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 \(\overrightarrow{P Q}\). $$P(-3,1,0), Q(-3,-4,1)$$

Short answer

Answer: The two unit vectors parallel to \(\overrightarrow{P Q}\) are \(\langle 0, \frac{-5}{\sqrt{26}}, \frac{1}{\sqrt{26}} \rangle\) and \(\langle 0, \frac{5}{\sqrt{26}}, \frac{-1}{\sqrt{26}} \rangle\).

Step-by-step solution

Find the components of \(\overrightarrow{P Q}\).

Subtract the coordinates of P from the coordinates of Q to find the components of the vector \(\overrightarrow{P Q}\). \(\overrightarrow{P Q} = \langle Q_x - P_x, Q_y - P_y, Q_z-P_z\rangle\) \(\overrightarrow{P Q} = \langle -3 - (-3), -4 - 1, 1-0 \rangle\) ##Step 2: Calculate the vector components##

Calculate the components of \(\overrightarrow{P Q}\).

Now, we can calculate the values of the components of \(\overrightarrow{P Q}\). \(\overrightarrow{P Q} = \langle 0, -5, 1 \rangle\) ##Step 3: Express the vector in two forms##

Express \(\overrightarrow{P Q}\) in the two required forms.

We can now write the vector \(\overrightarrow{P Q}\) in both \(\langle a, b, c\rangle\) form and \(a \mathbf{i}+b \mathbf{j}+c \mathbf{k}\) form. \(\overrightarrow{P Q} = \langle 0, -5, 1 \rangle\) and \(\overrightarrow{P Q} = 0\mathbf{i}-5\mathbf{j}+1\mathbf{k}\) ##Step 4: Find the magnitude of the vector##

Calculate the magnitude of \(\overrightarrow{P Q}\).

To find the magnitude of \(\overrightarrow{P Q}\), we use the formula: \(|\overrightarrow{P Q}| = \sqrt{a^2 + b^2 + c^2}\) By plugging in the values, we get: \(|\overrightarrow{P Q}| = \sqrt{0^2 + (-5)^2 + 1^2} = \sqrt{26}\) ##Step 5: Calculate unit vectors##

Find two unit vectors parallel to \(\overrightarrow{P Q}\).

To find the unit vector parallel to \(\overrightarrow{P Q}\), we will divide each component of \(\overrightarrow{P Q}\) by its magnitude. Then the opposite of that unit vector is also parallel to \(\overrightarrow{P Q}\). Unit Vector \(1 = \frac{\overrightarrow{P Q}}{|\overrightarrow{P Q}|} = \frac{\langle 0, -5, 1 \rangle}{\sqrt{26}} = \langle 0, \frac{-5}{\sqrt{26}}, \frac{1}{\sqrt{26}} \rangle\) Unit Vector \(2 = -\langle 0, \frac{-5}{\sqrt{26}}, \frac{1}{\sqrt{26}} \rangle = \langle 0, \frac{5}{\sqrt{26}}, \frac{-1}{\sqrt{26}} \rangle\) So, the two unit vectors parallel to vector \(\overrightarrow{P Q}\) are \(\langle 0, \frac{-5}{\sqrt{26}}, \frac{1}{\sqrt{26}} \rangle\) and \(\langle 0, \frac{5}{\sqrt{26}}, \frac{-1}{\sqrt{26}} \rangle\).

Key concepts

Unit Vector

A unit vector is a vector that has a magnitude of 1. It is often used to indicate direction without regard to size or length. In vector calculus, finding a unit vector involves dividing each component of a vector by its magnitude. This process effectively scales the original vector down to have a magnitude of 1 while preserving its direction.

To find a unit vector parallel to a given vector \(\overrightarrow{PQ}\), we use the formula:

• \( \text{Unit vector} = \frac{\overrightarrow{PQ}}{\|\overrightarrow{PQ}\|} \)

where \(\|\overrightarrow{PQ}\|\) is the magnitude of the vector. In our example, the magnitude of \(\overrightarrow{PQ}\) was found to be \(\sqrt{26}\), and thus unit vectors parallel to \(\overrightarrow{PQ}\) are \(\langle 0, \frac{-5}{\sqrt{26}}, \frac{1}{\sqrt{26}} \rangle \) and its opposite.

Unit vectors are crucial in mechanics and physics, where specific directions are important, but actual distances or speeds are not necessary.

Vector Magnitude

Vector magnitude, or the length of a vector, measures how long the vector is in a geometric space. It is obtained using the Euclidean norm. The formula for the magnitude of a 3-dimensional vector \(\langle a, b, c \rangle\) is given by:

• \( \| \overrightarrow{v} \| = \sqrt{a^2 + b^2 + c^2} \)

Understanding vector magnitude is essential because it allows us to scale vectors and find unit vectors. In our specific example, after calculating the components of \(\overrightarrow{PQ} = \langle 0, -5, 1 \rangle\), we found its magnitude to be \(\sqrt{26}\).

This measurement helps in determining the length between two points in space when dealing with coordinate vectors. Knowing the magnitude also aids in various applications like physics, where the displacement vector's magnitude helps to calculate distances covered.

Coordinate Vectors

Coordinate vectors aid in describing positions and directions within a 3-dimensional space using a set of numbers. Each vector is represented by its component along the axes, typically noted as \(\langle a, b, c \rangle\) or in vector component form like \(a \mathbf{i} + b \mathbf{j} + c \mathbf{k}\).

In our example, the points \(P(-3, 1, 0)\) and \(Q(-3, -4, 1)\) give rise to the vector \(\overrightarrow{PQ}\). By subtracting the coordinates, the vector is found to be \(\langle 0, -5, 1 \rangle \) and can also be expressed as \(0\mathbf{i} - 5\mathbf{j} + 1\mathbf{k}\). This showcases different ways to visualize vectors.

Coordinate vectors are a fundamental concept used in various fields such as engineering, physics, and computer graphics, aiding in tasks ranging from designing models to understanding forces acting on a point.