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Chapter 10: Problem 37

Find a unit vector orthogonal to both \(\vec{u}\) and \(\vec{v} .\) \(\vec{u}=\langle 5,0,2\rangle, \quad \vec{v}=\langle-3,0,7\rangle\)

Short Answer

Expert verified
The unit vector orthogonal to both \(\vec{u}\) and \(\vec{v}\) is \(\langle 0, -1, 0 \rangle\).

Step by step solution

01

Calculate the Cross Product

To find a vector orthogonal to both \( \vec{u} \) and \( \vec{v} \), find the cross product \( \vec{u} \times \vec{v} \). Use the formula: \(\vec{u} \times \vec{v} = \langle u_2v_3 - u_3v_2, u_3v_1 - u_1v_3, u_1v_2 - u_2v_1 \rangle\)Substitute the given vectors:\[ \vec{u} \times \vec{v} = \langle 0\cdot7 - 2\cdot0, 2\cdot(-3) - 5\cdot7, 5\cdot0 - 0\cdot(-3) \rangle = \langle 0, -41, 0 \rangle \]
02

Find the Magnitude of the Cross Product

Calculate the magnitude of the cross product vector \( \vec{u} \times \vec{v} = \langle 0, -41, 0 \rangle \) using the formula:\[ ||\vec{u} \times \vec{v}|| = \sqrt{0^2 + (-41)^2 + 0^2} = \sqrt{1681} = 41 \]
03

Normalize the Cross Product

To find the unit vector, divide the cross product by its magnitude. Normalize the vector \( \vec{u} \times \vec{v} = \langle 0, -41, 0 \rangle \):\[ \text{Unit vector} = \frac{1}{41} \langle 0, -41, 0 \rangle = \langle 0, -1, 0 \rangle \]

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Key Concepts

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

Vectors
In mathematics, vectors are fundamental elements used to describe quantities that have both a magnitude and direction. Many real-world concepts like force, velocity, and acceleration are represented as vectors. A vector is often denoted using arrow notation or brackets, such as \( \vec{u} = \langle a, b, c \rangle \) in three dimensions, representing a vector with components along the x, y, and z axes.
  • The magnitude of a vector indicates the size or "length" of the vector, calculated using the Pythagorean theorem.
  • The direction of a vector is shown by the orientation of the arrow or the sign of its components.
Vectors can be easily manipulated using operations like addition, subtraction, and multiplication. They are crucial in physics and engineering as they simplify the representation of physical quantities.
Orthogonality
Orthogonality is a key concept often used when discussing vectors, especially in the realm of linear algebra and geometry. Two vectors are orthogonal if they meet at a right angle, meaning their dot product is zero.
Consider vectors \( \vec{a} \) and \( \vec{b} \). These vectors are orthogonal if \( \vec{a} \cdot \vec{b} = 0 \). However, finding a vector specifically orthogonal to two other vectors involves calculating a cross product.
  • Cross product results in a vector that is perpendicular to the plane formed by the initial two vectors.
  • This is particularly useful in physics where determining a perpendicular vector aids in describing torque or rotational motion.
Understanding orthogonality aids in grasping more advanced concepts such as vector spaces and eigenvalues.
Unit Vector
A unit vector is a vector with a magnitude of exactly one unit and serves to indicate direction without affecting magnitude.
To convert any given vector into a unit vector, you "normalize" it, which means dividing the original vector by its own magnitude. For instance, given a vector \( \vec{v} = \langle x, y, z \rangle \), the unit vector \( \vec{u} \) is calculated as:
\[ \vec{u} = \frac{\vec{v}}{||\vec{v}||} \]
  • Unit vectors are crucial in vector algebra as they allow the representation of directional information purely.
  • They provide a standardized way to indicate direction, which is foundational in defining axes in coordinate systems like i, j, and k.
Magnitude of a Vector
The magnitude of a vector determines its length and is calculated by taking the square root of the sum of the squares of its components. It is always a non-negative value. If you have a vector \( \vec{v} = \langle a, b, c \rangle \), the magnitude is found using:
\[ ||\vec{v}|| = \sqrt{a^2 + b^2 + c^2} \]
  • The magnitude function maps a vector to a scalar value reflecting how much of the vector is present, akin to vector length.
  • Understanding the magnitude is vital for scaling vectors and transforming them into unit vectors.
Knowing how to compute the magnitude helps in various applications, including normalizing vectors and solving physics problems involving distance and forces.

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Most popular questions from this chapter

Give the equation of the described plane in standard and general forms. Contains the intersecting lines \(\vec{\ell}_{1}(t)=\langle 5,0,3\rangle+t\langle-1,1,1\rangle\) and $$ \vec{\ell}_{2}(t)=\langle 1,4,7\rangle+t\langle 3,0,-3\rangle $$

Write the vector, parametric and symmetric equations of the lines described. Passes through the point of intersection of \(\vec{\ell}_{1}(t)\) and \(\vec{\ell}_{2}(t)\) and orthogonal to both lines, where \(\vec{\ell}_{1}(t)=\langle 2,1,1\rangle+t\langle 5,1,-2\rangle\) and \(\overrightarrow{\ell_{2}}(t)=\langle-2,-1,2\rangle+t\langle 3,1,-1\rangle\)

To find an equation of a line, what two pieces of information are needed?

Find the distance from the point to the line. \(Q=(2,5,6), \quad \vec{\ell}(t)=\langle-1,1,1\rangle+t\langle 1,0,1\rangle\)

Find the point of intersection between the line and the plane. line: \(\langle 1,2,3\rangle+t\langle 3,5,-1\rangle\) plane: \(3 x-2 y-z=4\)
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