/*! 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 23 Compute the dot product of the v... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Compute the dot product of the vectors \(\mathbf{u}\) and \(\mathbf{v},\) and find the angle between the vectors. \(\mathbf{u}=2 \mathbf{i}-3 \mathbf{k}\) and \(\mathbf{v}=\mathbf{i}+4 \mathbf{j}+2 \mathbf{k}\)

Short Answer

Expert verified
Answer: The dot product of the vectors \(\mathbf{u}\) and \(\mathbf{v}\) is -4, and the angle between them is given by \(\theta = \arccos{\left( \frac{-4}{\sqrt{13 \cdot 21}} \right)}\).

Step by step solution

01

Write the given vectors in component form

Represent the given vectors \(\mathbf{u}\) and \(\mathbf{v}\) in component form as follows: $$\mathbf{u} = \begin{pmatrix} 2 \\ 0 \\ -3 \end{pmatrix} \text{ and } \mathbf{v} = \begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix}$$
02

Compute the dot product of the vectors

Using the component form of the vectors, compute the dot product of \(\mathbf{u}\) and \(\mathbf{v}\) as follows: $$ \mathbf{u}\cdot\mathbf{v} = \begin{pmatrix} 2 \\ 0 \\ -3 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix} = (2 \times 1) + (0 \times 4) + (-3 \times 2) = 2 + 0 - 6 = -4 $$ So, the dot product of \(\mathbf{u}\) and \(\mathbf{v}\) is -4.
03

Find the magnitudes of the vectors

Compute the magnitudes of \(\mathbf{u}\) and \(\mathbf{v}\), using the formula \(||\mathbf{u}||=\sqrt{u_x^2 + u_y^2 + u_z^2}\) and \(||\mathbf{v}||=\sqrt{v_x^2 + v_y^2 + v_z^2}\), respectively: $$ ||\mathbf{u}|| = \sqrt{(2)^2 + (0)^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13} $$ $$ ||\mathbf{v}|| = \sqrt{(1)^2 + (4)^2 + (2)^2} = \sqrt{1 + 16 + 4} = \sqrt{21} $$ The magnitudes of the vectors are \(||\mathbf{u}||=\sqrt{13}\) and \(||\mathbf{v}||=\sqrt{21}\).
04

Find the angle between the vectors

Using the dot product formula, which states that \(\mathbf{u}\cdot\mathbf{v} = ||\mathbf{u}||\cdot ||\mathbf{v}|| \cdot \cos{\theta}\), we can compute the angle between the vectors \(\mathbf{u}\) and \(\mathbf{v}\) as follows: $$ \cos{\theta} = \frac{\mathbf{u}\cdot\mathbf{v}}{||\mathbf{u}|| \cdot ||\mathbf{v}||} = \frac{-4}{\sqrt{13} \cdot \sqrt{21}} $$ Now, take the inverse cosine (arc cosine) to find the angle between the vectors: $$ \theta = \arccos{\left( \frac{-4}{\sqrt{13 \cdot 21}} \right)} $$ So, the angle between vectors \(\mathbf{u}\) and \(\mathbf{v}\) is given by \(\theta = \arccos{\left( \frac{-4}{\sqrt{13 \cdot 21}} \right)}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Vector Magnitude
Understanding the magnitude of a vector is essential when working with vectors in mathematics. The magnitude of a vector, often noted as \(||\mathbf{v}||\), represents its length in Euclidean space. It's calculated using the vector's components: if you have a vector \(\mathbf{v} = \begin{pmatrix} v_x \ v_y \ v_z \end{pmatrix}\), the magnitude can be found using the formula:\[||\mathbf{v}|| = \sqrt{v_x^2 + v_y^2 + v_z^2}\]This formula derives from the Pythagorean theorem. It simply sums up the squares of the components and takes the square root of the result. In our example, the vectors were \(\mathbf{u} = \begin{pmatrix} 2 \ 0 \ -3 \end{pmatrix}\) and \(\mathbf{v} = \begin{pmatrix} 1 \ 4 \ 2 \end{pmatrix}\). If we calculate:
  • \(||\mathbf{u}|| = \sqrt{2^2 + 0^2 + (-3)^2} = \sqrt{13}\)
  • \(||\mathbf{v}|| = \sqrt{1^2 + 4^2 + 2^2} = \sqrt{21}\)
These magnitudes help find relationships between vectors, such as the angle between them.
Angle Between Vectors
The angle between two vectors can tell us how they are oriented relative to each other. To find this angle, we rely on the dot product and magnitudes of the vectors. The dot product is a key part of this calculation and is computed as follows:If you have vectors \(\mathbf{u} = \begin{pmatrix} u_x \ u_y \ u_z \end{pmatrix}\) and \(\mathbf{v} = \begin{pmatrix} v_x \ v_y \ v_z \end{pmatrix}\), their dot product is:\[\mathbf{u} \cdot \mathbf{v} = u_x v_x + u_y v_y + u_z v_z\]Once the dot product is known, the angle \(\theta\) between two vectors can be found. Use the relationship:\[\cos{\theta} = \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{u}|| \cdot ||\mathbf{v}||}\]In this exercise, the dot product is \(-4\). Using the magnitudes found earlier, we insert these values into the formula to solve for \(\cos{\theta}\). This makes it easier to find \(\theta\) by taking the inverse cosine in the final step.
Inverse Cosine
Inverse cosine, also known as arc cosine, helps us find the actual angle when we know the cosine value. It's often represented as \(\arccos\). Given the cosine of an angle \(\cos{\theta}\), the inverse cosine operation returns the angle \(\theta\) itself. For vectors \(\mathbf{u}\) and \(\mathbf{v}\), having calculated \(\cos{\theta} = \frac{-4}{\sqrt{13 \cdot 21}}\), the next step is to determine the angle \(\theta\) using:\[\theta = \arccos{\left( \frac{-4}{\sqrt{13 \cdot 21}} \right)}\]To perform this operation, many use calculators or software that handle trigonometric functions. It gives the angle in radians or degrees, connecting the geometric interpretation of vectors to real-world dimensions. Understanding inverse cosine allows students to interpret angular relationships accurately within vector mathematics.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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.

The definition \(\mathbf{u} \cdot \mathbf{v}=|\mathbf{u}||\mathbf{v}| \cos \theta\) implies that \(|\mathbf{u} \cdot \mathbf{v}| \leq|\mathbf{u}||\mathbf{v}|(\text {because}|\cos \theta| \leq 1) .\) This inequality, known as the Cauchy-Schwarz Inequality, holds in any number of dimensions and has many consequences. Show that for real numbers \(u_{1}, u_{2},\) and \(u_{3},\) it is true that \(\left(u_{1}+u_{2}+u_{3}\right)^{2} \leq 3\left(u_{1}^{2}+u_{2}^{2}+u_{3}^{2}\right)\). (Hint: Use the Cauchy-Schwarz Inequality in three dimensions with \(\mathbf{u}=\left\langle u_{1}, u_{2}, u_{3}\right\rangle\) and choose \(\mathbf{v}\) in the right way.)

Carry out the following steps to determine the (smallest) distance between the point \(P\) and the line \(\ell\) through the origin. a. Find any vector \(\mathbf{v}\) in the direction of \(\ell\) b. Find the position vector u corresponding to \(P\). c. Find \(\operatorname{proj}_{\mathbf{v}} \mathbf{u}\). d. Show that \(\mathbf{w}=\mathbf{u}-\) projy \(\mathbf{u}\) is a vector orthogonal to \(\mathbf{v}\) whose length is the distance between \(P\) and the line \(\ell\) e. Find \(\mathbf{w}\) and \(|\mathbf{w}| .\) Explain why \(|\mathbf{w}|\) is the distance between \(P\) and \(\ell\). \(P(1,1,-1) ; \ell\) has the direction of $$\langle-6,8,3\rangle$$.

Consider the ellipse \(\mathbf{r}(t)=\langle a \cos t, b \sin t\rangle\) for \(0 \leq t \leq 2 \pi,\) where \(a\) and \(b\) are real numbers. Let \(\theta\) be the angle between the position vector and the \(x\) -axis. a. Show that \(\tan \theta=(b / a) \tan t\) b. Find \(\theta^{\prime}(t)\) c. Note that the area bounded by the polar curve \(r=f(\theta)\) on the interval \([0, \theta]\) is \(A(\theta)=\frac{1}{2} \int_{0}^{\theta}(f(u))^{2} d u\) Letting \(f(\theta(t))=|\mathbf{r}(\theta(t))|,\) show that \(A^{\prime}(t)=\frac{1}{2} a b\) d. Conclude that as an object moves around the ellipse, it sweeps out equal areas in equal times.

Alternative derivation of the curvature Derive the computational formula for curvature using the following steps. a. Use the tangential and normal components of the acceleration to show that \(\left.\mathbf{v} \times \mathbf{a}=\kappa|\mathbf{v}|^{3} \mathbf{B} . \text { (Note that } \mathbf{T} \times \mathbf{T}=\mathbf{0} .\right)\) b. Solve the equation in part (a) for \(\kappa\) and conclude that \(\kappa=\frac{|\mathbf{v} \times \mathbf{a}|}{\left|\mathbf{v}^{3}\right|},\) as shown in the text.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.