/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 9 For the following exercises, con... [FREE SOLUTION] | 91Ó°ÊÓ

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For the following exercises, consider points \(P(-1,3)\) \(Q(1,5),\) and \(R(-3,7) .\) Determine the requested vectors and express each of them a. in component form and b. by using the standard unit vectors. The unit vector in the direction of \(\overrightarrow{P Q}\)

Short Answer

Expert verified
The unit vector in the direction of \(\overrightarrow{PQ}\) is \(\frac{1}{\sqrt{2}}\mathbf{i} + \frac{1}{\sqrt{2}}\mathbf{j}\).

Step by step solution

01

Calculate the Component Form of \(\overrightarrow{PQ}\)

To find the component form of \(\overrightarrow{PQ}\), subtract the coordinates of point \(P\) from the coordinates of point \(Q\). This gives us the difference in each coordinate:\[\overrightarrow{PQ} = (x_2 - x_1, y_2 - y_1) = (1 - (-1), 5 - 3) = (2, 2)\]So the component form of \(\overrightarrow{PQ}\) is \((2, 2)\).
02

Express \(\overrightarrow{PQ}\) Using Standard Unit Vectors

Standard unit vectors are expressed in terms of \(\mathbf{i}\) and \(\mathbf{j}\), which represent the unit vectors in the direction of the x-axis and y-axis, respectively. From the component form \((2, 2)\), \(\overrightarrow{PQ}\) can be expressed as:\[\overrightarrow{PQ} = 2\mathbf{i} + 2\mathbf{j}\]
03

Calculate the Magnitude of \(\overrightarrow{PQ}\)

The magnitude of a vector \((a, b)\) is calculated using the formula:\[\|\overrightarrow{PQ}\| = \sqrt{a^2 + b^2} = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}\]
04

Find the Unit Vector in the Direction of \(\overrightarrow{PQ}\)

A unit vector in the direction of a given vector is obtained by dividing each component of the vector by its magnitude. Thus, the unit vector \(\overrightarrow{u}\) in the direction of \(\overrightarrow{PQ}\) is:\[\overrightarrow{u} = \frac{1}{\|\overrightarrow{PQ}\|}(2, 2) = \left(\frac{2}{2\sqrt{2}}, \frac{2}{2\sqrt{2}}\right) = \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\]Expressing this using standard unit vectors:\[\overrightarrow{u} = \frac{1}{\sqrt{2}}\mathbf{i} + \frac{1}{\sqrt{2}}\mathbf{j}\]

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Key Concepts

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

Component Form
When working with vectors, the component form is a simplified way to describe a vector in terms of its horizontal and vertical components. Given two points, say \(P(x_1, y_1)\) and \(Q(x_2, y_2)\), you can find the vector \(\overrightarrow{PQ}\) by subtracting the coordinates of \(P\) from \(Q\):
  • Horizontal component: \(x_2 - x_1\)
  • Vertical component: \(y_2 - y_1\)
Thus, the component form \(\overrightarrow{PQ}\) is expressed as \((x_2 - x_1, y_2 - y_1)\). This method gives you not only the direction but also how far the vector extends in each direction. For instance, in the example \(\overrightarrow{PQ} = (2, 2)\), this means that the vector moves 2 units horizontally and 2 units vertically from point \(P\) to point \(Q\).
This representation is essential because it illuminates the vector's path in a coordinate system.
Unit Vector
A unit vector is a vector that has a magnitude of 1, pointing in a specific direction. It is often used to express the direction of a vector without considering its magnitude. To find a unit vector in the direction of a given vector \(\overrightarrow{v}\), you divide the vector by its magnitude.
  • Unit vector formula: \(\overrightarrow{u} = \frac{\overrightarrow{v}}{\|\overrightarrow{v}\|}\)
This procedure allows you to keep the direction steady while normalizing the length to 1. For example, the unit vector \(\overrightarrow{u}\) in the direction of \(\overrightarrow{PQ} = (2, 2)\) is calculated as:
  • \(\overrightarrow{u} = \left(\frac{2}{2\sqrt{2}}, \frac{2}{2\sqrt{2}}\right) = \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\)
This process transforms any vector, regardless of its size, into a vector with a consistent magnitude of one, which is particularly helpful in simplifying calculations and visualizations in various applications.
Standard Unit Vectors
Standard unit vectors are predefined vectors that are used as a basis to express any vector in a two-dimensional plane. They are typically denoted as \(\mathbf{i}\) and \(\mathbf{j}\), where:
  • \(\mathbf{i}\) is the unit vector along the x-axis: \((1, 0)\)
  • \(\mathbf{j}\) is the unit vector along the y-axis: \((0, 1)\)
Any vector in the plane can be expressed as a combination of these standard unit vectors. For a vector in component form, say \((a, b)\), it can be rewritten using these unit vectors as \(a\mathbf{i} + b\mathbf{j}\). This method is particularly useful because it breaks down vectors into standard directional components that can be easily manipulated and understood. For example, the component form \((2, 2)\) translates to \(2\mathbf{i} + 2\mathbf{j}\), indicating 2 units in both the x and y directions. This clear representation simplifies computations involving multiple vectors.
Magnitude of a Vector
The magnitude of a vector represents its length and is a measure of how far it extends from its initial point. Calculating the magnitude of a vector \((a, b)\) involves using the Pythagorean theorem, as the vector forms a right triangle with the axes.
  • Magnitude formula: \(\|\overrightarrow{v}\| = \sqrt{a^2 + b^2}\)
This formula essentially combines the horizontal and vertical components into a single scalar value, indicating the size of the vector. In practice, knowing the magnitude is crucial when normalizing vectors to create unit vectors. For instance, for the vector \(\overrightarrow{PQ} = (2, 2)\), the magnitude is:
  • \(\|\overrightarrow{PQ}\| = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}\)
Understanding magnitude provides insights into not just the size of the vector, but also facilitates transitions between different forms and uses of vectors in various fields such as physics and engineering.

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Most popular questions from this chapter

Determine whether the line of parametric equations \(x=1+2 t, y=-2 t, z=2+t, \quad t \in \mathbb{R}\) intersects the plane with equation \(3 x+4 y+6 z-7=0\) . If it does intersect, find the point of intersection.

For the following exercises, lines \(L_{1}\) and \(L_{2}\) are given. Determine whether the lines are equal, parallel but not equal, skew, or intersecting. \(L_{1} : 3 x=y+1=2 z\) and \(L_{2} : x=6+2 t, y=17+6 t, z=9+3 t, \quad t \in \mathbb{R}\)

Let \(\mathbf{r}(t)=\langle x(t), y(t), z(t)\rangle\) be the position vector of a particle at the time \(t \in[0, T],\) where \(x, y,\) and \(z\) are smooth functions on \([0, T] .\) The instantaneous velocity of the particle at time \(t\) is defined by vector \(\mathbf{v}(t)=\left\langle x^{\prime}(t), y^{\prime}(t), z^{\prime}(t)\right\rangle, \quad\) with components that are the derivatives with respect to \(t,\) of the functions \(x, y,\) and z, respectively. The magnitude \(\|\mathbf{v}(t)\| \quad\) of the instantaneous velocity vector is called the speed of the particle at time t. Vector \(\mathbf{a}(t)=\left\langle x^{\prime \prime}(t), y^{\prime \prime}(t), z^{\prime \prime}(t)\right\rangle\) with components that are the second derivatives with respect to \(t,\) of the functions \(x, y,\) and \(z\) , respectively, gives the acceleration of the particle at time \(t\) . Consider \(\mathbf{r}(t)=\langle\cos t, \sin t, 2 t\rangle\) the position vector of a particle at time \( t \in[0,30],\) where the components of \(\mathbf{r}\) are expressed in centimeters and time is expressed in seconds. a. Find the instantaneous velocity, speed, and acceleration of the particle after the first second. Round your answer to two decimal places. b. Use a CAS to visualize the path of the particle - that is, the set of all points of coordinates \((\cos t, \sin t, 2 t),\) where \(t \in[0,30] .\)

[T] Consider \(\mathbf{r}(t)=\langle\sin t, \cos t, 2 t\rangle\) the position vector of a particle at time \(t \in[0,3], \) where the components of \(\mathbf{r}\) are expressed in centimeters and time is measured in seconds. Let \(\overrightarrow{O P}\) be the position vector of the particle after 1 sec. a. Determine the velocity vector \(\mathbf{v}(1)\) of the particle after 1 sec. b. Find the scalar equation of the plane that is perpendicular to \(\mathbf{v}(1)\) and passes through point \(P\) This plane is called the normal plane to the path of the particle at point \(P\) . c. Use a CAS to visualize the path of the particle along with the velocity vector and normal plane at point \(P .\)

Show there is no plane perpendicular to \(\mathbf{n}=\mathbf{i}+\mathbf{j}\) that passes through points \(P(1,2,3)\) and \(Q(2,3,4)\)

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