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Chapter 11: Problem 40

In Exercises \(32-52,\) for the given vector \(\vec{v}\), find the magnitude \(\|\vec{v}\|\) and an angle \(\theta\) with \(0 \leq \theta<360^{\circ}\) so that \(\vec{v}=\|\vec{v}\|\langle\cos (\theta), \sin (\theta)\rangle\) (See Definition 11.8.) Round approximations to two decimal places. $$ \vec{v}=\langle 0, \sqrt{7}\rangle $$

Short Answer

Expert verified
The magnitude is \( \sqrt{7} \) and the angle \( \theta \) is \(90^\circ\).

Step by step solution

01

Understand the vector components

The vector \( \vec{v} \) is given as \( \langle 0, \sqrt{7} \rangle \). This means the vector is directed only along the y-axis.
02

Calculate the magnitude of the vector

The magnitude of a vector \( \vec{v} = \langle a, b \rangle \) is given by the formula \( \|\vec{v}\| = \sqrt{a^2 + b^2} \). For the vector \( \vec{v} = \langle 0, \sqrt{7} \rangle \), this becomes \( \|\vec{v}\| = \sqrt{0^2 + (\sqrt{7})^2} = \sqrt{7} \).
03

Find the angle \(\theta\)

Since the vector is along the positive y-axis, the angle \(\theta\) with the positive x-axis is \(90^\circ\). This is derived from the properties of vectors where a vector along the y-axis with positive direction has an angle of \(90^\circ\) from the positive x-axis.
04

Verification

To verify, we check with \( \|\vec{v}\|\langle \cos(\theta), \sin(\theta) \rangle \). Here, \( \cos(90^\circ) = 0 \) and \( \sin(90^\circ) = 1 \), so the vector becomes \( \sqrt{7} \langle 0, 1 \rangle = \langle 0, \sqrt{7} \rangle \), which matches the original vector \( \vec{v} \).

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Key Concepts

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

Vector Components
In mathematics, a vector is often represented in terms of its components, which describe its direction and magnitude in a coordinate system. Understanding vector components is essential in precalculus and physics.

For a two-dimensional vector \( \vec{v} \) represented as \( \langle a, b \rangle\), \( a \) and \( b \) are its components along the x-axis and y-axis, respectively. These components can be thought of as coordinates in the vector space.
For instance, in the given vector \( \vec{v} = \langle 0, \sqrt{7} \rangle\), it shows:
  • \(<0>\) along x-axis 鈥� indicating no movement in this direction.
  • \(<\sqrt{7>}\) along y-axis 鈥� showing the vector's actual movement is upward in the y-direction.
In essence, vector components allow us to break down a vector into manageable parts to better understand its overall direction and magnitude.
Angle Between Vectors
Understanding the angle between vectors is crucial when working with direction and orientation in a plane. When we talk about the angle \( \theta \) between a vector and an axis, we're referring to its direction relative to that axis.

In our exercise, the angle is defined with respect to the positive x-axis. The vector \( \vec{v} = \langle 0, \sqrt{7} \rangle \) is entirely along the positive y-axis. Thus, there is a right angle or \( 90^\circ \) gap from the positive x-axis.
This concept is crucial when:
  • Mapping the direction of vectors.
  • Determining fields such as mechanical physics where directionality is key.
Remember, the angle between vectors can provide insights into their potential interaction and influence in a given space.
Trigonometric Functions
Trigonometric functions are vital tools used in understanding vector directions and rotations. These functions, namely sine, cosine, and tangent, relate angles to side ratios in right triangles, crucial for vector orientation.

For a vector \( \vec{v} = \|\vec{v}\|\langle \cos(\theta), \sin(\theta) \rangle \), the use of sine and cosine helps represent its direction in a coordinate system. In our example:
  • \(\cos(90^\circ) = 0\)
  • \(\sin(90^\circ) = 1\)
As such, the vector simplifies to \( \sqrt{7} \langle 0, 1 \rangle\), matching its component form. Understanding these functions allows for:
  • Accurate conversion of angular data into linear components.
  • Analysis of confined space problems in physics and engineering.
  • Better grasp on vector manipulation for rotation and movement.
Mastering trigonometric functions enhances the ability to model and predict rotational dynamics efficiently.
Calculation of Vector
Calculating various properties of vectors involves a systematic approach using algebraic and geometric principles. To find a vector's magnitude or length, use the formula:

\[ \|\vec{v}\| = \sqrt{a^2 + b^2} \]
For our vector \( \vec{v} = \langle 0, \sqrt{7} \rangle \), the calculation simplifies to \( \|\vec{v}\| = \sqrt{0^2 + (\sqrt{7})^2} = \sqrt{7} \).
This concise process involves:
  • Identifying vector components \(a\) and \(b\).
  • Applying the magnitude formula to compute the length.
The ability to calculate a vector's magnitude enables deeper understanding of:
  • Force and motion description in physics.
  • Spatial dynamics in engineering tasks.
  • General problem-solving involving directions and rates of change.
Mastering these calculations is critical in developing foundational knowledge useful across various advanced mathematical and scientific disciplines.
Precalculus
Precalculus serves as the bridge to understanding more complex mathematical concepts encountered in calculus and beyond. It establishes essential principles involving vectors, such as vector operations and their applications.

Precalculus introduces vectors as arrows indicating direction and magnitude, essential for physics and engineering. Understanding involves:
  • Breaking down a vector into components.
  • Applying trigonometric principles to determine angles.
These concepts form the groundwork for further mathematical exploration by:
  • Enabling learners to predict motions and forces analytically.
  • Facilitating computation of vector-related problems.
In this way, precalculus is invaluable in transforming abstract mathematical ideas into practical, real-world applications.

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Most popular questions from this chapter

Recall from Section 3.4 that given a complex number \(z=a+b i\) its complex conjugate, denoted \(\bar{z},\) is given by \(\bar{z}=a-b i .\) (a) Prove that \(|\bar{z}|=|z|\). (b) Prove that \(|z|=\sqrt{z \bar{z}}\) (c) Show that \(\operatorname{Re}(z)=\frac{z+\bar{z}}{2}\) and \(\operatorname{Im}(z)=\frac{z-\bar{z}}{2 i}\) (d) Show that if \(\theta \in \arg (z)\) then \(-\theta \in \arg (\bar{z}) .\) Interpret this result geometrically. (e) Is it always true that \(\operatorname{Arg}(z)=-\operatorname{Arg}(z) ?\)

Find a parametric description for the given oriented curve. the curve \(x=y^{2}-9\) from (-5,-2) to (0,3) .

60 through #63. In Exercises \(60-63,\) you need to solve for \(\theta\) 鈥� # Convert the equation from rectangular coordinates into polar coordinates. Solve for \(r\) in all but #60 through #63. In Exercises \(60-63,\) you need to solve for \(\theta\) \(y=x \sqrt{3}\)

Find a parametric description for the given oriented curve. the circle \((x-3)^{2}+(y+1)^{2}=117\), oriented counter-clockwise

Convert the equation from polar coordinates into rectangular coordinates. \(r=-\sqrt{5} \csc (\theta)\)
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