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Equality of vectors Assume \(\overrightarrow{P Q}\) equals \(\overrightarrow{R S}\). Does it follow that \(\overrightarrow{P R}\) is equal to \(\overrightarrow{Q S} ?\) Explain your answer.

Short Answer

Expert verified
Answer: Yes, if \(\overrightarrow{P Q}\) and \(\overrightarrow{R S}\) are equal, then \(\overrightarrow{P R}\) and \(\overrightarrow{Q S}\) are also equal.

Step by step solution

01

Understand the given vectors equality

We are given that the vectors \(\overrightarrow{P Q}\) and \(\overrightarrow{R S}\) are equal. This means that their magnitudes and directions are the same.
02

Establish the relationship between the vectors components

Let \(\overrightarrow{P Q} = \begin{pmatrix} p_1 - q_1 \\ p_2 - q_2 \end{pmatrix}\) and \(\overrightarrow{R S} = \begin{pmatrix} r_1 - s_1 \\ r_2 - s_2 \end{pmatrix}\). Given that \(\overrightarrow{P Q} = \overrightarrow{R S}\), their components must be equal which gives us the following equations: 1. \(p_1 - q_1 = r_1 - s_1\) 2. \(p_2 - q_2 = r_2 - s_2\)
03

Determine whether \(\overrightarrow{P R}\) and \(\overrightarrow{Q S}\) are equal

Now, let's find \(\overrightarrow{P R}\) and \(\overrightarrow{Q S}\): \(\overrightarrow{P R} = \begin{pmatrix} p_1 - r_1 \\ p_2 - r_2 \end{pmatrix}\) \(\overrightarrow{Q S} = \begin{pmatrix} q_1 - s_1 \\ q_2 - s_2 \end{pmatrix}\) For these vectors to be equal, the following conditions must hold: 1. \(p_1 - r_1 = q_1 - s_1\) 2. \(p_2 - r_2 = q_2 - s_2\) Now we have two sets of simultaneous equations. Let's compare the equations in each set: From equations 1 and 2 in Step 2, and using the transitive property, we have: \(p_1 - q_1 = r_1 - s_1 \Rightarrow p_1 - r_1 = q_1 - s_1\) \(p_2 - q_2 = r_2 - s_2 \Rightarrow p_2 - r_2 = q_2 - s_2\) These equations are the same as the conditions we derived in Step 3 for \(\overrightarrow{P R}\) and \(\overrightarrow{Q S}\) to be equal. Hence, as \(\overrightarrow{P Q} = \overrightarrow{R S}\), \(\overrightarrow{P R}\) must equal \(\overrightarrow{Q S}\).

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Key Concepts

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

Vector Components
When discussing vectors, an essential aspect is their components. These components help us understand how a vector is expressed in terms of its basic directional parts. For two-dimensional vectors, we represent it as
  • A horizontal component, denoted as the first coordinate (usually marked as \(x\) or \(p_1\)),
  • A vertical component, marked as the second coordinate (often labeled as \(y\) or \(p_2\)).
Vectors like \(\overrightarrow{PQ}\) can be expressed with the components \(p_1 - q_1\) for the x-direction, and \(p_2 - q_2\) for the y-direction. Knowing the vector components gives us precise detail of each vector's direction and magnitude.
By comparing the components of vectors \(\overrightarrow{PQ}\) and \(\overrightarrow{RS}\), where equality of vector components indicates that the two vectors are equal, you establish a clear understanding of their nature.
Learning to translate vector equality into equations like \(p_1 - q_1 = r_1 - s_1\) is a fundamental skill for solving problems related to vector equality.
Magnitude and Direction
Magnitude and direction are two critical attributes describing a vector fully. The magnitude is essentially the "size" or "length" of the vector, calculated using the Pythagorean theorem for two-dimensional vectors. For a vector like \(\overrightarrow{PQ} = \begin{pmatrix} p_1 - q_1 \ p_2 - q_2 \end{pmatrix}\), the magnitude \(|\overrightarrow{PQ}|\) is given as
  • \( |\overrightarrow{PQ}| = \sqrt{(p_1 - q_1)^2 + (p_2 - q_2)^2}\).
The direction relates to the angle this vector makes with a reference axis, providing the tilt or orientation of the line it forms.
When two vectors possess the same magnitude and direction, they are identical by definition. Hence, the given vectors \(\overrightarrow{PQ}\) and \(\overrightarrow{RS}\) being equal means they have not only the same length but also point in the same direction. This property is crucial when identifying if any two vectors are indeed equivalent.
If a vector's components are equal, then by default, its magnitude and direction must align as well, leading to vector equality.
Simultaneous Equations
Simultaneous equations are a powerful tool that can be used to solve problems involving vector components. They come into play when multiple equations need to be solved together, offering a method to find values that satisfy all conditions at once.
In vector equality problems, like comparing \(\overrightarrow{PR}\) and \(\overrightarrow{QS}\), simultaneous equations help verify equality. We derive these equations from vector components' equality:
  • First equation: \(p_1 - r_1 = q_1 - s_1\),
  • Second equation: \(p_2 - r_2 = q_2 - s_2\).
These equations represent the idea that both the horizontal and vertical displacements must match for the vectors to be equal.
Solving these simultaneous equations confirms the relationships between different points, such as proving that \(\overrightarrow{PR}\) equals \(\overrightarrow{QS}\) when \(\overrightarrow{PQ}\) and \(\overrightarrow{RS}\) are equal.
Understanding simultaneous equations allows students to isolate variables and solve diverse mathematical challenges, further reinforcing their grasp of vector equality.

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Most popular questions from this chapter

Consider the ellipse \(\mathbf{r}(t)=\langle a \cos t, b \sin t\rangle\) for \(0 \leq t \leq 2 \pi,\) where \(a\) and \(b\) are real numbers. Let \(\theta\) be the angle between the position vector and the \(x\) -axis. a. Show that \(\tan \theta=(b / a) \tan t\) b. Find \(\theta^{\prime}(t)\) c. Note that the area bounded by the polar curve \(r=f(\theta)\) on the interval \([0, \theta]\) is \(A(\theta)=\frac{1}{2} \int_{0}^{\theta}(f(u))^{2} d u\) Letting \(f(\theta(t))=|\mathbf{r}(\theta(t))|,\) show that \(A^{\prime}(t)=\frac{1}{2} a b\) d. Conclude that as an object moves around the ellipse, it sweeps out equal areas in equal times.

Let \(\mathbf{u}=\left\langle u_{1}, u_{2}, u_{3}\right\rangle\) \(\mathbf{v}=\left\langle v_{1}, v_{2}, v_{3}\right\rangle\), and \(\mathbf{w}=\) \(\left\langle w_{1}, w_{2}, w_{3}\right\rangle\). Let \(c\) be a scalar. Prove the following vector properties. \(\mathbf{u} \cdot \mathbf{v}=\mathbf{v} \cdot \mathbf{u}\)

Carry out the following steps to determine the (smallest) distance between the point \(P\) and the line \(\ell\) through the origin. a. Find any vector \(\mathbf{v}\) in the direction of \(\ell\) b. Find the position vector u corresponding to \(P\). c. Find \(\operatorname{proj}_{\mathbf{v}} \mathbf{u}\). d. Show that \(\mathbf{w}=\mathbf{u}-\) projy \(\mathbf{u}\) is a vector orthogonal to \(\mathbf{v}\) whose length is the distance between \(P\) and the line \(\ell\) e. Find \(\mathbf{w}\) and \(|\mathbf{w}| .\) Explain why \(|\mathbf{w}|\) is the distance between \(P\) and \(\ell\). \(P(-12,4) ; \ell: y=2 x\)

Let \(\mathbf{v}=\langle a, b, c\rangle\) and let \(\alpha, \beta\) and \(\gamma\) be the angles between \(\mathbf{v}\) and the positive \(x\) -axis, the positive \(y\) -axis, and the positive \(z\) -axis, respectively (see figure). a. Prove that \(\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1\) b. Find a vector that makes a \(45^{\circ}\) angle with i and \(\mathbf{j}\). What angle does it make with k? c. Find a vector that makes a \(60^{\circ}\) angle with i and \(\mathbf{j}\). What angle does it make with k? d. Is there a vector that makes a \(30^{\circ}\) angle with i and \(\mathbf{j}\) ? Explain. e. Find a vector \(\mathbf{v}\) such that \(\alpha=\beta=\gamma .\) What is the angle?

The definition \(\mathbf{u} \cdot \mathbf{v}=|\mathbf{u}||\mathbf{v}| \cos \theta\) implies that \(|\mathbf{u} \cdot \mathbf{v}| \leq|\mathbf{u}||\mathbf{v}|(\text {because}|\cos \theta| \leq 1) .\) This inequality, known as the Cauchy-Schwarz Inequality, holds in any number of dimensions and has many consequences. Use the vectors \(\mathbf{u}=\langle\sqrt{a}, \sqrt{b}\rangle\) and \(\mathbf{v}=\langle\sqrt{b}, \sqrt{a}\rangle\) to show that \(\sqrt{a b} \leq(a+b) / 2,\) where \(a \geq 0\) and \(b \geq 0\).

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