Problem 36 Find a unit vector orthogonal to... [FREE SOLUTION]

Chapter 10: Problem 36

Find a unit vector orthogonal to both \(\vec{u}\) and \(\vec{v} .\) \(\vec{u}=\langle 1,-2,1\rangle, \quad \vec{v}=\langle 3,2,1\rangle\)

Short Answer

Expert verified

The unit vector orthogonal to both is \( \left\langle \frac{-2}{\sqrt{21}}, \frac{1}{\sqrt{21}}, \frac{4}{\sqrt{21}} \right\rangle \).

Step by step solution

Understand Orthogonal Vectors

Two vectors are orthogonal if their dot product is zero. To find a vector orthogonal to both given vectors, we will use the cross product.

Calculate the Cross Product

The cross product of vectors \( \vec{u} = \langle 1, -2, 1 \rangle \) and \( \vec{v} = \langle 3, 2, 1 \rangle \) is calculated using the determinant of a matrix:\[ \vec{u} \times \vec{v} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \ 1 & -2 & 1 \ 3 & 2 & 1 \end{vmatrix} = \vec{i}(( -2) \cdot 1 - 1 \cdot 2) - \vec{j}(1 \cdot 1 - 1 \cdot 3) + \vec{k}(1 \cdot 2 - (-2) \cdot 3) \]Simplifying each component:\[ = \vec{i}(-2 - 2) - \vec{j}(1 - 3) + \vec{k}(2 + 6) \]\[ = \vec{i}(-4) + \vec{j}(2) + \vec{k}(8) \]The cross product is \( \vec{w} = \langle -4, 2, 8 \rangle \).

Find the Magnitude of the Cross Product

The magnitude of \( \vec{w} = \langle -4, 2, 8 \rangle \) is calculated using the formula:\[ |\vec{w}| = \sqrt{(-4)^2 + 2^2 + 8^2} \]\[ = \sqrt{16 + 4 + 64} \]\[ = \sqrt{84} \]\[ = 2\sqrt{21} \]

Calculate the Unit Vector

A unit vector \( \vec{u} \) is found by dividing \( \vec{w} \) by its magnitude:\[ \vec{u}_{unit} = \frac{1}{|\vec{w}|} \times \langle -4, 2, 8 \rangle \]\[ = \frac{1}{2\sqrt{21}} \times \langle -4, 2, 8 \rangle \]\[ = \left\langle \frac{-4}{2\sqrt{21}}, \frac{2}{2\sqrt{21}}, \frac{8}{2\sqrt{21}} \right\rangle \]The unit vector is:\[ \left\langle \frac{-2}{\sqrt{21}}, \frac{1}{\sqrt{21}}, \frac{4}{\sqrt{21}} \right\rangle \]

Simplify the Unit Vector (if necessary)

The unit vector is already in its simplest form using the factors of \( \sqrt{21} \). We can also write each component in terms of decimal approximation for clarity, but the fractions provide an exact solution.

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 91影视!

Key Concepts

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

Unit Vector

A **unit vector** is a vector that has a magnitude of exactly one. It is often used to indicate direction rather than magnitude and is denoted by the vector symbol with a hat, such as \( \hat{u} \). To create a unit vector from any vector, we divide each component of the vector by the vector's magnitude.

For example, if you have a vector \( \vec{a} = \langle a_1, a_2, a_3 \rangle \), the unit vector \( \hat{a} \) would be computed as \( \hat{a} = \frac{1}{|\vec{a}|} \times \langle a_1, a_2, a_3 \rangle \).
This transformation ensures that the resulting unit vector maintains the same direction as the original vector, but with a length of 1.
Unit vectors are particularly useful in physics and engineering to describe a direction vector independent of its magnitude.

Understanding how to compute and apply unit vectors allows for the simplification and analysis of directional properties in vector calculus without changing the inherent direction of the vectors involved.

Cross Product

The **cross product** is a binary operation on two vectors in three-dimensional space. Unlike the dot product, the result of a cross product is a vector rather than a scalar. The cross product vector is orthogonal (perpendicular) to both of the original vectors.

For vectors \( \vec{u} = \langle u_1, u_2, u_3 \rangle \) and \( \vec{v} = \langle v_1, v_2, v_3 \rangle \), the cross product is given by:
\[ \vec{u} \times \vec{v} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \ u_1 & u_2 & u_3 \ v_1 & v_2 & v_3 \end{vmatrix} \]
The result is: \( (u_2v_3 - u_3v_2)\vec{i} - (u_1v_3 - u_3v_1)\vec{j} + (u_1v_2 - u_2v_1)\vec{k} \), which is a new vector perpendicular to the original pair.
One important use of the cross product is to find a normal vector to a plane defined by the two vectors.

The cross product can also determine the area of a parallelogram defined by the vectors.

Magnitude of a Vector

The **magnitude of a vector** represents the length or size of the vector, which is akin to the distance from the origin point to the vector's tip in Euclidean space. It is always a non-negative number.

For a vector \( \vec{v} = \langle v_1, v_2, v_3 \rangle \), the magnitude \( |\vec{v}| \) is calculated by the formula:
\[ |\vec{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2} \]
This formula is derived from the Pythagorean theorem and measures the vector's length in the three-dimensional space.
Understanding the magnitude is crucial because it provides insight into how large the vector is relative to other vectors.

Vector magnitude is foundational in determining unit vectors, comparing vectors, and applying to real-world problems where vector lengths signify quantities such as force or velocity.

91影视

Find a unit vector orthogonal to both \(\vec{u}\) and \(\vec{v} .\) \(\vec{u}=\langle 1,-2,1\rangle, \quad \vec{v}=\langle 3,2,1\rangle\)

Short Answer

Step by step solution

Understand Orthogonal Vectors

Calculate the Cross Product

Find the Magnitude of the Cross Product

Calculate the Unit Vector

Simplify the Unit Vector (if necessary)

Key Concepts

Unit Vector

Cross Product

Magnitude of a Vector

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Statistics

Applied Mathematics

Probability and Statistics

Discrete Mathematics

Calculus

Theoretical and Mathematical Physics

Study anywhere. Anytime. Across all devices.