Calculating Vectors: AB And Point C

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Calculating Vectors: AB and Point C

Hey math enthusiasts! Let's dive into some vector calculations. We've got a couple of points, A and B, and a vector, and we're going to use them to find another vector and the coordinates of a mystery point. Sounds fun, right? Don't worry, it's not as scary as it sounds. We'll break it down step by step, so even if you're new to vectors, you'll be able to follow along. So, let's get started!

Finding the Column Vector AB

Alright, first things first, we need to find the column vector AB. This vector represents the displacement from point A to point B. Think of it like this: if you're standing at point A and want to get to point B, how much do you have to move in the x-direction and the y-direction? That movement is what the column vector describes. To calculate this, we use the following formula:

AB = B - A

Where A and B are the position vectors of points A and B respectively. Remember that a position vector is just a column vector that points from the origin (0, 0) to a specific point in space. Let's get down to the details. We're given that point A is (-4, 5) and point B is (6, 1). First, let's write down the position vectors for A and B:

  • A = (−4 5)\begin{pmatrix}-4 \ 5\end{pmatrix}
  • B = (6 1)\begin{pmatrix}6 \ 1\end{pmatrix}

Now, we just need to subtract the components of vector A from the corresponding components of vector B: The calculation is as simple as subtracting the x-coordinate of A from the x-coordinate of B and doing the same for the y-coordinates. Thus:

AB = (6 1)\begin{pmatrix}6 \ 1\end{pmatrix} - (−4 5)\begin{pmatrix}-4 \ 5\end{pmatrix} = (6−(−4) 1−5)\begin{pmatrix}6 - (-4) \ 1 - 5\end{pmatrix} = (10 −4)\begin{pmatrix}10 \ -4\end{pmatrix}

So, the column vector AB is (10 −4)\begin{pmatrix}10 \ -4\end{pmatrix}. This tells us that to get from point A to point B, we move 10 units in the positive x-direction (to the right) and 4 units in the negative y-direction (downwards). Pretty straightforward, eh? We successfully calculated the column vector AB. Now that we have that figured out, let's move on to the next part of the problem!

To make sure we have understood the topic of calculating column vector AB let's do another example. Let's say we have points P(1, 2) and Q(4, 6). The column vector PQ would be calculated as follows:

  • P = (1 2)\begin{pmatrix}1 \ 2\end{pmatrix}
  • Q = (4 6)\begin{pmatrix}4 \ 6\end{pmatrix}

PQ = (4 6)\begin{pmatrix}4 \ 6\end{pmatrix} - (1 2)\begin{pmatrix}1 \ 2\end{pmatrix} = (4−1 6−2)\begin{pmatrix}4 - 1 \ 6 - 2\end{pmatrix} = (3 4)\begin{pmatrix}3 \ 4\end{pmatrix}

Thus, the column vector PQ is (3 4)\begin{pmatrix}3 \ 4\end{pmatrix}. This means, in order to get from point P to point Q, we need to move 3 units right, and 4 units up.

Now we're ready for the second part of our problem!

Finding the Coordinates of Point C

Okay, guys, now we're moving on to a slightly different challenge: finding the coordinates of point C. We're given that vector BC = (−3 −4)\begin{pmatrix}-3 \ -4\end{pmatrix}. Remember, a vector tells us how to move from one point to another. In this case, vector BC tells us how to move from point B to point C. The question is, how do we find the actual coordinates of point C? We know the coordinates of B, and we know how to move from B to C, so this is quite easy. The trick is to use the same logic as before, but in reverse. We use the formula:

C = B + BC

Basically, to find the position vector of C, we add the vector BC to the position vector of B. We know:

  • B = (6 1)\begin{pmatrix}6 \ 1\end{pmatrix}
  • BC = (−3 −4)\begin{pmatrix}-3 \ -4\end{pmatrix}

So, let's plug these values into our equation and calculate the coordinates of C: C = (6 1)\begin{pmatrix}6 \ 1\end{pmatrix} + (−3 −4)\begin{pmatrix}-3 \ -4\end{pmatrix} = (6+(−3) 1+(−4))\begin{pmatrix}6 + (-3) \ 1 + (-4)\end{pmatrix} = (3 −3)\begin{pmatrix}3 \ -3\end{pmatrix}.

Therefore, the coordinates of point C are (3, -3). Easy peasy, right? We've successfully found both the column vector AB and the coordinates of point C. You see, with a little bit of knowledge about vectors and some simple calculations, these problems become quite manageable. The most important thing is to understand what each vector represents and how they relate to the points in space.

Let's do another quick example. Suppose we have the point D(2, 7) and vector DE = (1 5)\begin{pmatrix}1 \ 5\end{pmatrix}. What are the coordinates of point E? We use the same formula as before:

E = D + DE

  • D = (2 7)\begin{pmatrix}2 \ 7\end{pmatrix}
  • DE = (1 5)\begin{pmatrix}1 \ 5\end{pmatrix}

E = (2 7)\begin{pmatrix}2 \ 7\end{pmatrix} + (1 5)\begin{pmatrix}1 \ 5\end{pmatrix} = (2+1 7+5)\begin{pmatrix}2 + 1 \ 7 + 5\end{pmatrix} = (3 12)\begin{pmatrix}3 \ 12\end{pmatrix}. Thus, the coordinates of point E are (3, 12).

And that's it! We have successfully solved both parts of the problem. Remember, practice makes perfect. The more you work with vectors, the more comfortable you'll become. So, keep practicing, keep learning, and don't be afraid to ask for help if you get stuck. Keep up the amazing work.

Conclusion

So, there you have it! We've successfully navigated the world of vectors, finding the column vector AB and determining the coordinates of point C. Remember, the key takeaways are:

  1. Column vectors represent displacement between points.
  2. To find a column vector AB, subtract the position vector of A from the position vector of B ( B - A ).
  3. To find the coordinates of a point C given vector BC and point B, add the vector BC to the position vector of B ( B + BC ).

Mastering these concepts is crucial for a solid foundation in mathematics, especially when dealing with topics like linear algebra and physics. Keep practicing, and you'll become a vector whiz in no time. If you enjoyed this explanation, or if you have any questions, feel free to leave a comment below. Happy vectoring, and keep exploring the fascinating world of mathematics!