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Chapter 2: Problem 133

For the following exercises, the two-dimensional vectors a and \(\mathbf{b}\) are given. a. Find the measure of the angle \(\theta\) between a and b. Express the answer in radians rounded to two decimal places, if it is not possible to express it exactly. b. Is \(\theta\) an acute angle? $$ \mathbf{u}=3 \mathbf{i}, \quad \mathbf{v}=4 \mathbf{i}+4 \mathbf{j} $$

Short Answer

Expert verified
a. \( \theta = \frac{\pi}{4} \) radians b. Yes, the angle is acute.

Step by step solution

01

Find the Dot Product

First, compute the dot product of vectors \( \mathbf{u} = 3\mathbf{i} \) and \( \mathbf{v} = 4\mathbf{i} + 4\mathbf{j} \). The dot product is calculated using the formula: \( \mathbf{u} \cdot \mathbf{v} = 3 \times 4 + 0 \times 4 = 12. \) Since \( \mathbf{u} \) has no \( \mathbf{j} \)-component, this simplifies our calculation.
02

Compute the Magnitudes of the Vectors

Calculate the magnitude of each vector. The magnitude of \( \mathbf{u} \) is \( |\mathbf{u}| = \sqrt{3^2} = 3. \) For \( \mathbf{v} \), \( |\mathbf{v}| = \sqrt{4^2 + 4^2} = \sqrt{32} = 4\sqrt{2}. \)
03

Formula for Cosine of Theta

Use the dot product and magnitudes to find \( \cos\theta \) using the formula: \( \cos\theta = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}| |\mathbf{v}|}. \) Substituting the previously computed values, we get \( \cos\theta = \frac{12}{3 \times 4\sqrt{2}} = \frac{1}{\sqrt{2}}. \)
04

Determine Theta in Radians

Use the inverse cosine function to find \( \theta \). \( \theta = \cos^{-1}\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4} \text{ radians}. \)
05

Check If Theta is Acute

An angle is acute if it is less than \(\frac{\pi}{2}\) radians. Since \( \frac{\pi}{4} \text{ radians} \) is less than \(\frac{\pi}{2}\), the angle \( \theta \) is indeed acute.

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Key Concepts

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

Dot Product
The dot product, also known as the scalar product, is a fundamental operation for two-dimensional vectors. It's primarily used to find the angle between vectors and determine if they are perpendicular (orthogonal). Given vectors \( \mathbf{u} = 3\mathbf{i} \) and \( \mathbf{v} = 4\mathbf{i} + 4\mathbf{j} \), the dot product is calculated as \( \mathbf{u} \cdot \mathbf{v} = 3 \times 4 + 0 \times 4 = 12 \). Here, the dot product is efficiently computed because \( \mathbf{u} \) lacks a \( \mathbf{j} \)-component, simplifying the process.
  • The dot product is zero if vectors are perpendicular.
  • The dot product can also reveal the similarity in direction.
Dot product results in a scalar (a single number), which can indicate how much one vector extends in the direction of another.
Magnitude of a Vector
The magnitude of a vector represents its length or size. It's crucial for calculating the angle between vectors, as it helps normalize the vectors. For the vector \( \mathbf{u} = 3\mathbf{i} \), its magnitude is calculated as \( |\mathbf{u}| = \sqrt{3^2} = 3 \). Similarly, for vector \( \mathbf{v} = 4\mathbf{i} + 4\mathbf{j} \), the magnitude is \( |\mathbf{v}| = \sqrt{4^2 + 4^2} = \sqrt{32} = 4\sqrt{2} \).
  • A vector's magnitude is always a non-negative scalar.
  • It鈥檚 essential for understanding vector properties and interactions.
Calculating the magnitude uses the Pythagorean theorem in two dimensions.
Angle Between Vectors
The angle \( \theta \) between two vectors is determined using the dot product and the magnitudes of the vectors. The formula \( \cos\theta = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}| |\mathbf{v}|} \) relates the dot product to the magnitudes. For vectors \( \mathbf{u} \) and \( \mathbf{v} \), \( \cos\theta = \frac{12}{3 \times 4\sqrt{2}} = \frac{1}{\sqrt{2}} \). By calculating the arc cosine, \( \theta = \cos^{-1}\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4} \) radians is obtained.
  • Angles can show the orientation between vectors.
  • An angle of zero denotes vectors pointing in the same direction.
Understanding vector angles aids in vector alignment analysis.
Acute Angle
An acute angle is one that measures less than \( \frac{\pi}{2} \) radians or 90 degrees. In vector calculations, determining if an angle is acute helps in understanding the spatial relationship between vectors. In our example, the angle \( \theta = \frac{\pi}{4} \) radians qualifies as acute since it is less than \( \frac{\pi}{2} \).
  • An acute angle ensures that vectors are closer to aligning directly rather than being perpendicular.
  • Acute angles can suggest similar directionality between vectors.
Identifying acute angles is a fundamental component in vector geometry.

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Most popular questions from this chapter

For the following exercises, the equation of a plane is given. a. Find normal vector \(\mathbf{n}\) to the plane. Express \(\mathbf{n}\) using standard unit vectors. b. Find the intersections of the plane with the axes of coordinates. c. Sketch the plane. \(3 x-2 y+4 z=0\)

Two soccer players are practicing for an upcoming game. One of them runs 10 \(\mathrm{m}\) from point \(A\) to point \(B\) . She then turns left at \(90^{\circ}\) and runs 10 \(\mathrm{m}\) until she reaches point C. Then she kicks the ball with a speed of 10 \(\mathrm{m} / \mathrm{sec}\) at an upward angle of \(45^{\circ}\) to her teammate, who is located at point \(A\) . Write the velocity of the ball in component form.

The following problems consider your unsuccessful attempt to take the tire off your car using a wrench to loosen the bolts. Assume the wrench is 0.3 \(\mathrm{m}\) long and you are able to apply a 200 -N force. Because your tire is flat, you are only able to apply your force at a \(60^{\circ}\) angle. What is the torque at the center of the bolt? Assume this force is not enough to loosen the bolt.

In the following exercises, points \(P\) and \(Q\) are given. Let \(L\) be the line passing through points \(P\) and \(Q .\) a. Find the vector equation of line \(L\) b. Find parametric equations of line \(L\) c. Find symmetric equations of line \(L\) d. Find parametric equations of the line segment determined by \(P\) and \(Q\) . \(P(-3,5,9), Q(4,-7,2)\)

[T]Let \(\mathbf{r}(t)=\left\langle t, 2 t^{2}, 4 t^{2}\right\rangle\) be the position vector of a particle at time \(t\) (in seconds), where \(t \in[0,10]\) (here the components of \(\mathbf{r}\) are expressed in centimeters). a. Find the instantaneous velocity, speed, and acceleration of the particle after the first two seconds. Round your answer to two decimal places. b. Use a CAS to visualize the path of the particle defined by the points \(\left(t, 2 t^{2}, 4 t^{2}\right), \quad\) where \(t \in[0,60] .\)
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