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

Short Answer

Expert verified
The resultant vector is \( (4, 6) \) with magnitude \( 2\sqrt{13} \).

Step by step solution

01

Define the Vectors

The vector \(\overrightarrow{OP}\) is the vector from point \(O\) to point \(P\), so its coordinates are \((-1, 3)\). The vector \(\overrightarrow{PQ}\) is the vector from point \(P\) to point \(Q\), so to find it, we subtract the coordinates of \(P\) from \(Q\): \(\overrightarrow{PQ} = (4-(-1), 6-3) = (5, 3)\).
02

Calculate Vector Sum

Using the vectors from Step 1, we add \(\overrightarrow{OP}\) and \(\overrightarrow{PQ}\) by adding their respective components: \((x_1 + x_2, y_1 + y_2) = (-1 + 5, 3 + 3) = (4, 6)\). Thus, \(\overrightarrow{OP} + \overrightarrow{PQ} = (4, 6)\).
03

Compute Magnitude of the Resultant Vector

The magnitude of the vector \((4, 6)\) is calculated using the formula \(\|\mathbf{v}\| = \sqrt{x^2 + y^2}\): \[ \|\mathbf{v}\| = \sqrt{4^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13} \].
04

Confirm using the Parallelogram Law

The parallelogram law of vector addition involves constructing a parallelogram with \(\overrightarrow{OP}\) and \(\overrightarrow{PQ}\) as adjacent sides. The resultant vector, \(\overrightarrow{OA}\), from \(O\) to the opposite corner of the parallelogram, confirms the addition: \(\overrightarrow{OA} = \overrightarrow{OP} + \overrightarrow{PQ} = (4, 6)\), confirming the previous result.

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Key Concepts

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

Coordinate Geometry
Coordinate geometry is a branch of mathematics that helps us deal with geometry using a coordinate system. A coordinate system uses numbers as coordinates to describe the position of points on a plane.

In our exercise, we work in a two-dimensional plane with pairs of numbers, \(x, y\), representing each point like \((x, y)\). For example, point \(P(-1, 3)\) tells us that \(-1\) is how far 'left or right' it is from the origin (0,0), and \(3\) is how far 'up or down'.
  • The negative sign with \(-1\) means it is to the left of the origin.
  • The positive \(3\) means it is above the origin.
Understanding the position of each point in coordinate geometry is fundamental to determining characteristics like distance or direction.
Magnitude of a Vector
The magnitude of a vector tells us about its length. It's a bit like figuring out how long a stick would measure if it was positioned between two points. In math, we find the magnitude by using the Pythagorean theorem.
For a vector with components \(x\) and \(y\), its magnitude is represented by \(\|\mathbf{v}\|\) and calculated as: \[\|\mathbf{v}\| = \sqrt{x^2 + y^2}\].

In our problem, the resultant vector \(\overrightarrow{OP} + \overrightarrow{PQ}\) has the coordinates \(4,6\). To find its magnitude, we plug \(4\) and \(6\) into our formula:
\[\|\mathbf{v}\| = \sqrt{4^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13} .\]
This magnitude tells us the straight-line distance from point \(O\) to the resultant point in the plane. It's an important property in physics and engineering where vectors represent quantities like velocity or force.
Parallelogram Law
The parallelogram law is a visual way to understand vector addition. Imagine drawing a parallelogram where two vectors serve as adjacent sides. The diagonal of this parallelogram represents the sum of the two vectors.

In our specific case, \(\overrightarrow{OP}\) and \(\overrightarrow{PQ}\) create adjacent sides. The diagonal from \(O\) represents \(\overrightarrow{OP} + \overrightarrow{PQ}\). This diagonal or resultant vector should match our earlier calculated sum.
  • Start at \(O\), draw \(\overrightarrow{OP}\).
  • Then from \(P\), draw \(\overrightarrow{PQ}\).
  • Complete the parallelogram from these points.
  • The vector from \(O\) to the opposite corner of the parallelogram is \(\overrightarrow{OA}\).
By applying the parallelogram law, we confirm both visually and mathematically that \(\overrightarrow{OA} = (4, 6)\), providing an intuitive check.
Vector Components
Vector components break a vector down into parts based on the coordinate directions. Think of a vector as having two fingers pointing: one horizontally along the x-axis, and the other vertically along the y-axis.
  • The x-component tells how far along the x-direction the vector is.
  • The y-component shows how far along the y-direction.
In our problem, vector \(\overrightarrow{OP}\) is made up of components \(-1\) (for the x-direction) and \(3\) (for the y-direction). When adding vectors, it's helpful to sum their respective components separately, like we did with \(\overrightarrow{OP}\) and \(\overrightarrow{PQ}\).

For example, the sum vector \(\overrightarrow{OP} + \overrightarrow{PQ} = (4, 6)\) was calculated by adding the x-components together and the y-components together: \((-1 + 5, 3 + 3) = (4, 6)\). This method makes the math straightforward and logical while following the rules of coordinate geometry thoroughly.

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Most popular questions from this chapter

Let \(\theta\) (where \(0 \leq \theta \leq \pi\) ) denote the angle between the two nonzero vectors \(\mathbf{A}\) and \(\mathbf{B}\). Then it can be shown that the cosine of \(\theta\) is given by the formula $$\cos \theta=\frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}||\mathbf{B}|}$$ (See Exercise 77 for the derivation of this result.) In Exercises \(65-70,\) sketch each pair of vectors as position vectors, then use this formula to find the cosine of the angle between the given pair of vectors. Also, in each case, use a calculator to compute the angle. Express the angle using degrees and using radians. Round the values to two decimal places. $$\mathbf{A}=\langle 4,1\rangle \text { and } \mathbf{B}=\langle 2,6\rangle$$

Let \(\theta\) (where \(0 \leq \theta \leq \pi\) ) denote the angle between the two nonzero vectors \(\mathbf{A}\) and \(\mathbf{B}\). Then it can be shown that the cosine of \(\theta\) is given by the formula $$\cos \theta=\frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}||\mathbf{B}|}$$ (See Exercise 77 for the derivation of this result.) In Exercises \(65-70,\) sketch each pair of vectors as position vectors, then use this formula to find the cosine of the angle between the given pair of vectors. Also, in each case, use a calculator to compute the angle. Express the angle using degrees and using radians. Round the values to two decimal places. $$\mathbf{A}=\langle 3,0\rangle \text { and } \mathbf{B}=\langle 1,4\rangle$$

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=\frac{3 t}{1+t^{3}}, y=\frac{3 t^{2}}{1+t^{3}}, \quad-\infty

Round each answer to one decimal place. Two trains leave the railroad station at noon. The first train travels along a straight track at 90 mph. The second train travels at 75 mph along another straight track that makes an angle of \(130^{\circ}\) with the first track. At what time are the trains 400 miles apart? Round your answer to the nearest minute.

Verify that \(\mathbf{v}+(\mathbf{u}-\mathbf{v})=\mathbf{u}\).

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