Vectors: Finding K For Perpendicularity & Angle Calculation

Hey guys! Let's dive into some vector problems today. We're going to figure out how to find a constant that makes two vectors perpendicular and then calculate the angle between vectors. This is a fundamental concept in linear algebra and has tons of applications in physics, engineering, and computer graphics. So, buckle up, and let's get started!

Finding the Value of 'k' for Perpendicular Vectors

Our first task is to find the value of the constant k that makes two given vectors, ${\bar{U}}$ and ${\bar{V}}$, perpendicular. Remember, two vectors are perpendicular (or orthogonal) if their dot product is zero. This is a crucial concept, so let's break it down. The dot product essentially measures how much two vectors point in the same direction. If they are perfectly perpendicular, they don't point in the same direction at all, hence a dot product of zero.

We're given the vectors:

${\begin{aligned} \bar{U} & = (2, k^2, -4, -1) \\ \bar{V} & = (1, 1, 2, k) \end{aligned}}$

To find the value of k that makes these vectors perpendicular, we need to calculate their dot product and set it equal to zero. The dot product of two vectors is calculated by multiplying corresponding components and then summing the results. So, for our vectors ${\bar{U}}$ and ${\bar{V}}$, the dot product is:

${\bar{U} \cdot \bar{V} = (2)(1) + (k^2)(1) + (-4)(2) + (-1)(k)}$

Simplifying this expression, we get:

${\bar{U} \cdot \bar{V} = 2 + k^2 - 8 - k}$

Now, we combine like terms:

${\bar{U} \cdot \bar{V} = k^2 - k - 6}$

Since we want the vectors to be perpendicular, we set the dot product equal to zero:

${k^2 - k - 6 = 0}$

This is a quadratic equation, and we can solve it by factoring. We're looking for two numbers that multiply to -6 and add up to -1. Those numbers are -3 and 2. So, we can factor the equation as:

${(k - 3)(k + 2) = 0}$

This equation is satisfied if either ${k - 3 = 0}$ or ${k + 2 = 0}$. Solving for k in each case, we get:

${k = 3 \text{ or } k = -2}$

Therefore, the values of k that make the vectors ${\bar{U}}$ and ${\bar{V}}$ perpendicular are 3 and -2. We've successfully found the values of k! This involves understanding the concept of the dot product, its relation to perpendicularity, and solving a quadratic equation. This skill is very important in various fields. For example, in computer graphics, ensuring vectors are perpendicular is crucial for lighting calculations and creating realistic shadows. In physics, it can be used to analyze forces acting at right angles. So, mastering this concept is super beneficial.

Finding the Angle Between Vectors

Now that we've tackled perpendicularity, let's shift gears and learn how to find the angle between two vectors. This is another fundamental concept in vector algebra, and it builds upon our understanding of the dot product. The angle between vectors gives us a measure of how much they align or diverge. A small angle means they point in roughly the same direction, while a large angle means they point in opposite directions.

The formula to find the angle ${\theta}$ between two vectors ${\bar{A}}$ and ${\bar{B}}$ is derived from the definition of the dot product:

${\bar{A} \cdot \bar{B} = ||\bar{A}|| \cdot ||\bar{B}|| \cdot \cos(\theta)}$

Where:

• ${||\bar{A}||}$ and ${||\bar{B}||}$ represent the magnitudes (or lengths) of vectors ${\bar{A}}$ and ${\bar{B}}$, respectively.
• ${\cos(\theta)}$ is the cosine of the angle between the vectors.

To find the angle ${\theta}$, we rearrange the formula:

${\cos(\theta) = \frac{\bar{A} \cdot \bar{B}}{||\bar{A}|| \cdot ||\bar{B}||}}$

And then take the inverse cosine (also known as arccosine) to find ${\theta}$:

${\theta = \arccos\left(\frac{\bar{A} \cdot \bar{B}}{||\bar{A}|| \cdot ||\bar{B}||}\right)}$

Let's break down how to apply this formula step-by-step. First, you need to calculate the dot product of the two vectors. We've already discussed how to do this – multiply the corresponding components and add them up. Second, you need to find the magnitude of each vector. The magnitude of a vector is its length, and it's calculated using the Pythagorean theorem. For a vector ${\bar{A} = (a_1, a_2, ..., a_n)}$, the magnitude is:

${||\bar{A}|| = \sqrt{a_1^2 + a_2^2 + ... + a_n^2}}$

So, you square each component, add them together, and then take the square root. Third, you plug the dot product and the magnitudes into the formula for ${\cos(\theta)}$. Finally, you use the arccosine function (usually found on your calculator or in programming libraries) to find the angle ${\theta}$. Remember that the angle ${\theta}$ will be in radians or degrees, depending on the setting of your calculator or software. Make sure you know which unit you're using!

For example, let's say we have two vectors:

${\bar{A} = (1, 2)}$

${\bar{B} = (3, -1)}$

First, we calculate the dot product:

${\bar{A} \cdot \bar{B} = (1)(3) + (2)(-1) = 3 - 2 = 1}$

Next, we find the magnitudes:

${||\bar{A}|| = \sqrt{1^2 + 2^2} = \sqrt{5}}$

${||\bar{B}|| = \sqrt{3^2 + (-1)^2} = \sqrt{10}}$

Now, we plug these values into the formula for ${\cos(\theta)}$:

${\cos(\theta) = \frac{1}{\sqrt{5} \cdot \sqrt{10}} = \frac{1}{\sqrt{50}}}$

Finally, we find the angle:

${\theta = \arccos\left(\frac{1}{\sqrt{50}}\right) \approx 1.4056 \text{ radians} \approx 80.54^\circ}$

So, the angle between vectors ${\bar{A}}$ and ${\bar{B}}$ is approximately 80.54 degrees. Understanding how to find the angle between vectors is super useful in many real-world scenarios. In physics, it's used to calculate the work done by a force. In computer graphics, it's used in lighting models and collision detection. This concept is also vital in navigation and robotics, where determining the relative orientation of objects is crucial.

Wrapping Up

Today, we've explored two important concepts in vector algebra: finding the value of a constant that makes vectors perpendicular and calculating the angle between vectors. These are fundamental skills that have wide-ranging applications. Remember, perpendicularity is linked to a dot product of zero, and the angle between vectors can be found using the arccosine of a specific formula involving the dot product and magnitudes. Keep practicing these concepts, and you'll become a vector whiz in no time! Keep exploring, guys, and happy calculating!