Find Vector C's Magnitude And Direction Angle

Hey guys, welcome back to Plastik Magazine! Today, we're diving deep into the awesome world of vectors in math. Specifically, we're going to tackle a problem that might seem a bit tricky at first glance, but trust me, it's totally doable and super satisfying once you nail it. We're talking about finding the magnitude and direction angle of a vector $c$, which is defined as the difference between two other vectors, $a$ and $b$. So, let's get our math hats on and break this down step-by-step.

Understanding Vectors: The Basics

Before we jump into solving for $c$, let's quickly refresh what vectors are all about. Think of a vector as an arrow that has both a magnitude (length) and a direction. In math, we often represent vectors using components, like $\langle x, y\rangle$. The first number, $x$, tells us how much the vector moves horizontally, and the second number, $y$, tells us how much it moves vertically. In our problem, we're given two vectors: $a = \langle 5, -9 \rangle$ and $b = \langle -3, 1 \rangle$. Vector $a$ points 5 units to the right and 9 units down. Vector $b$ points 3 units to the left and 1 unit up. Pretty straightforward, right?

Calculating Vector c: The Difference

Our problem states that $c = b - a$. Subtracting vectors is just like subtracting their corresponding components. So, to find $c$, we'll subtract the $x$-component of $a$ from the $x$-component of $b$, and do the same for the $y$-components. Let's do the math:

$c_x = b_x - a_x = -3 - 5 = -8$

$c_y = b_y - a_y = 1 - (-9) = 1 + 9 = 10$

So, our new vector $c$ is $\langle -8, 10 \rangle$. This means vector $c$ moves 8 units to the left and 10 units up from its starting point. Easy peasy!

Finding the Magnitude of Vector c

Now, let's talk about the magnitude. The magnitude of a vector is simply its length. We can find this using the Pythagorean theorem, because if you imagine the vector drawn on a graph, its components form the two shorter sides of a right-angled triangle, and the vector itself is the hypotenuse. The formula for the magnitude of a vector $\langle x, y \rangle$ is $|v| = \sqrt{x^2 + y^2}$.

For our vector $c = \langle -8, 10 \rangle$, the magnitude $|c|$ is calculated as follows:

$|c| = \sqrt{(-8)^2 + (10)^2}$

$|c| = \sqrt{64 + 100}$

$|c| = \sqrt{164}$

We can simplify $\sqrt{164}$ by finding perfect square factors. 164 is divisible by 4:

$|c| = \sqrt{4 \times 41}$

$|c| = 2\sqrt{41}$

So, the magnitude of vector $c$ is $2\sqrt{41}$. This is the length of our vector $c$. You can also get a decimal approximation if needed, which is about 12.8. This means the arrow representing vector $c$ has a length of approximately 12.8 units.

Determining the Direction Angle of Vector c

Next up is the direction angle. This is the angle the vector makes with the positive $x$-axis, usually measured counterclockwise. We use trigonometry to find this angle. If a vector is $\langle x, y \rangle$, the tangent of its direction angle, let's call it $\theta$, is given by $\tan(\theta) = \frac{y}{x}$.

For our vector $c = \langle -8, 10 \rangle$, we have:

$\tan(\theta) = \frac{10}{-8}$

$\tan(\theta) = -\frac{5}{4}$

To find the angle $\theta$, we need to take the inverse tangent (arctan) of $-5/4$:

$\theta = \arctan(-\frac{5}{4})$

Using a calculator, we find that $\arctan(-5/4) \approx -51.34^{\circ}$.

Important Note: Calculators often give an angle in the range of -90 to 90 degrees. We need to make sure our angle is in the correct quadrant. Our vector $c$ has a negative $x$-component ($-8$) and a positive $y$-component ($10$). This places vector $c$ in the second quadrant. The angle $-51.34^{\circ}$ is in the fourth quadrant. To get the correct angle in the second quadrant, we need to add $180^{\circ}$ to the calculator's result:

$\theta \approx -51.34^{\circ} + 180^{\circ}$

$\theta \approx 128.66^{\circ}$

So, the direction angle of vector $c$ is approximately $128.66^{\circ}$. This tells us that if you start pointing along the positive $x$-axis and rotate counterclockwise, you'll need to turn about $128.66^{\circ}$ to align with the direction of vector $c$.

Putting It All Together

Alright guys, we've successfully broken down this vector problem! We started with two vectors, $a = \langle 5, -9 \rangle$ and $b = \langle -3, 1 \rangle$. We calculated $c = b - a$ to be $\langle -8, 10 \rangle$. Then, we found the magnitude of $c$ using the Pythagorean theorem, which came out to be $|c| = 2\sqrt{41}$ (approximately 12.8). Finally, we determined the direction angle of $c$ by using the arctan function and adjusting for the correct quadrant, giving us $\theta \approx 128.66^{\circ}$.

Key Takeaways:

• Vector Subtraction: Subtract corresponding components. $c = b - a = \langle b_x - a_x, b_y - a_y \rangle$.
• Magnitude: Use the Pythagorean theorem: $|v| = \sqrt{x^2 + y^2}$.
• Direction Angle: Use $\tan(\theta) = \frac{y}{x}$, but always check the quadrant of the vector to ensure you have the correct angle.

Working with vectors is a fundamental part of so many areas in science and engineering, so getting a good handle on these concepts is super valuable. Keep practicing these types of problems, and don't be afraid to break them down into smaller, manageable steps. That's all for today, folks! Keep exploring the amazing world of mathematics with us here at Plastik Magazine. See you next time!