Finding a Unit Vector Perpendicular to Given Vectors A and B

Finding a Unit Vector Perpendicular to a Plane: A Fun Dive into Vector Calculus

Hey there! If you're delving into the world of engineering, you've probably bumped into concepts that make your head spin—like vectors, matrices, and cross products. It might sound like abstract math, but trust me, understanding these concepts is like learning the building blocks of the entire engineering universe. Today, we're going to unravel how to find a unit vector that’s perpendicular to a plane determined by two vectors. It’s more straightforward than it seems! So, grab that cup of coffee, and let’s jump in.

What Are We Working With?

Okay, let’s start with the vectors in question. We’ve got two vectors defined as follows:

• Vector A: (A = 2i + 4j)

• Vector B: (B = i + j - k)

For a clearer perspective, think of vectors like arrows in space. Vector A points towards coordinates (2, 4, 0), and Vector B points to (1, 1, -1). Together, these vectors actually define a plane in the three-dimensional space. But what if you want to find a vector that's at a right angle to this plane? That’s where the fun begins!

Enter the Cross Product

To find a vector that's perpendicular to the plane formed by vectors A and B, we need to calculate the cross product of A and B. Now, don’t let the jargon scare you. The cross product essentially gives us a new vector that’s orthogonal (a fancy word for “perpendicular”) to both vectors.

The formula for the cross product of vectors (A) and (B) can be computed using a determinant. Imagine we’ve got a little matrix ready to help us out:

[

A \times B = \begin{vmatrix}

\mathbf{i} & \mathbf{j} & \mathbf{k} \

2 & 4 & 0 \

1 & 1 & -1

\end{vmatrix}

]

This matrix looks complex at first glance, but let's break it down step by step.

Step-by-Step Calculation

1. Setting the Scene: You see the unit vectors (i), (j), and (k) sitting at the top. These represent the x, y, and z directions, respectively.

2. Expanding the Determinant:

• For (i) (the first element), we calculate:

[

(4 \cdot -1 - 0 \cdot 1) = -4

]

• For (j) (the second element), it’s a little tricky because of the negative sign in front:

[

-(2 \cdot -1 - 0 \cdot 1) = 2

]

• For (k) (the last element), we find:

[

(2 \cdot 1 - 4 \cdot 1) = -2

]

Putting it all together, we get:

[

A \times B = -4i + 2j - 2k

]

The Vector That Counts

Now that we have our cross product, it looks like this:

[

C = -4i + 2j - 2k

]

But wait! We need a unit vector that’s perpendicular to our plane. A unit vector is simply any vector with a length of 1. Sounds simple enough, right? To convert vector (C) into a unit vector, we need to divide it by its magnitude (or length).

Calculating the Magnitude

The magnitude of vector (C) is calculated like this:

[

|C| = \sqrt{(-4)^2 + (2)^2 + (-2)^2} = \sqrt{16 + 4 + 4} = \sqrt{24} = 2\sqrt{6}

]

Getting Our Unit Vector

Now, we have everything we need! To get our unit vector (U), we simply divide each component of (C) by its magnitude:

[

U = \frac{-4i + 2j - 2k}{2\sqrt{6}}

]

Simplifying that gives us:

[

U = \frac{-2i + j - k}{\sqrt{6}}

]

Aha! We have our unit vector that’s perpendicular to the plane defined by vectors A and B. But wait, let’s put it in a more convenient format, and we get:

[

U = \frac{-2i + j - k}{\sqrt{6}}

]

So, What’s the Final Answer?

If you refer back to the original options, the answer is:

C. ((-2i + j - k) / \sqrt{6}).

Wrapping It Up

So there you have it! Finding a unit vector perpendicular to a plane defined by two vectors isn’t just a robotic calculation. It’s like solving a puzzle, where each piece connects to create something robust and structured.

And while we’re at it, remember that the world of vectors extends far beyond just perpendicular lines. These fundamental concepts pave the way for understanding more complex ideas in fields like mechanics, structural analysis, and even computer graphics. Isn’t it fascinating how such abstract concepts can be so relevant?

Next time you tackle a vector problem, remember this method—you’ll build confidence along the way. It’s all about practice and understanding the why. Until next time, keep exploring those mathematical realms!