Vector Operations: Finding Coordinates And Lengths

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Hey guys! Let's dive into some vector math. We've got two vectors, a⃗\vec{a} and b⃗\vec{b}, and we're going to use them to find the coordinates and lengths of some new vectors. This is a super important concept in algebra and has tons of applications in physics, computer graphics, and a bunch of other fields. So, let's get started!

Understanding the Problem

First off, let's break down what the problem is asking us to do. We're given two vectors: aβƒ—=βˆ’2iβƒ—+6jβƒ—βˆ’4kβƒ—\vec{a} = -2\vec{i} + 6\vec{j} - 4\vec{k} and bβƒ—=βˆ’5iβƒ—+3jβƒ—+7kβƒ—\vec{b} = -5\vec{i} + 3\vec{j} + 7\vec{k}. Notice that these vectors are expressed in terms of their components along the x, y, and z axes, represented by the unit vectors iβƒ—\vec{i}, jβƒ—\vec{j}, and kβƒ—\vec{k}, respectively. This is a pretty standard way to define vectors in 3D space. The key is to remember that these unit vectors are just directions, and the coefficients tell us how much of each direction we have. The goal is to determine the coordinates of new vectors created by a combination of the original vectors, and also find their lengths, which represents the magnitude of the vectors. Specifically, we need to calculate two new vectors, cβƒ—\vec{c} and dβƒ—\vec{d}, based on the following operations: cβƒ—=aβƒ—+2bβƒ—\vec{c} = \vec{a} + 2\vec{b} and dβƒ—=2aβƒ—+bβƒ—\vec{d} = 2\vec{a} + \vec{b}.

So, essentially, we're going to add, subtract, and multiply vectors by scalars (regular numbers). Vector addition and scalar multiplication are the fundamental operations here, and the ability to do these operations correctly is critical. Scalar multiplication involves multiplying each component of a vector by a scalar. When we add vectors, we add their corresponding components. Vector addition is done component-wise. This means that to add two vectors, you add their respective i, j, and k components separately. For example, if uβƒ—=uxiβƒ—+uyjβƒ—+uzkβƒ—\vec{u} = u_x\vec{i} + u_y\vec{j} + u_z\vec{k} and vβƒ—=vxiβƒ—+vyjβƒ—+vzkβƒ—\vec{v} = v_x\vec{i} + v_y\vec{j} + v_z\vec{k}, then uβƒ—+vβƒ—=(ux+vx)iβƒ—+(uy+vy)jβƒ—+(uz+vz)kβƒ—\vec{u} + \vec{v} = (u_x + v_x)\vec{i} + (u_y + v_y)\vec{j} + (u_z + v_z)\vec{k}. To understand the length or magnitude of a vector is important. The length of a vector vβƒ—=vxiβƒ—+vyjβƒ—+vzkβƒ—\vec{v} = v_x\vec{i} + v_y\vec{j} + v_z\vec{k} is calculated using the formula: ∣vβƒ—βˆ£=vx2+vy2+vz2|\vec{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2}. This is derived from the Pythagorean theorem. Remember, vector operations and understanding their geometric interpretations are the cornerstones of many scientific and engineering fields. Mastering these fundamentals gives you a significant advantage.

Now, let's look at how we can do this step-by-step. Remember that each component of the resulting vectors represents its projection onto the respective axis. This geometric viewpoint is important for visualization. So, are you ready to get started with some vector calculations? Let's begin the exciting journey and solve this mathematical problem together. By the end, you'll be able to confidently handle vector operations.

Calculating Vector c⃗=a⃗+2b⃗\vec{c} = \vec{a} + 2\vec{b}

Alright, let's start with the first part of our problem: calculating cβƒ—\vec{c}. We know that cβƒ—=aβƒ—+2bβƒ—\vec{c} = \vec{a} + 2\vec{b}. This means we need to multiply vector bβƒ—\vec{b} by the scalar 2, and then add the result to vector aβƒ—\vec{a}. First, let's find 2bβƒ—2\vec{b}. We do this by multiplying each component of bβƒ—\vec{b} by 2. So, if bβƒ—=βˆ’5iβƒ—+3jβƒ—+7kβƒ—\vec{b} = -5\vec{i} + 3\vec{j} + 7\vec{k}, then 2bβƒ—=2(βˆ’5iβƒ—+3jβƒ—+7kβƒ—)=βˆ’10iβƒ—+6jβƒ—+14kβƒ—2\vec{b} = 2(-5\vec{i} + 3\vec{j} + 7\vec{k}) = -10\vec{i} + 6\vec{j} + 14\vec{k}. See? Pretty straightforward! We simply multiply the scalar by each component separately.

Now, to find cβƒ—\vec{c}, we add aβƒ—\vec{a} and 2bβƒ—2\vec{b}. We know aβƒ—=βˆ’2iβƒ—+6jβƒ—βˆ’4kβƒ—\vec{a} = -2\vec{i} + 6\vec{j} - 4\vec{k} and we just found that 2bβƒ—=βˆ’10iβƒ—+6jβƒ—+14kβƒ—2\vec{b} = -10\vec{i} + 6\vec{j} + 14\vec{k}. Adding the corresponding components, we get: cβƒ—=(βˆ’2βˆ’10)iβƒ—+(6+6)jβƒ—+(βˆ’4+14)kβƒ—=βˆ’12iβƒ—+12jβƒ—+10kβƒ—\vec{c} = (-2 - 10)\vec{i} + (6 + 6)\vec{j} + (-4 + 14)\vec{k} = -12\vec{i} + 12\vec{j} + 10\vec{k}. Therefore, the coordinates of vector cβƒ—\vec{c} are (-12, 12, 10). Awesome, we've found the coordinates of cβƒ—\vec{c}! That was not too hard, right? Now, let's calculate the length (magnitude) of cβƒ—\vec{c}. Remember, the length of a vector cβƒ—=cxiβƒ—+cyjβƒ—+czkβƒ—\vec{c} = c_x\vec{i} + c_y\vec{j} + c_z\vec{k} is given by ∣cβƒ—βˆ£=cx2+cy2+cz2|\vec{c}| = \sqrt{c_x^2 + c_y^2 + c_z^2}. In our case, cx=βˆ’12c_x = -12, cy=12c_y = 12, and cz=10c_z = 10. So, ∣cβƒ—βˆ£=(βˆ’12)2+(12)2+(10)2=144+144+100=388|\vec{c}| = \sqrt{(-12)^2 + (12)^2 + (10)^2} = \sqrt{144 + 144 + 100} = \sqrt{388}. This is the length of vector cβƒ—\vec{c}. We can simplify this to 2972\sqrt{97}, but it's perfectly fine to leave it as 388\sqrt{388}. So, we've successfully found both the coordinates and the length of vector cβƒ—\vec{c}! It is really exciting to see how all the calculations come together, isn't it? Practice is critical here, so try a few more examples. Understanding scalar multiplication and vector addition is so important and that’s what we've done here.

Now, let's summarize what we've learned for vector cβƒ—\vec{c}. The vector cβƒ—\vec{c} has coordinates (-12, 12, 10), and its length is 388\sqrt{388} or 2972\sqrt{97}. We are on a roll, and now let’s solve the next part.

Calculating Vector d⃗=2a⃗+b⃗\vec{d} = 2\vec{a} + \vec{b}

Okay, guys, let's move on to the second part of the problem and find vector d⃗\vec{d}. We know that d⃗=2a⃗+b⃗\vec{d} = 2\vec{a} + \vec{b}. This is similar to what we did before, but now we're multiplying vector a⃗\vec{a} by the scalar 2, and then adding vector b⃗\vec{b}.

First, let's find 2aβƒ—2\vec{a}. We multiply each component of aβƒ—\vec{a} by 2. Since aβƒ—=βˆ’2iβƒ—+6jβƒ—βˆ’4kβƒ—\vec{a} = -2\vec{i} + 6\vec{j} - 4\vec{k}, then 2aβƒ—=2(βˆ’2iβƒ—+6jβƒ—βˆ’4kβƒ—)=βˆ’4iβƒ—+12jβƒ—βˆ’8kβƒ—2\vec{a} = 2(-2\vec{i} + 6\vec{j} - 4\vec{k}) = -4\vec{i} + 12\vec{j} - 8\vec{k}. Simple enough, right? Scalar multiplication is a piece of cake now. Next, to find dβƒ—\vec{d}, we add 2aβƒ—2\vec{a} and bβƒ—\vec{b}. We've just calculated 2aβƒ—=βˆ’4iβƒ—+12jβƒ—βˆ’8kβƒ—2\vec{a} = -4\vec{i} + 12\vec{j} - 8\vec{k}, and we know that bβƒ—=βˆ’5iβƒ—+3jβƒ—+7kβƒ—\vec{b} = -5\vec{i} + 3\vec{j} + 7\vec{k}. Adding the corresponding components gives us: dβƒ—=(βˆ’4βˆ’5)iβƒ—+(12+3)jβƒ—+(βˆ’8+7)kβƒ—=βˆ’9iβƒ—+15jβƒ—βˆ’1kβƒ—\vec{d} = (-4 - 5)\vec{i} + (12 + 3)\vec{j} + (-8 + 7)\vec{k} = -9\vec{i} + 15\vec{j} - 1\vec{k}. Therefore, the coordinates of vector dβƒ—\vec{d} are (-9, 15, -1). We are close to the finish line and we can do it!

Now, let's find the length (magnitude) of vector dβƒ—\vec{d}. Remember the formula: ∣dβƒ—βˆ£=dx2+dy2+dz2|\vec{d}| = \sqrt{d_x^2 + d_y^2 + d_z^2}. In our case, dx=βˆ’9d_x = -9, dy=15d_y = 15, and dz=βˆ’1d_z = -1. So, ∣dβƒ—βˆ£=(βˆ’9)2+(15)2+(βˆ’1)2=81+225+1=307|\vec{d}| = \sqrt{(-9)^2 + (15)^2 + (-1)^2} = \sqrt{81 + 225 + 1} = \sqrt{307}. This is the length of vector dβƒ—\vec{d}. Since 307 is a prime number, we can't simplify this any further. So, we've found both the coordinates and the length of vector dβƒ—\vec{d}! See? Not too bad at all! Practice is what matters the most. We've gone through everything step by step. Congratulations, you've successfully completed the calculation of both vectors cβƒ—\vec{c} and dβƒ—\vec{d}! Understanding these calculations, step by step, is fundamental to your success. Vector operations are key to understanding more advanced concepts, so pat yourself on the back, you’ve done a great job!

Conclusion

So, there you have it, folks! We've successfully found the coordinates and lengths of both c⃗\vec{c} and d⃗\vec{d}. We've used the principles of scalar multiplication and vector addition to calculate the coordinates, and the magnitude formula to find their lengths. This is a fundamental concept in vector algebra. This demonstrates how vectors behave under different operations and provides a foundation for more complex mathematical concepts.

  • Coordinates of cβƒ—\vec{c}: (-12, 12, 10)
  • Length of cβƒ—\vec{c}: 388\sqrt{388} or 2972\sqrt{97}
  • Coordinates of dβƒ—\vec{d}: (-9, 15, -1)
  • Length of dβƒ—\vec{d}: 307\sqrt{307}

Keep practicing these problems and you'll become a pro in no time! Remember to always break down the problem into smaller steps. Make sure to understand the fundamental concepts of vector addition, scalar multiplication and vector length calculation, and you’ll be set for success! Keep up the great work and thanks for following along! And remember to always double-check your calculations to avoid any errors. Hope this was helpful!