Vector Magnitude And Distance Calculation: A Physics Problem

Hey guys! Ever get those physics problems that look like a jumble of letters, arrows, and numbers? Well, we're tackling one today that involves vectors, magnitudes, and distances. Don't worry, we'll break it down step by step so it's super clear. This type of problem is crucial for understanding concepts in electromagnetism, mechanics, and many other fields. So, let's dive into this vector problem and make it crystal clear!

Understanding the Problem

Let's start by stating the problem clearly. We're given a vector field $\vec{A}$ defined as $\vec{A} = x^2y\vec{a}_x - yz\vec{a}_y + yz^2\vec{a}_z$. This vector field exists in three-dimensional space, where the components are expressed in terms of the Cartesian coordinates $x$, $y$, and $z$. We have two main tasks here:

1. Find the magnitude of $\vec{A}$ at point $M(2, -1, 3)$: This means we need to calculate the length or absolute value of the vector $\vec{A}$ at a specific location in space. Basically, how strong is the vector at that point?
2. Determine the distance vector from $M$ to $N$ if $N$ is 5.6 units away: Here, we need to figure out the vector that points from point $M$ to another point $N$, given that the distance between them is 5.6 units. This involves understanding direction and magnitude in 3D space. Think of it like drawing an arrow from one point to another, where the arrow's length is 5.6 units.

Why is This Important?

You might be thinking, "Okay, cool problem, but why should I care?" Well, these types of calculations are fundamental in physics and engineering. For example:

• Electromagnetism: Vector fields are used to represent electric and magnetic fields. Calculating their magnitude and direction is essential for understanding how these fields interact with charged particles and materials.
• Mechanics: Vectors are used to represent forces, velocities, and accelerations. Knowing how to find magnitudes and distances helps us analyze motion and interactions between objects.
• Computer Graphics: Vectors are used to define the position and orientation of objects in 3D space. Calculating distances and magnitudes is crucial for rendering realistic scenes.

So, mastering these concepts opens the door to a deeper understanding of the physical world and numerous applications. Now, let's get into the nitty-gritty of solving this problem!

1. Magnitude of $\vec{A}$ at Point $M$

Alright, let's tackle the first part: finding the magnitude of the vector $\vec{A}$ at the point $M(2, -1, 3)$. Remember, the magnitude of a vector is its length, and it's always a scalar quantity (just a number, no direction). Here's how we do it:

Step 1: Substitute Coordinates into $\vec{A}$

First, we need to plug in the coordinates of point $M$ into the expression for $\vec{A}$: $\vec{A} = x^2y\vec{a}_x - yz\vec{a}_y + yz^2\vec{a}_z$. So, we substitute $x = 2$, $y = -1$, and $z = 3$:

$\vec{A}(M) = (2^2)(-1)\vec{a}_x - (-1)(3)\vec{a}_y + (-1)(3^2)\vec{a}_z$

Let's simplify this:

$\vec{A}(M) = -4\vec{a}_x + 3\vec{a}_y - 9\vec{a}_z$

Okay, now we have the vector $\vec{A}$ specifically at point $M$. It has components in the $x$, $y$, and $z$ directions.

Step 2: Calculate the Magnitude

The magnitude of a vector is calculated using the Pythagorean theorem in 3D. If we have a vector $\vec{V} = V_x\vec{a}_x + V_y\vec{a}_y + V_z\vec{a}_z$, then its magnitude, denoted as $|\vec{V}|$, is:

$|\vec{V}| = \sqrt{V_x^2 + V_y^2 + V_z^2}$

In our case, $\vec{A}(M) = -4\vec{a}_x + 3\vec{a}_y - 9\vec{a}_z$, so $A_x = -4$, $A_y = 3$, and $A_z = -9$. Let's plug these values into the formula:

$|\vec{A}(M)| = \sqrt{(-4)^2 + (3)^2 + (-9)^2}$

$|\vec{A}(M)| = \sqrt{16 + 9 + 81}$

$|\vec{A}(M)| = \sqrt{106}$

So, the magnitude of $\vec{A}$ at point $M$ is $\sqrt{106}$, which is approximately 10.3. That's it for the first part! We've found the "strength" of the vector field at that specific location.

2. Distance Vector from $M$ to $N$

Now, let's move on to the second part of the problem: finding the distance vector from point $M$ to point $N$, given that the distance between them is 5.6 units. This is where things get a little trickier because we need to consider direction as well as magnitude.

Understanding the Challenge

We know the position of point $M$ (2, -1, 3), and we know the distance between $M$ and $N$ (5.6 units). However, we don't know the exact coordinates of point $N$. There are infinitely many points that are 5.6 units away from $M$, forming a sphere around $M$. So, we need a way to specify the direction from $M$ to $N$.

The Unit Vector Approach

The key to solving this is to use a unit vector. A unit vector is a vector with a magnitude of 1. It points in a specific direction and is super useful for scaling distances. Here's the general strategy:

1. Assume a direction: We'll need to either be given a direction or assume one. In a real-world problem, you'd likely have more information about the direction. For now, let's assume we have a unit vector $\vec{a}_{MN}$ that points in the direction from $M$ to $N$.
2. Scale the unit vector: We'll multiply the unit vector $\vec{a}_{MN}$ by the distance between $M$ and $N$ (which is 5.6 units) to get the distance vector $\vec{MN}$.
3. Find the coordinates of N: We'll add the distance vector $\vec{MN}$ to the position vector of $M$ to find the position vector of $N$, and then extract the coordinates of $N$.

Let's Get Specific (with an assumption)

Since we don't have a specific direction given in the problem, let's assume for the sake of demonstration that the direction from $M$ to $N$ is along the vector $\vec{A}(M)$ that we calculated earlier. This means we'll use the direction of the vector field at point $M$ as our guide. This is a common approach when you want to move a certain distance in the direction of a field.

So, our assumption is: $\vec{a}_{MN}$ is in the same direction as $\vec{A}(M) = -4\vec{a}_x + 3\vec{a}_y - 9\vec{a}_z$.

Step 1: Find the Unit Vector $\vec{a}_{MN}$

To find the unit vector in the direction of $\vec{A}(M)$, we need to divide $\vec{A}(M)$ by its magnitude:

$\vec{a}_{MN} = \frac{\vec{A}(M)}{|\vec{A}(M)|} = \frac{-4\vec{a}_x + 3\vec{a}_y - 9\vec{a}_z}{\sqrt{106}}$

So,

$\vec{a}_{MN} = \frac{-4}{\sqrt{106}}\vec{a}_x + \frac{3}{\sqrt{106}}\vec{a}_y + \frac{-9}{\sqrt{106}}\vec{a}_z$

This is a vector with a magnitude of 1, pointing in the direction we want.

Step 2: Calculate the Distance Vector $\vec{MN}$

Now, we multiply the unit vector by the distance (5.6 units) to get the distance vector:

$\vec{MN} = 5.6 \vec{a}_{MN} = 5.6 \left( \frac{-4}{\sqrt{106}}\vec{a}_x + \frac{3}{\sqrt{106}}\vec{a}_y + \frac{-9}{\sqrt{106}}\vec{a}_z \right)$

Let's simplify this:

$\vec{MN} = \frac{-22.4}{\sqrt{106}}\vec{a}_x + \frac{16.8}{\sqrt{106}}\vec{a}_y + \frac{-50.4}{\sqrt{106}}\vec{a}_z$

This is the vector that points from $M$ to $N$. We can approximate the components:

$\vec{MN} \approx -2.17 \vec{a}_x + 1.63 \vec{a}_y - 4.89 \vec{a}_z$

Step 3: Find the Coordinates of Point $N$

To find the coordinates of $N$, we add the distance vector $\vec{MN}$ to the position vector of $M$. The position vector of $M$ is simply $\vec{OM} = 2\vec{a}_x - 1\vec{a}_y + 3\vec{a}_z$.

So, the position vector of $N$ is:

$\vec{ON} = \vec{OM} + \vec{MN}$

$\vec{ON} = (2\vec{a}_x - 1\vec{a}_y + 3\vec{a}_z) + (-2.17 \vec{a}_x + 1.63 \vec{a}_y - 4.89 \vec{a}_z)$

Combine the components:

$\vec{ON} = (2 - 2.17)\vec{a}_x + (-1 + 1.63)\vec{a}_y + (3 - 4.89)\vec{a}_z$

$\vec{ON} = -0.17\vec{a}_x + 0.63\vec{a}_y - 1.89\vec{a}_z$

Therefore, the approximate coordinates of point $N$ are (-0.17, 0.63, -1.89).

Conclusion

Woohoo! We made it through a pretty meaty physics problem. Let's recap what we did:

• We calculated the magnitude of a vector field $\vec{A}$ at a specific point by substituting the coordinates and using the Pythagorean theorem in 3D.
• We found the distance vector from one point to another by using the concept of a unit vector and scaling it by the given distance. We also had to make an assumption about the direction, which highlights the importance of having sufficient information in real-world problems.
• We determined the coordinates of the second point by adding the distance vector to the position vector of the first point.

This problem illustrates several important concepts in vector algebra and their applications in physics. Remember, the key is to break down complex problems into smaller, manageable steps. And don't be afraid to make assumptions when necessary, but always be aware of the limitations of those assumptions!

I hope this explanation was helpful, guys. Keep practicing, and you'll become vector ninjas in no time!