Video Transcript
True or false: If π is the point two, zero, two; π is the point five, six, seven; and π is the point nine, 10, eight, then the vector from π to π minus the vector from π to π is equal to the vector negative four, negative four, negative one.
In this question, weβre given the coordinates of three points in three dimensions. We need to use this to determine whether an equation involving vectors is true. There are a few different ways we could answer this question. The first way will be to evaluate the left-hand side of our equation. Thatβs the vector from π to π minus the vector from π to π.
Weβll start by finding the vector from π to π. Remember, thatβs the position vector of π minus the position vector of π. And remember, the position vector of a point will have components equal to the coordinates of the point. So the position vector of the point π is the vector five, six, seven, and the position vector of the point π is two, zero, two. So we need to evaluate the vector five, six, seven minus the vector two, zero, two.
To subtract two vectors of equal dimension, we just need to subtract the corresponding components of the two vectors. This gives us the vector five minus two, six minus zero, seven minus two, which we can simplify to give us the vector three, six, five. We could do exactly the same to find the vector from π to π. We need to subtract the position vector of π from the position vector of π. Thatβs the vector nine, 10, eight minus the vector two, zero, two. Once again, we evaluate this component-wise. We get the vector nine minus two, 10 minus zero, eight minus two, which simplifies to give us the vector seven, 10, six.
Weβre now ready to evaluate the left-hand side of the equation given to us in the question. Substituting in the expressions we found for our vectors, we get the vector from π to π minus the vector from π to π is equal to the vector three, six, five minus the vector seven, 10, six. And once again, we subtract these vectors component-wise. We get the vector three minus seven, six minus 10, five minus six, which simplifies to give us the vector negative four, negative four, negative one, which we can see is exactly equal to the vector given to us in the question. Therefore, we can conclude that the statement given to us in the question is true.
We could end the question here. However, itβs also worth pointing out there is a second method of answering this question. To do this, we need to notice we can simplify the vector equation given to us. Weβll do this by remembering the vector from π to π is equal to negative the vector from π to π. And this is true for any points π and π.
So applying this property with the points π and π, we get the vector from π to π is equal to negative the vector from π to π. We can then substitute this into our expression. We get the vector from π to π minus the vector from π to π is equal to the vector from π to π minus negative the vector from π to π. Subtracting the negative of a vector is the same as adding the vector. So this simplifies to give us the vector from π to π added to the vector from π to π.
We can then simplify this even further. Weβll start by switching the order of the two vectors around. This gives us the vector from π to π added to the vector from π to π. And then we can simplify this by recalling, for any points π, π, and π, the vector from π to π added to the vector from π to π is just equal to the vector from π to π. Applying this to our two vectors, we just get that this simplifies to give us the vector from π to π. Then, we can evaluate this in the same way we did previously. Itβs the position vector of π minus the position vector of π, which gives us the vector negative four, negative four, negative one.
Therefore, we were able to show if π is the point two, zero, two and π is the point five, six, seven and π is the point nine, 10, eight, then the vector from π to π minus the vector from π to π will be equal to the vector negative four, negative four, negative one.
