Problem 8 - Precalculus

Question

Assume that the coordinates of the points \(P\) \(Q, R, S,\) and \(O\) are as follows: \(P(-1,3) \quad Q(4,6) \quad R(4,3) \quad S(5,9) \quad O(0,0)\) Draw the indicated vector (using graph paper) and compute its magnitude. Compute the sums using the definition. Use the parallelogram law to compute the sums. $$\overrightarrow{S Q}+\overrightarrow{Q P}$$

Step-by-Step Solution

Answer

The magnitude of \(\overrightarrow{SQ} + \overrightarrow{QP}\) is \(6\sqrt{2}\).

Step 1: Identify Vectors

Identify the vectors we need to work with. First, calculate vectors \(\overrightarrow{SQ}\) and \(\overrightarrow{QP}\) from the given points:- \(S(5,9)\) to \(Q(4,6): \overrightarrow{SQ} = Q - S = (4-5, 6-9) = (-1, -3)\)- \(Q(4,6)\) to \(P(-1,3): \overrightarrow{QP} = P - Q = (-1-4, 3-6) = (-5, -3)\)

Step 2: Add the Vectors Using Definition

Add the vectors \(\overrightarrow{SQ}\) and \(\overrightarrow{QP}\) by summing their components:- \(\overrightarrow{SQ} + \overrightarrow{QP} = (-1, -3) + (-5, -3) = (-1-5, -3-3) = (-6, -6)\)

Step 3: Calculate Magnitude

Calculate the magnitude of the resulting vector \((-6, -6)\):- The magnitude \(|\overrightarrow{SQ} + \overrightarrow{QP}|\) is given by the formula \(\sqrt{x^2 + y^2}\).- \(|(-6, -6)| = \sqrt{(-6)^2 + (-6)^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2}\).

Step 4: Parallelogram Law Application

According to the parallelogram law, the sum of two vectors forms the diagonal of a parallelogram. The vectors \(\overrightarrow{SQ}\) and \(\overrightarrow{QP}\) can be represented as adjacent sides of a parallelogram. The diagonal from the origin (or the starting point of the first vector) will be \(\overrightarrow{SQ} + \overrightarrow{QP}\). The calculation in step 2 represents the diagonal vector by vector addition.

Key Concepts

Magnitude Calculation

The magnitude of a vector is essentially its length. Calculating this magnitude helps us understand just how "long" a vector is in the graphical representation of vectors.
It’s important in solving problems involving vectors because it gives us a scalar representation of the vector’s size, regardless of its direction.

To find the magnitude of a two-dimensional vector, such as \((-6, -6)\), you can utilize the Pythagorean theorem. Imagine if the vector forms the diagonal of a right triangle with its components acting as perpendicular sides:

• The formula used here is \(\sqrt{x^2 + y^2}\), where x and y are the horizontal and vertical components of the vector, respectively.
• For our vector, this results in \(\sqrt{(-6)^2 + (-6)^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2}\).

This shows that \(6\sqrt{2}\) is the length of vector \((-6, -6)\). Recognizing and using the formula for magnitude is crucial for further vector calculations.

Parallelogram Law

The parallelogram law is a simple and elegant way to visualize vector addition. When you have two vectors, \(\overrightarrow{SQ}\) and \(\overrightarrow{QP}\) for example, and you place them so that their tails coincide, they form two sides of a parallelogram.
The resulting sum of these vectors can be visualized as the diagonal of this parallelogram.

• Placing \(\overrightarrow{SQ}\) and \(\overrightarrow{QP}\) with a common starting point illustrates how they form a parallelogram shape.
• The diagonal that starts from this common point represents \(\overrightarrow{SQ} + \overrightarrow{QP}\).

In our exercise, the parallelogram law neatly sums \(\overrightarrow{SQ}\) and \(\overrightarrow{QP}\) to determine their combined effect as a vector. This geometrical approach provides a clear method for understanding vector addition, securing its spot as a fundamental principle in physics and engineering.

Vector Components

Vectors are typically described in component form, as components allow us to handle vectors algebraically and simplify many calculations.
Each vector has unique horizontal (x-axis) and vertical (y-axis) components, which makes comparisons and additions straightforward.

If you consider a vector from point \(A(x_1, y_1)\) to point \(B(x_2, y_2)\), its components are given by:

• The horizontal component is the difference in x-coordinates: \((x_2-x_1)\) or simply the change in horizontal distance.
• The vertical component is the difference in y-coordinates: \((y_2-y_1)\) or the change in vertical distance.

For instance, given vectors \(\overrightarrow{SQ}\) with positions \((5,9)\) to \((4,6)\) and \(\overrightarrow{QP}\) from \((4,6)\) to \((-1,3)\), their respective transformations into \((-1, -3)\) and \((-5, -3)\) show how each movement is distinct yet describe the vectors comprehensively. Working with these components simplifies addition, subtraction, and comparison of vectors greatly, which is why learning about vector components is so vital.