represented by this arrow. And our task is to identify which of the other vectors is equal to π¨ plus π©. So letβs recall how to add two vectors when they are represented by arrows. An arrow representing a vector has a head, thatβs the pointy end, a tail, thatβs the other end. And a straight line connects the tail to the head. Now, when we want to add two vectors, we draw the tail of one vector at the head of the other vector. If we connect the other tail to the other head with a straight line, this new vector is exactly equal to the sum of the two vectors we had previously drawn.

In fact, we can also see from this diagram that vector addition is commutative; that is, π plus π is equal to π plus π. By drawing the tail of π at the head of π instead of the other way around, we have effectively reversed the order of the addition. But the tail of π plus π is still at the remaining tail and the head of π plus π is still at the remaining head. So π plus π is equal to π plus π. So to find the vector that is equal to π¨ plus π©, we just need to redraw π¨ at the head of π© or vice versa, redraw π© at the head of π¨. Letβs do both just to confirm that we get the same answer both ways.

The arrow representing π¨ extends two units to the right and three units upward. Starting with the tail at π© and drawing two units to the right and three units upward, we see that the head is at the head of π. Now the tail of π© is at the origin, the same as all the other vectors, so it looks like π is our answer. But letβs double-check this by drawing the arrow for π© at the head of π¨. π© extends five units to the left and two units downward. So starting at the head of π¨ and drawing five units to the left and two units downward, we see that the head is again the head of π. And again the tails of all of the vectors are at the same point, the origin. So whichever way we draw our picture, the vector π is equal to π¨ plus π©.