# Question Video: Calculating a Conditional Probability from a Two-Way Frequency Table Mathematics • 12th Grade

The additive inverse of the vector ๐ฉ๐ is the vector ๏ผฟ. [A] ๐ฉ๐ [B] ๐๐ฉ [C] the zero vector [D] ๐๐ฉ

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### Video Transcript

The additive inverse of the vector ๐ฉ๐ is the vector blank. Option (A) the vector ๐ฉ๐, option (B) the vector negative ๐๐ฉ, option (C) the zero vector, or option (D) the vector ๐๐ฉ.

Letโs try and construct the same property, but this time with vectors. First, weโre going to need to think what do we mean by the additive identity of a vector. Either by thinking graphically or in terms of components, we know that this will be the zero vector. We know for any vector ๐ฏ added to the zero vector of the same dimension will just be equal to this vector ๐ฏ. Therefore, to find the additive inverse of any vector ๐ฏ, we want to find the vector ๐ฎ we add to ๐ฏ to get the zero vector. In our question, we want to find the additive inverse of the vector ๐ฉ๐.

Thereโs a few different ways of finding the additive inverse of ๐ฉ๐. Weโll start by doing this graphically. Letโs start with two points, ๐ฉ and ๐, and letโs assume that these two points are not the same. We can then graphically represent the vector ๐ฉ๐ by connecting ๐ฉ and ๐ with a line segment, and we know our vector starts at ๐ฉ and ends at ๐. We want to find the vector which when we add to this vector of ๐ฉ๐ we get the zero vector. To do this, we need to recall how we add two vectors together graphically. We need the terminal point of our first vector to be equal to the initial point of our second vector. Then, graphically, traveling along both of these vectors will be the same as adding the two vectors together. We want to use this to find the additive inverse of our vector ๐ฉ๐.

So for ๐ฎ to be the additive inverse of ๐ฉ๐, ๐ฉ๐ plus ๐ฎ must be equal to zero vector. So in our diagram, the pink vector, or the vector ๐ฉ๐ plus ๐ฎ, must be the zero vector. It must have zero magnitude. And the only way this can happen is if our vector ๐ฎ has exactly the same magnitude as vector ๐ฉ๐ but points in the opposite direction. And we can represent this as the vector ๐๐ฉ. Therefore, weโve shown if ๐ฎ is the additive inverse of the vector ๐ฉ๐, then ๐ฎ should be equal to the vector ๐๐ฉ, which was our option (D).

But itโs worth pointing out this is not the only way we couldโve answered this question. If we go back to our definition of the additive inverse of two vectors, then we could subtract the vector ๐ฎ from both sides of the equation. Since we know the vector ๐ฎ subtracted from itself is equal to the zero vector, we get that ๐ฏ is equal to negative ๐ฎ. We can then just apply this to the vector given to us in the question to find its additive inverse. We know if ๐ฎ is the additive inverse of the vector ๐ฉ๐, then ๐ฉ๐ should be equal to negative ๐ฎ.

Of course, we can solve for ๐ฎ by just multiplying through by negative one. We get that the additive inverse of the vector ๐ฉ๐ is the vector negative ๐ฉ๐. And it is worth pointing out all this is really saying is that ๐ฉ๐ plus negative ๐ฉ๐ is the zero vector. And this is the exact statement you would get if you tried to think about this component-wise.

But of course, this is not quite enough to answer our question because in this case we get the answer negative ๐ฉ๐. However, we know the answer weโre supposed to get is the vector ๐๐ฉ. Instead, we need to use one more piece of information we know about vectors. When you multiply a vector by negative one, you switch its directions. So negative ๐ฉ๐ is going to be the vector ๐๐ฉ.

Therefore, in this question, we were able to determine exactly what is meant by the additive inverse of a vector. And we were able to use this to find multiple different methods of finding an expression for the additive inverse of the vector ๐ฉ๐. In our case, we showed this was the vector ๐๐ฉ, which was our option (D).