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Chapter 12: Problem 26

Define the points \(P(-3,-1)\) \(Q(-1,2), R(1,2), S(3,5), T(4,2),\) and \(U(6,4)\). Find the equal vectors among \(\overrightarrow{P Q}, \overrightarrow{R S},\) and \(\overrightarrow{T U}\).

Short Answer

Expert verified
The equal vectors are \(\overrightarrow{P Q}\) and \(\overrightarrow{R S}\), as both have the same components, \(\langle 2, 3 \rangle\).

Step by step solution

01

Converting coordinates to vectors

Given the coordinates of points P, Q, R, S, T, and U, we can represent them as vectors. \(P = \langle -3, -1 \rangle\) \(Q = \langle -1, 2 \rangle\) \(R = \langle 1, 2 \rangle\) \(S = \langle 3, 5 \rangle\) \(T = \langle 4, 2 \rangle\) \(U = \langle 6, 4 \rangle\)
02

Computing the vectors \(\overrightarrow{P Q}, \overrightarrow{R S},\) and \(\overrightarrow{T U}\)

Now that we have the vector representation of the points, we can compute the difference between the terminal and initial points for each vector. \(\overrightarrow{P Q} = Q - P = \langle -1 - (-3), 2 - (-1) \rangle = \langle 2, 3 \rangle\) \(\overrightarrow{R S} = S - R = \langle 3 - 1, 5 - 2 \rangle = \langle 2, 3 \rangle\) \(\overrightarrow{T U} = U - T = \langle 6 - 4, 4 - 2 \rangle = \langle 2, 2 \rangle\)
03

Comparing the vectors

Now that we have computed the vectors, we can easily compare them to find the equal ones. \(\overrightarrow{P Q} = \langle 2, 3 \rangle\) \(\overrightarrow{R S} = \langle 2, 3 \rangle\) \(\overrightarrow{T U} = \langle 2, 2 \rangle\) From the comparison, we can see that \(\overrightarrow{P Q}\) is equal to \(\overrightarrow{R S}\), as they have the same components. The vector \(\overrightarrow{T U}\) is different from the other two vectors. So the equal vectors among \(\overrightarrow{P Q}, \overrightarrow{R S},\) and \(\overrightarrow{T U}\) are \(\overrightarrow{P Q}\) and \(\overrightarrow{R S}\).

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Key Concepts

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

Equal Vectors
Equal vectors are vectors that have the same magnitude and direction. This means their components should be identical. In the context of this exercise, you are asked to determine which of the vectors, \(\overrightarrow{P Q}\), \(\overrightarrow{R S}\), and \(\overrightarrow{T U}\), are equal.
To do this, we compute each vector by subtracting the coordinates of the starting point from the ending point.

  • For \(\overrightarrow{P Q} = \langle 2, 3 \rangle \)
  • For \(\overrightarrow{R S} = \langle 2, 3 \rangle \)
  • For \(\overrightarrow{T U} = \langle 2, 2 \rangle \)

The vectors \(\overrightarrow{P Q}\) and \(\overrightarrow{R S}\) have exactly the same components, meaning they are equal vectors. Vector \(\overrightarrow{T U}\) has different components and therefore is not equal to the other two vectors.
Coordinate Geometry
Coordinate geometry, also known as analytic geometry, involves the study of geometry using a coordinate system. This allows for precise definitions and calculations of geometric concepts, such as points, lines and vectors.
By placing geometric figures in a coordinate plane, we can use algebraic techniques to solve geometric problems.

In the given exercise, each point is represented by specific coordinates. For example, point \(P\) corresponds to the coordinates \((-3, -1)\) while point \(Q\) corresponds to \((-1, 2)\). These coordinates allow us to perform calculations such as vector addition and subtraction by treating the points as vectors. This makes it possible to determine other geometric relationships and properties systematically.
Vector Subtraction
Vector subtraction is a key operation in vector mathematics. It involves finding the difference between two vectors and is crucial for calculating vectors from points.'
To subtract one vector \(\mathbf{a} = \langle x_1, y_1 \rangle\) from another vector \(\mathbf{b} = \langle x_2, y_2 \rangle\), you compute \(\mathbf{b} - \mathbf{a} = \langle x_2 - x_1, y_2 - y_1 \rangle\)

In our exercise:
  • For \(\overrightarrow{P Q} = Q - P = \langle -1 - (-3), 2 - (-1) \rangle = \langle 2, 3 \rangle\)
  • For \(\overrightarrow{R S} = S - R = \langle 3 - 1, 5 - 2 \rangle = \langle 2, 3 \rangle\)
  • For \(\overrightarrow{T U} = U - T = \langle 6 - 4, 4 - 2 \rangle = \langle 2, 2 \rangle\)

Through vector subtraction, we determine the direction and magnitude of a vector that arises when transitioning from one point to another on the plane. Understanding this concept is pivotal for various applications in physics and engineering.

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Most popular questions from this chapter

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?

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. \(\mathbf{u} \cdot \mathbf{v}=\mathbf{v} \cdot \mathbf{u}\)

Consider the trajectory given by the position function $$\mathbf{r}(t)=\left\langle 50 e^{-t} \cos t, 50 e^{-t} \sin t, 5\left(1-e^{-t}\right)\right), \quad \text { for } t \geq 0$$ a. Find the initial point \((t=0)\) and the "terminal" point \(\left(\lim _{t \rightarrow \infty} \mathbf{r}(t)\right)\) of the trajectory. b. At what point on the trajectory is the speed the greatest? c. Graph the trajectory.

\(\mathbb{R}^{2}\) Consider the vectors \(\mathbf{I}=\langle 1 / \sqrt{2}, 1 / \sqrt{2}\rangle\) and \(\mathbf{J}=\langle-1 / \sqrt{2}, 1 / \sqrt{2}\rangle\). Express \(\mathbf{I}\) and \(\mathbf{J}\) in terms of the usual unit coordinate vectors i and j. Then, write i and \(\mathbf{j}\) in terms of \(\mathbf{I}\) and \(\mathbf{J}\).

Let \(D\) be a solid heat-conducting cube formed by the planes \(x=0, x=1, y=0, y=1, z=0,\) and \(z=1 .\) The heat flow at every point of \(D\) is given by the constant vector \(\mathbf{Q}=\langle 0,2,1\rangle\) a. Through which faces of \(D\) does \(Q\) point into \(D ?\) b. Through which faces of \(D\) does \(\mathbf{Q}\) point out of \(D ?\) c. On which faces of \(D\) is \(Q\) tangential to \(D\) (pointing neither in nor out of \(D\) )? d. Find the scalar component of \(\mathbf{Q}\) normal to the face \(x=0\). e. Find the scalar component of \(\mathbf{Q}\) normal to the face \(z=1\). f. Find the scalar component of \(\mathbf{Q}\) normal to the face \(y=0\).
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