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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. $$\overrightarrow{S Q}+\overrightarrow{Q P}$$

Short Answer

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

Step by step solution

01

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 \(P(-1,3): \overrightarrow{QP} = P - Q = (-1-4, 3-6) = (-5, -3)\)
02

Add the Vectors Using Definition

Add the vectors \(\overrightarrow{SQ}\) and \(\overrightarrow{QP}\) by summing their components:- \(\overrightarrow{SQ} + \overrightarrow{QP} = (-1, -3) + (-5, -3) = (-1-5, -3-3) = (-6, -6)\)
03

Calculate Magnitude

Calculate the magnitude of the resulting vector \((-6, -6)\):- The magnitude \(|\overrightarrow{SQ} + \overrightarrow{QP}|\) is given by the formula \(\sqrt{x^2 + y^2}\).- \(|(-6, -6)| = \sqrt{(-6)^2 + (-6)^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2}\).
04

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Round each answer to one decimal place. A regular pentagon is inscribed in a circle of radius 1 unit. Find the perimeter of the pentagon. Hint: First find the length of a side using the law of cosines.

You are given the weight of a block on an inclined plane, along with the angle \(\theta\) that the inclined plane makes with the horizontal. In each case, determine the components of the weight perpendicular to and parallel to the plane. (Round your answers to two decimal places where necessary.) $$12 \mathrm{lb} ; \theta=10^{\circ}$$

Graph the parametric equations using the given range for the parameter t. In each case, begin with the standard viewing rectangle and then make adjustments, as necessary, so that the graph utilizes as much of the viewing screen as possible. For example, in graphing the circle given by \(x=\cos t\) and \(y=\sin t,\)it would be natural to choose a viewing rectangle extending from -1 to 1 in both the \(x\) - and \(y\) -directions. \(x=4 \cos t, y=-3 \sin t, 0 \leq t \leq 2 \pi(\) the same as the ellipse in Exercise 25 but traced out in the opposite direction)

Graph the parametric equations using the given range for the parameter t. In each case, begin with the standard viewing rectangle and then make adjustments, as necessary, so that the graph utilizes as much of the viewing screen as possible. For example, in graphing the circle given by \(x=\cos t\) and \(y=\sin t,\)it would be natural to choose a viewing rectangle extending from -1 to 1 in both the \(x\) - and \(y\) -directions. \(x=8 \cos t+\cos 8 t, y=8 \sin t-\sin 8 t, \quad 0 \leq t \leq 2 \pi\) (hypocycloid with nine cusps)

Determine a polar equation for the circle satisfying the given conditions. The radius is \(\sqrt{6},\) and the polar coordinates of the center are: (a) \((2, \pi)\) (b) \(\left(2, \frac{3 \pi}{4}\right)\) (c) (0,0)

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