Question

Consider the following points \(P\) and \(Q\) a. Find \(\overrightarrow{P Q}\) and state your answer in two forms: \(\langle a, b, c\rangle\) and \(a \mathbf{i}+b \mathbf{j}+c \mathbf{k}\). b. Find the magnitude of \(\overrightarrow{P Q}\). c. Find two unit vectors parallel to \(\overrightarrow{P Q}\). $$P(1,5,0), Q(3,11,2)$$

Short answer

Question: Find two unit vectors parallel to the vector connecting points P(1, 5, 0) and Q(3, 11, 2). Answer: The two unit vectors parallel to the vector connecting points P and Q are: 1. \(\frac{1}{\sqrt{44}}(2 \mathbf{i}+6 \mathbf{j}+2 \mathbf{k})\) 2. \(-\frac{1}{\sqrt{44}}(2 \mathbf{i}+6 \mathbf{j}+2 \mathbf{k})\)

Step-by-step solution

Find the components of the vector PQ

Subtract the coordinates of point P from the coordinates of point Q: $$\overrightarrow{P Q}=\langle3-1,11-5,2-0\rangle=\langle 2,6,2 \rangle$$

Express the result in the desired forms

The vector \(\overrightarrow{P Q}\) can be written in the following two forms: $$\langle 2, 6, 2 \rangle\quad \text{and}\quad 2 \mathbf{i}+6 \mathbf{j}+2 \mathbf{k}$$

Find the magnitude of the vector PQ

Use the Pythagorean theorem to calculate the magnitude of the vector \(\overrightarrow{P Q}\): $$\Vert \overrightarrow{P Q} \Vert = \sqrt{(2)^2 + (6)^2 + (2)^2} = \sqrt{44}$$

Find two unit vectors parallel to the vector PQ

To find two unit vectors parallel to \(\overrightarrow{P Q}\), we first calculate their components by dividing the components of \(\overrightarrow{P Q}\) by its magnitude: $$u_1 = \frac{\overrightarrow{P Q}}{\Vert \overrightarrow{P Q} \Vert}=\frac{1}{\sqrt{44}}\langle 2, 6, 2 \rangle = \frac{1}{\sqrt{44}}(2 \mathbf{i}+6 \mathbf{j}+2 \mathbf{k})$$ The second unit vector, \(u_2\), has the same direction as \(u_1\), but with the opposite sign: $$u_2 = -u_1 = -\frac{1}{\sqrt{44}} (2 \mathbf{i}+6 \mathbf{j}+2 \mathbf{k})$$ So, the two unit vectors parallel to \(\overrightarrow{P Q}\) are \(u_1\) and \(u_2\) as computed above.

Key concepts

Vector Components

Vector components are at the heart of understanding vectors in any dimension. They are a way of breaking down a vector into parts that align with the familiar x, y, and z axes. Just like a traveler uses directions to break a journey into stretches of east-west and north-south, a vector can be described by how far it goes along each axis.

In our exercise, the vector \(\overrightarrow{PQ}\) goes from point P to point Q. To find the components of \(\overrightarrow{PQ}\), we subtract the coordinates of point P from point Q, giving us \(\langle 2, 6, 2 \rangle\). This means \(\overrightarrow{PQ}\) extends 2 units in the x-direction (\(\bf{i}\) component), 6 units in the y-direction (\(\bf{j}\) component), and 2 units in the z-direction (\(\bf{k}\) component). Understanding vector components is essential for all subsequent calculations involving vectors.

Unit Vector

A unit vector has a magnitude (length) of 1 and points in a particular direction. It is the standard measure for direction in vector mathematics. To find a unit vector in the same direction as another vector, you divide each of the original vector's components by its magnitude.

This process doesn't change the direction—think of it as a scaling down of the vector to a friendly 'unit' size while keeping the direction intact. In our exercise, we took the vector \(\overrightarrow{PQ}\) and divided it by its magnitude, yielding the unit vector \(u_1\). The second unit vector, \(u_2\), simply reverses the direction.

Pythagorean Theorem

The Pythagorean theorem is not just for right-angled triangles on a flat plane—it's a tool for finding distances in space! The theorem tells us that the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides.

When dealing with vectors, this principle helps us calculate the magnitude of a vector (which you can think of as the hypotenuse of a space triangle), given its components. For our three-dimensional vector \(\overrightarrow{PQ}\), we added the squares of its components (2, 6, and 2) and took the square root to find the magnitude. Using \(\sqrt{44}\) is how we apply the Pythagorean theorem in three dimensions, allowing us to quantify how 'long' the vector is in space.

Vectors in Three Dimensions

Vectors in three dimensions extend our understanding of direction and magnitude into the spatial world we live in. Unlike two-dimensional vectors that lay flat on a plane, three-dimensional vectors can point anywhere in space, like an arrow flying through the air. They have three components (x, y, and z), and we use them to describe movement and forces that aren't confined to a single plane.

Vectors in three dimensions are not only fundamental in physics and engineering, but also in computer graphics, where they describe objects' orientations and movements in 3D space. In our exercise, the vector \(\overrightarrow{PQ}\) demonstrates this three-dimensional behavior perfectly, extending from point P to point Q through the space defined by our xyz-coordinate system.