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For the following exercises, the two-dimensional vectors a and b are given. Find the measure of the angle ? between a and b. Express the answer in radians rounded to two decimal places, if it is not possible to express it exactly. Is ? an acute angle? $$ \mathbf{u}=5 \mathbf{i}, \mathbf{v}=-6 \mathbf{i}+6 \mathbf{j} $$

Short Answer

Expert verified
The angle is approximately 2.36 radians and it is not acute.

Step by step solution

01

Understand the Vectors

First, identify the given vectors: \( \mathbf{u} = 5 \mathbf{i} \) and \( \mathbf{v} = -6 \mathbf{i} + 6 \mathbf{j} \). This means \( \mathbf{u} = (5, 0) \) and \( \mathbf{v} = (-6, 6) \).
02

Calculate the Dot Product

The dot product \( \mathbf{u} \cdot \mathbf{v} \) is calculated as follows: \( \mathbf{u} \cdot \mathbf{v} = 5 \times (-6) + 0 \times 6 = -30 \).
03

Calculate the Magnitudes of the Vectors

Calculate the magnitudes: \[ \|\mathbf{u}\| = \sqrt{5^2 + 0^2} = 5 \]\[ \|\mathbf{v}\| = \sqrt{(-6)^2 + 6^2} = \sqrt{72} = 6\sqrt{2} \]
04

Apply the Cosine Formula

The angle between vectors can be found using the formula: \[ \cos(\theta) = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\| \times \|\mathbf{v}\|} \]Substituting the values found: \[ \cos(\theta) = \frac{-30}{5 \times 6\sqrt{2}} = \frac{-1}{\sqrt{2}} \].
05

Calculate the Angle

Compute \( \theta \) using the inverse cosine:\[ \theta = \cos^{-1}\left(\frac{-1}{\sqrt{2}}\right) \approx 2.36 \text{ radians} \]
06

Determine if the Angle is Acute

An angle \( \theta \) is acute if \( 0 < \theta < \frac{\pi}{2} \). Since \( 2.36 \) radians is greater than \( \frac{\pi}{2} \approx 1.57 \), the angle is not acute.

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Key Concepts

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

Understanding the Dot Product
The dot product, also known as the scalar product, is an essential operation in vector mathematics. It provides a way to multiply two vectors that results in a scalar rather than a vector. This value helps to determine how "aligned" two vectors are with respect to each other. The dot product is calculated by multiplying the corresponding components of two vectors and then summing those products. For vectors \( \mathbf{a} = (a_1, a_2) \) and \( \mathbf{b} = (b_1, b_2) \), the dot product \( \mathbf{a} \cdot \mathbf{b} \) is: \[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \]This operation has various applications, one of the most prominent being to find the angle between two vectors through its role in the cosine formula. The value of the dot product can also indicate if vectors are orthogonal (perpendicular); if the dot product equals zero, the vectors are orthogonal.
Calculating Magnitude of Vectors
The magnitude of a vector essentially describes its length. This concept is crucial when working with vectors because it gives us the scale of influence a vector has in a given space. To find the magnitude of a two-dimensional vector \( \mathbf{a} = (a_1, a_2) \), the following formula is used:\[ \|\mathbf{a}\| = \sqrt{a_1^2 + a_2^2} \]It's akin to applying the Pythagorean theorem, considering the vector's components as a right triangle's base and height. Knowing both the magnitude and direction is important in numerous applications, such as physics, where they describe forces and velocity.
The Cosine Formula for Angle Between Vectors
The cosine formula is a powerful tool for finding the angle between two vectors. It builds upon the dot product and the magnitudes of each vector. This is expressed as:\[ \cos(\theta) = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\| \times \|\mathbf{v}\|} \]Where \( \theta \) is the angle between vectors \( \mathbf{u} \) and \( \mathbf{v} \).To solve for the angle, calculate \( \cos^{-1} \) (also known as the arc cosine) of the resulting value. The cosine formula derives from the innate relationship between dot products and magnitudes, translating the "projection" of one vector on another into a tangible angle measure. This translates well into real-world problems like determining the angle between directional headings or forces.
Understanding Radians
Radians are a natural way to express angles because they derive directly from the geometry of circles. While degrees are more common in everyday language, radians simplify many mathematical equations. One complete circle is \( 2\pi \) radians, analogous to \( 360^\circ \). Conversion between degrees and radians is common:
  • \( 180^\circ = \pi \; \text{radians} \)
  • To convert from degrees to radians, multiply by \( \frac{\pi}{180} \)
  • To convert from radians to degrees, multiply by \( \frac{180}{\pi} \)
When expressing angles for mathematical computations or deeper geometrical understanding, radians often provide a cleaner and more consistent approach. Radians naturally integrate with calculus and trigonometry, enhancing clarity in the relationship between angles and arc lengths.

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Most popular questions from this chapter

Three forces with magnitudes \(80 \mathrm{lb}, 120 \mathrm{lb}\), and \(60 \mathrm{lb}\) act on an object at angles of \(45^{\circ}, 60^{\circ}\) and \(30^{\circ}\), respectively, with the positive \(x\) -axis. Find the magnitude and direction angle from the positive \(x\) -axis of the resultant force. (Round to two decimal places.)

A 1500 -lb boat is parked on a ramp that makes an angle of \(30^{\circ}\) with the horizontal. The boat's weight vector points downward and is a sum of two vectors: a horizontal vector \(\mathbf{v}_{1}\) that is parallel to the ramp and a vertical vector \(\mathbf{v}_{2}\) that is perpendicular to the inclined surface. The magnitudes of vectors \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}\) are the horizontal and vertical component, respectively, of the boat's weight vector. Find the magnitudes of \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}\). (Round to the nearest integer.)

The ring torus symmetric about the \(z\) -axis is a special type of surface in topology and its equation is given by \(\left(x^{2}+y^{2}+z^{2}+R^{2}-r^{2}\right)^{2}=4 R^{2}\left(x^{2}+y^{2}\right)\), where \(R>r>0 .\) The numbers \(R\) and \(r\) are called are the major and minor radii, respectively, of the surface. The following figure shows a ring torus for which \(R=2\) and \(r=1\). a. Write the equation of the ring torus with \(R=2\) and \(r=1\), and use a CAS to graph the surface. Compare the graph with the figure given. b. Determine the equation and sketch the trace of the ring torus from a. on the \(x y\) -plane. c. Give two examples of objects with ring torus shapes.

Find the trace of the given quadric surface in the specified plane of coordinates and sketch it. $$ -4 x^{2}+25 y^{2}+z^{2}=100, x=0 $$

Rewrite the given equation of the quadric surface in standard form. Identify the surface. $$ -x^{2}+36 y^{2}+36 z^{2}=9 $$

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