/*! 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 193 Determine the real number \(\alp... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Determine the real number \(\alpha\) such that \(\mathbf{u} \times \mathbf{v}\) and \(\mathbf{i}\) are orthogonal, where \(\quad \mathbf{u}=3 \mathbf{i}+\mathbf{j}-5 \mathbf{k} \quad\) and \(\mathbf{v}=4 \mathbf{i}-2 \mathbf{j}+\alpha \mathbf{k}.\)

Short Answer

Expert verified
\(\alpha = 10\).

Step by step solution

01

Compute the Cross Product

First, we must compute the cross product \( \mathbf{u} \times \mathbf{v} \). The vectors \( \mathbf{u} \) and \( \mathbf{v} \) are given by:\[ \mathbf{u} = 3 \mathbf{i} + \mathbf{j} - 5 \mathbf{k} \]\[ \mathbf{v} = 4 \mathbf{i} - 2 \mathbf{j} + \alpha \mathbf{k} \]The cross product \( \mathbf{u} \times \mathbf{v} \) can be computed using the determinant method:\[ \mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 3 & 1 & -5 \ 4 & -2 & \alpha \end{vmatrix} \]This expands to:\[ \mathbf{u} \times \mathbf{v} = \left( (1)(\alpha) - (-5)(-2) \right) \mathbf{i} - \left( (3)(\alpha) - (-5)(4) \right) \mathbf{j} + \left( (3)(-2) - (1)(4) \right) \mathbf{k} \]
02

Simplify the Determinant

Now, simplify the computation for each component:- The \( \mathbf{i} \) component: \( \alpha - 10 \)- The \( \mathbf{j} \) component: \( - (3\alpha - 20) = 20 - 3\alpha \)- The \( \mathbf{k} \) component: \( -6 - 4 = -10 \)Thus, \( \mathbf{u} \times \mathbf{v} = (\alpha - 10)\mathbf{i} + (20 - 3\alpha)\mathbf{j} - 10\mathbf{k} \).
03

Determining Orthogonality

The condition for orthogonality with \( \mathbf{i} \) is that the dot product of \( \mathbf{u} \times \mathbf{v} \) and \( \mathbf{i} \) is zero:\[ (\alpha - 10)(1) + (20 - 3\alpha)(0) + (-10)(0) = 0 \]Simplifying this, we have:\[ \alpha - 10 = 0 \]
04

Solve for \(\alpha\)

From \( \alpha - 10 = 0 \), solve for \( \alpha \) by isolating it on one side:\[ \alpha = 10 \]

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.

Orthogonal Vectors
Orthogonal vectors are at the heart of many fundamental concepts in vector algebra. When two vectors are orthogonal, they are perpendicular to each other, forming a 90-degree angle. This means the dot product of the two vectors equals zero. Orthogonal vectors are important because they simplify calculations and are used in various scientific and engineering fields.
For example, in the original exercise, the cross product of vectors \( \mathbf{u} \) and \( \mathbf{v} \) is orthogonal to vector \( \mathbf{i} \). This orthogonality provides a condition where the dot product of our resulting vector with \( \mathbf{i} \) is zero. Thus, orthogonal vectors make it easier to solve problems involving directions and magnitudes in a 3D space.
Determinant Method
The determinant method is a reliable way to compute the cross product of two vectors. You arrange the components of the vectors in a 3x3 matrix format, with the unit vectors \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \) forming the first row.
The original exercise shows this as\[\mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 3 & 1 & -5 \ 4 & -2 & \alpha \end{vmatrix}\]
To find the cross product, you calculate the determinant to derive each component of the resulting vector. Expanding the determinant involves calculating products and subtractions specific to each component. This method is powerful because it systematically handles the complexity of cross product calculations.
Dot Product
A dot product, also known as a scalar product, is an algebraic operation that takes two equal-length sequences of numbers and returns a single number. This operation is integral when determining orthogonality between vectors.
In the exercise, the dot product of the cross product of vectors \( \mathbf{u} \) and \( \mathbf{v} \) with \( \mathbf{i} \) is set to zero to check for orthogonality. Specifically, the calculation is:\[ (\alpha - 10)(1) + (20 - 3\alpha)(0) + (-10)(0) = 0 \]
The key point in a dot product, like in this exercise, is that if the result is zero between two vectors, those vectors are orthogonal.
Vector Calculation
Vector calculation involves manipulating vectors using various operations to solve problems, such as finding magnitudes, directions, or resultant vectors. In our exercise, several vector operations come into play.
- **Cross Product**: This is used to find a vector perpendicular to two given vectors.
- **Dot Product**: This helps verify orthogonality between vectors.
Overall, vector calculation is an essential tool in physics, engineering, and graphics, allowing for the precise control of direction and magnitude within a multidimensional space. Understanding how to perform these calculations accurately leads to effective problem-solving.

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

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. Someone lends you a tire jack and you are now able to apply a 200 -N force at an \(80^{\circ}\) angle. Is your resulting torque going to be more or less? What is the new resulting torque at the center of the bolt? Assume this force is not enough to loosen the bolt.

For the following exercises, convert the given equations from cylindrical or spherical coordinates to rectangular coordinates. Identify the given surface. $$\rho^{2}\left(\sin ^{2}(\varphi)-\cos ^{2}(\varphi)\right)=1$$

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. \(x+z=0\)

For the following exercises, determine whether the statement is true or false. Justify the answer with a proof or a counterexample. For vectors a and \(\mathbf{b}\) and any given scalar \(c,\) \(c(\mathbf{a} \times \mathbf{b})=(c \mathbf{a}) \times \mathbf{b}\).

For the following exercises, point \(P\) and vector \(\mathbf{n}\) are given. a. Find the scalar equation of the plane that passes through \(P\) and has normal vector \(\mathbf{n} .\) b. Find the general form of the equation of the plane that passes through \(P\) and has normal vector \(\mathbf{n} .\) \(P(1,2,3), \quad \mathbf{n}=\langle 1,2,3\rangle\)

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.