Cosine Of Angle Between Vectors AB And DC: A Step-by-Step Guide

Hey guys! Today, we're diving into a fun geometry problem: finding the cosine of the angle between two vectors in 3D space. Specifically, we'll tackle a problem where we're given points A(3; 5; 7), B(6; 4; 2), C(1; 2; 3), and D(5; 4; 3), and our mission is to determine the cosine of the angle between vectors ${\vec{AB}}$ and ${\vec{DC}}$. Buckle up, because we're about to make some math magic happen!

Understanding the Fundamentals of Vectors and Cosines

Before we jump into the solution, let's quickly refresh our understanding of vectors and how cosines fit into the picture. Vectors, in simple terms, are quantities that have both magnitude (length) and direction. Think of them as arrows pointing from one point to another. In 3D space, we represent vectors using three components: x, y, and z, which tell us how much the vector moves along each axis. The cosine of an angle, on the other hand, is a trigonometric function that relates an angle in a right triangle to the ratio of the adjacent side to the hypotenuse. But here's the cool part: we can also use cosines to find the angle between any two vectors, not just those in right triangles! The formula we'll be using is:

${\cos(\theta) = \frac{\vec{AB} \cdot \vec{DC}}{|\vec{AB}| |\vec{DC}|}}$

Where:

• ${\theta}$ is the angle between the vectors.
• ${\vec{AB} \cdot \vec{DC}}$ is the dot product of the vectors.
• |${\vec{AB}}$| and |${\vec{DC}}$| are the magnitudes (lengths) of the vectors.

This formula is a cornerstone in vector algebra and allows us to find the angle between any two vectors, regardless of their orientation in space. It elegantly connects the geometric concept of an angle with the algebraic operations on vectors. Understanding this formula is crucial, as it forms the basis for solving a wide range of problems in physics, engineering, and computer graphics. For example, in physics, it can be used to calculate the work done by a force acting at an angle. In computer graphics, it helps in determining the lighting and shading of 3D objects. So, with this foundational knowledge, we're well-equipped to tackle our problem and find the cosine of the angle between vectors ${\vec{AB}}$ and ${\vec{DC}}$.

Step 1: Finding the Vectors ${\vec{AB}}$ and ${\vec{DC}}$

Okay, first things first, we need to determine the vectors ${\vec{AB}}$ and ${\vec{DC}}$. Remember, a vector represents the displacement from one point to another. To find the components of a vector, we subtract the coordinates of the initial point from the coordinates of the terminal point. So, for ${\vec{AB}}$, we subtract the coordinates of point A from the coordinates of point B:

${\vec{AB} = B - A = (6 - 3, 4 - 5, 2 - 7) = (3, -1, -5)}$

Similarly, for ${\vec{DC}}$, we subtract the coordinates of point C from the coordinates of point D:

${\vec{DC} = D - C = (5 - 1, 4 - 2, 3 - 3) = (4, 2, 0)}$

Now we have our vectors! ${\vec{AB}}$ is (3, -1, -5), which means if you start at point A and move 3 units in the x-direction, -1 unit in the y-direction, and -5 units in the z-direction, you'll end up at point B. Similarly, ${\vec{DC}}$ is (4, 2, 0), indicating the displacement from point C to point D. These vectors are the building blocks for the next steps in our calculation. We've successfully translated the given points into vector form, which is a crucial step in solving geometric problems in 3D space. With these vectors in hand, we're ready to move on to the next phase: calculating the dot product and the magnitudes.

Step 2: Calculating the Dot Product of ${\vec{AB}}$ and ${\vec{DC}}$

Next up, we need to calculate the dot product of our two vectors, ${\vec{AB}}$ and ${\vec{DC}}$. The dot product is a scalar (a single number) that tells us something about the relationship between the vectors – specifically, how much they point in the same direction. To calculate the dot product, we multiply the corresponding components of the vectors and then add the results together:

${\vec{AB} \cdot \vec{DC} = (3)(4) + (-1)(2) + (-5)(0) = 12 - 2 + 0 = 10}$

So, the dot product of ${\vec{AB}}$ and ${\vec{DC}}$ is 10. What does this number tell us? Well, a positive dot product indicates that the angle between the vectors is acute (less than 90 degrees). If the dot product were negative, the angle would be obtuse (greater than 90 degrees), and if it were zero, the vectors would be orthogonal (perpendicular). The magnitude of the dot product also gives us information about the strength of the alignment between the vectors. A larger magnitude (positive or negative) suggests a stronger alignment or opposition. In our case, the dot product of 10 suggests that the vectors have a significant component pointing in the same general direction. This calculation is a key step in finding the cosine of the angle, as it appears in the numerator of the formula we discussed earlier. Now that we have the dot product, we're one step closer to our final answer! Next, we'll need to calculate the magnitudes of the vectors.

Step 3: Finding the Magnitudes of ${\vec{AB}}$ and ${\vec{DC}}$

Now, let's find the magnitudes (or lengths) of the vectors ${\vec{AB}}$ and ${\vec{DC}}$. The magnitude of a vector is essentially its length, and we calculate it using the Pythagorean theorem in 3D space. We square each component of the vector, add them up, and then take the square root. For ${\vec{AB}}$, which is (3, -1, -5), the magnitude is:

${|\vec{AB}| = \sqrt{(3)^2 + (-1)^2 + (-5)^2} = \sqrt{9 + 1 + 25} = \sqrt{35}}$

And for ${\vec{DC}}$, which is (4, 2, 0), the magnitude is:

${|\vec{DC}| = \sqrt{(4)^2 + (2)^2 + (0)^2} = \sqrt{16 + 4 + 0} = \sqrt{20}}$

So, the magnitude of ${\vec{AB}}$ is ${\sqrt{35}}$, and the magnitude of ${\vec{DC}}$ is ${\sqrt{20}}$. These values represent the lengths of our vectors in 3D space. The magnitude is always a non-negative value, as it represents a physical length. In the context of our problem, these magnitudes are crucial for normalizing the dot product, which is essential for calculating the cosine of the angle. Think of it this way: the dot product tells us about the alignment of the vectors, but the magnitudes scale this alignment by the lengths of the vectors. By dividing the dot product by the product of the magnitudes, we get a pure measure of the angle between the vectors, which is exactly what the cosine provides. With the magnitudes calculated, we have all the pieces we need to assemble our final answer. Let's move on to the exciting conclusion!

Step 4: Calculating the Cosine of the Angle

Alright, guys, we're in the home stretch! We've found the vectors, calculated their dot product, and determined their magnitudes. Now, we just need to plug these values into our cosine formula:

${\cos(\theta) = \frac{\vec{AB} \cdot \vec{DC}}{|\vec{AB}| |\vec{DC}|}}$

We know that ${\vec{AB} \cdot \vec{DC} = 10}$, |${\vec{AB}}$| = ${\sqrt{35}}$, and |${\vec{DC}}$| = ${\sqrt{20}}$. Plugging these in, we get:

${\cos(\theta) = \frac{10}{\sqrt{35} \sqrt{20}} = \frac{10}{\sqrt{700}}}$

We can simplify this a bit further. Since ${700 = 100 \times 7}$, we have ${\sqrt{700} = \sqrt{100 \times 7} = 10\sqrt{7}}$. So,

${\cos(\theta) = \frac{10}{10\sqrt{7}} = \frac{1}{\sqrt{7}}}$

And there you have it! The cosine of the angle between vectors ${\vec{AB}}$ and ${\vec{DC}}$ is ${\frac{1}{\sqrt{7}}}$. We can also rationalize the denominator to get ${\frac{\sqrt{7}}{7}}$, which is the same value expressed in a slightly different form. This is our final answer, and it represents the cosine of the angle between the two vectors in 3D space. Remember, the cosine value tells us about the angle itself – a value close to 1 means the angle is small, a value close to -1 means the angle is close to 180 degrees, and a value close to 0 means the angle is close to 90 degrees. We've successfully navigated this geometry problem by breaking it down into smaller, manageable steps. We started with the fundamentals of vectors and cosines, calculated the necessary vector components and magnitudes, and finally, plugged everything into the formula to arrive at our solution. Great job, guys!

Conclusion

So, there you have it! We've successfully found the cosine of the angle between vectors ${\vec{AB}}$ and ${\vec{DC}}$. We broke down the problem into manageable steps: finding the vectors, calculating the dot product, determining the magnitudes, and finally, plugging everything into the cosine formula. This approach is super helpful for tackling all sorts of geometry problems. Remember, the key is to understand the underlying concepts and then apply them systematically. Keep practicing, and you'll become a vector math whiz in no time! Geometry can seem daunting at first, especially in 3D space, but with a solid understanding of the fundamentals and a step-by-step approach, even complex problems become solvable. The beauty of vector algebra is its ability to represent geometric concepts algebraically, allowing us to manipulate and solve problems using the powerful tools of algebra. This example showcases how vectors can be used to describe directions and magnitudes in space, and how the dot product and magnitudes can be combined to find the angle between them. These techniques are not only useful in pure mathematics but also have wide-ranging applications in physics, engineering, computer graphics, and many other fields. So, by mastering these concepts, you're not just learning math; you're also equipping yourself with valuable problem-solving skills that can be applied in a variety of contexts. Keep exploring, keep learning, and most importantly, keep having fun with math!