Problem 16
Question
Suppose a vector \(\left[\begin{array}{l}x_{1} \\ x_{2}\end{array}\right]\) has length 1 and is \(90^{\circ}\) counterclockwise from the negative \(x_{2}\) -axis. Find \(x_{1}\) and \(x_{2}\).
Step-by-Step Solution
Verified Answer
The vector is [1, 0], with x₁ = 1 and x₂ = 0.
1Step 1: Understand the Position of the Vector
The vector is said to be 90 degrees counterclockwise from the negative x₂-axis. This means if we start from the negative x₂-axis and move 90 degrees counterclockwise, we reach the positive x₁-axis. As a result, the vector would align with the direction of the positive x₁-axis.
2Step 2: Recognize the Property of the Vector
The vector is given to have a length of 1. Since it points along the positive x₁-axis, its component form would be [1, 0]. Here, x₁ is 1 and x₂ is 0.
3Step 3: Verify the Length of the Vector
We need to confirm that the vector with components x₁ = 1 and x₂ = 0 has a length of 1. The length of a vector \([x₁, x₂]\) is computed using the formula \(\sqrt{x_1^2 + x_2^2}\). For [1, 0], the length is \(\sqrt{1^2 + 0^2} = \sqrt{1} = 1\). Thus, the length requirement is satisfied.
Key Concepts
Length of a VectorCoordinate GeometryUnit Vector
Length of a Vector
The length of a vector, also known as its magnitude, is a crucial concept in both geometry and physics. It describes how long a vector is, regardless of its direction.
To determine the length of a vector \([x_1, x_2]\), you use the formula \(\sqrt{x_1^2 + x_2^2}\). This formula is much like the Pythagorean theorem, which helps us understand lengths within right triangles.
In most coordinate systems, a vector with a length of 1 is termed a "unit vector". Unit vectors are unique because they purely describe direction without any regard to distance.
To determine the length of a vector \([x_1, x_2]\), you use the formula \(\sqrt{x_1^2 + x_2^2}\). This formula is much like the Pythagorean theorem, which helps us understand lengths within right triangles.
In most coordinate systems, a vector with a length of 1 is termed a "unit vector". Unit vectors are unique because they purely describe direction without any regard to distance.
- Example: Given a vector \[ [3, 4] \], calculate its length: \[p\ = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\]
Coordinate Geometry
Coordinate geometry, also known as analytic geometry, provides a link between algebra and geometry.
It uses a coordinate system to evaluate geometric properties using algebraic equations, thereby making spatial concepts more intuitive.
It uses a coordinate system to evaluate geometric properties using algebraic equations, thereby making spatial concepts more intuitive.
- Coordinates describe a point's position in space. For example, \[ (x_1, x_2) \]
- With a vector \[ [x_1, x_2] \], the coordinates determine its point of origin and direction.
- This field enables us to calculate distances, slopes, and understand shapes algebraically.
Unit Vector
A unit vector is a vector with a length of exactly 1. It is often used to indicate direction, as it lacks any dimensional length.
The term can arise from scaling any non-zero vector. By dividing a vector by its magnitude, you adjust it to become a unit vector.
The term can arise from scaling any non-zero vector. By dividing a vector by its magnitude, you adjust it to become a unit vector.
- Formula to Find a Unit Vector: If you have a vector \[ [x_1, x_2] \], its unit vector is \[ \left[ \frac{x_1}{||v||}, \frac{x_2}{||v||} \right] \]where \[ ||v|| \] is the original vector's length.
- This concept is essential in physics and engineering for simplifying calculations and is also used in computer graphics to conserve computational resources.
Other exercises in this chapter
Problem 16
Find the dot product of \(\mathbf{x}=[1,2]^{\prime}\) and \(\mathbf{y}=[3,-1]^{\prime}\).
View solution Problem 16
Suppose that $$ L=\left[\begin{array}{ll} 0 & 5 \\ 0.9 & 0 \end{array}\right] $$ is the Leslie matrix for a population with two age classes. (a) Determine both
View solution Problem 16
Suppose \(A\) is a \(2 \times 2\) matrix. Find conditions on the entries of \(A\) such that $$ A-A^{\prime}=\mathbf{0} $$
View solution Problem 16
In Problems 9-16, reduce the system of equations to upper triangular form and find all the solutions. $$ \begin{array}{r} x-2 y=2 \\ 4 y-2 x=-4 \end{array} $$
View solution