Question

Consider the following points \(P\) and \(Q\) a. Find \(\overrightarrow{P Q}\) and state your answer in two forms: \((a, b, c)\) and \(a i+b j+c \mathbf{k}\) 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

Question: Given points P(-3,1,0) and Q(-3,-4,1), find the vector $\overrightarrow{PQ}$ and express it in two different forms. Then find its magnitude and determine two unit vectors parallel to it. Answer: The vector $\overrightarrow{PQ}$ can be expressed in two forms: (a) (0, -5, 1) and (b) 0i - 5j + 1k. The magnitude of $\overrightarrow{PQ}$ is $\sqrt{26}$. Two unit vectors parallel to $\overrightarrow{PQ}$ are (0, -$\frac{5}{\sqrt{26}}$, $\frac{1}{\sqrt{26}}$) and (-0, $\frac{5}{\sqrt{26}}$, -$\frac{1}{\sqrt{26}}$).

Step-by-step solution

Find the vector PQ

To find the vector \(\overrightarrow{P Q}\), we must subtract the coordinates of point P from the coordinates of point Q. This can be done as follows: $$\overrightarrow{P Q} = Q - P = (-3 - (-3), -4 - 1, 1 - 0)$$

Calculate the coordinates of vector PQ

Now, let's calculate the actual coordinates of the vector \(\overrightarrow{P Q}\): $$\overrightarrow{P Q} = (0, -5, 1)$$

Write the vector in both forms

We can now express the vector \(\overrightarrow{P Q}\) in both required forms: a) \((0, -5, 1)\) b) \(0\textbf{i} - 5\textbf{j} + 1\textbf{k}\)

Find the magnitude of PQ

To find the magnitude of vector \(\overrightarrow{P Q}\), we use the formula: $$||\overrightarrow{P Q}|| = \sqrt{(a^2 + b^2 + c^2)}$$ Substitute the values a, b, and c with the coordinates of the vector \(\overrightarrow{P Q}\): $$||\overrightarrow{P Q}|| = \sqrt{(0^2 + (-5)^2 + 1^2)} = \sqrt{26}$$

Find two unit vectors parallel to PQ

To find the unit vectors parallel to vector \(\overrightarrow{P Q}\), we must first find the vector with a magnitude of 1 by dividing the vector \(\overrightarrow{P Q}\) by its magnitude: Unit vector in the direction of \(\overrightarrow{P Q}\): $$\frac{1}{\sqrt{26}} (0, -5, 1) = (0, -\frac{5}{\sqrt{26}}, \frac{1}{\sqrt{26}})$$ Now, we must find a unit vector opposite to the direction of \(\overrightarrow{P Q}\). This can be found by simply negating the values of the unit vector in the direction of \(\overrightarrow{P Q}\): Unit vector opposite to the direction of \(\overrightarrow{P Q}\): $$(-0, \frac{5}{\sqrt{26}}, -\frac{1}{\sqrt{26}})$$

Key concepts

Unit Vector

A unit vector is a vector that has a magnitude of exactly one. Unit vectors are used to indicate direction without concern for length. They are particularly useful in vector calculus and physics to express directions in three-dimensional space. If you have a vector, you can find a unit vector pointing in the same direction by dividing each component of the vector by its magnitude.

In the context of our exercise where we have vector \(\overrightarrow{P Q} = (0, -5, 1)\), we want to convert this into a unit vector. The magnitude of \(\overrightarrow{P Q}\) is \(\sqrt{26}\). Hence, the unit vector \(\mathbf{u}\) in the direction of \(\overrightarrow{P Q}\) will be:

\[ \mathbf{u} = \left( 0, -\frac{5}{\sqrt{26}}, \frac{1}{\sqrt{26}} \right) \]

This unit vector maintains the direction of \(\overrightarrow{P Q}\) while having a magnitude of 1. If we want a unit vector in the opposite direction, we simply negate each component of \(\mathbf{u}\). This gives us another unit vector pointing in the opposite direction and is expressed as:

\[ \mathbf{u}_{\text{opposite}} = \left(0, \frac{5}{\sqrt{26}}, -\frac{1}{\sqrt{26}} \right) \]

Vector Magnitude

Vector magnitude is a measure of a vector's length. For any vector \((a, b, c)\) in three-dimensional space, you can calculate its magnitude with the formula:

\[ ||\mathbf{v}|| = \sqrt{a^2 + b^2 + c^2} \]

This formula is derived from the Pythagorean theorem and reflects the "straight-line" distance from the start to the end of the vector.

In our scenario, the vector \(\overrightarrow{P Q}\) has components \((0, -5, 1)\). Using the formula, the magnitude is calculated as:

\[ ||\overrightarrow{P Q}|| = \sqrt{0^2 + (-5)^2 + 1^2} = \sqrt{26} \]

Knowing the magnitude is crucial when you need to normalize the vector or find the unit vector, as it helps to scale the vector to have a magnitude of one while preserving its direction.

Coordinate Subtraction

Coordinate subtraction is a technique used to find the vector between two points, like from point \(P\) to point \(Q\) in our exercise. To do this, you subtract the corresponding coordinates of the two points. If \(P(x_1, y_1, z_1)\) and \(Q(x_2, y_2, z_2)\), then the vector from \(P\) to \(Q\) is represented as:

\[ \overrightarrow{PQ} = (x_2 - x_1, y_2 - y_1, z_2 - z_1) \]

This step essentially "moves" the starting point of the vector to the origin, making calculations more straightforward.

In the document's context, we were given \(P(-3, 1, 0)\) and \(Q(-3, -4, 1)\). By employing coordinate subtraction, we calculated:

\[ \overrightarrow{PQ} = ((-3) - (-3), (-4) - 1, 1 - 0) = (0, -5, 1) \]

Coordinate subtraction is vital for obtaining vectors in various applications of vector calculus, physics, and engineering.