Problem 8
Question
Represent each given vector \(\mathrm{x}=\left[\begin{array}{l}x_{1} \\\ x_{2}\end{array}\right]\) in the \(x_{1}-x_{2}\) plane, and determine its length and the angle that it forms with the positive \(x_{1}\) -axis (measured counterclockwise). $$\mathbf{x}=\left[\begin{array}{r}1 \\ -\sqrt{3}\end{array}\right]$$
Step-by-Step Solution
Verified Answer
The vector length is 2, and it forms a \(300^\circ\) angle with the positive \(x_1\)-axis.
1Step 1: Understanding the Vector Components
The given vector \( \mathbf{x} = \begin{bmatrix} 1 \ -\sqrt{3} \end{bmatrix} \) has components \( x_1 = 1 \) and \( x_2 = -\sqrt{3} \). In an \( x_1-x_2 \) plane, this corresponds to the point (1, -\sqrt{3}).
2Step 2: Calculating the Length of the Vector
The length (or magnitude) of a vector \( \mathbf{x} = \begin{bmatrix} x_1 \ x_2 \end{bmatrix} \) is calculated using the formula \( \| \mathbf{x} \| = \sqrt{x_1^2 + x_2^2} \). For our vector, this becomes \( \| \mathbf{x} \| = \sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \).
3Step 3: Finding the Angle with the Positive x1-axis
The angle \( \theta \) that the vector forms with the positive \( x_1 \)-axis is determined using the arctangent function: \( \theta = \tan^{-1}\left(\frac{x_2}{x_1}\right) \). Here, \( \theta = \tan^{-1}\left(\frac{-\sqrt{3}}{1}\right) = \tan^{-1}(-\sqrt{3}) \).
4Step 4: Adjusting for the Correct Quadrant
Since \( x_1 = 1\) and \( x_2 = -\sqrt{3} \), our vector lies in the fourth quadrant of the coordinate plane. The principal value of \( \tan^{-1}(-\sqrt{3}) \) is \( -\frac{\pi}{3} \) radians (or \(-60^\circ\)). To represent the angle counterclockwise from the positive \( x_1 \)-axis, convert this to a positive angle in the standard range, \( 2\pi - \frac{\pi}{3} = \frac{5\pi}{3} \) radians (or \(300^\circ\)).
Key Concepts
Vector ComponentsLength of a VectorAngle with Axis
Vector Components
In the world of vectors, understanding the components is essential. A vector in a 2D space is typically expressed with two components, which denote its position along the x and y axes in a coordinate plane. For the vector \( \mathbf{x} = \begin{bmatrix} 1 \ -\sqrt{3} \end{bmatrix} \), the components are \( x_1 = 1 \) and \( x_2 = -\sqrt{3} \).
These components tell us that the vector points 1 unit in the positive direction along the x-axis and \(-\sqrt{3}\) units in the negative direction along the y-axis.
This can be visualized as the point (1, -\sqrt{3}) in the \( x_1-x_2 \) plane, helping us see how the vector behaves in space.
Understanding this layout is crucial in interpreting how vectors influence directions and positions in the plane.
These components tell us that the vector points 1 unit in the positive direction along the x-axis and \(-\sqrt{3}\) units in the negative direction along the y-axis.
This can be visualized as the point (1, -\sqrt{3}) in the \( x_1-x_2 \) plane, helping us see how the vector behaves in space.
Understanding this layout is crucial in interpreting how vectors influence directions and positions in the plane.
Length of a Vector
To measure how long a vector is extends in space, we calculate its length, often called the magnitude. It gives us a quantitative value representing its size. For a vector \( \mathbf{x} = \begin{bmatrix} x_1 \ x_2 \end{bmatrix} \), the length is found using the equation:
\[ \| \mathbf{x} \| = \sqrt{x_1^2 + x_2^2} \] Applying this to our vector \( \mathbf{x} = \begin{bmatrix} 1 \ -\sqrt{3} \end{bmatrix} \), we calculate:
\[ \| \mathbf{x} \| = \sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \]
This result, 2, indicates our vector stretches 2 units from the origin to its point in space.
This concept is fundamental in determining the extent of vectors in both physics and engineering.
\[ \| \mathbf{x} \| = \sqrt{x_1^2 + x_2^2} \] Applying this to our vector \( \mathbf{x} = \begin{bmatrix} 1 \ -\sqrt{3} \end{bmatrix} \), we calculate:
\[ \| \mathbf{x} \| = \sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \]
This result, 2, indicates our vector stretches 2 units from the origin to its point in space.
This concept is fundamental in determining the extent of vectors in both physics and engineering.
Angle with Axis
The angle a vector makes with the x-axis is a crucial piece of information, especially when analyzing direction or performing rotational transformations.
To find this angle for the vector \( \mathbf{x} \), use the arctangent of the ratio of the y-component to the x-component:
\[ \theta = \tan^{-1}\left(\frac{x_2}{x_1}\right) \] In our example,
\( \theta = \tan^{-1}\left(\frac{-\sqrt{3}}{1}\right) = \tan^{-1}(-\sqrt{3}) \).
The principal value here is \(-\frac{\pi}{3}\) radians (or \(-60^\circ\)). Since this places the vector in the fourth quadrant, where angles are measured clockwise from the positive x-axis, we convert this to a positive counterclockwise standard angle:
\( 2\pi - \frac{\pi}{3} = \frac{5\pi}{3} \) radians (or \(300^\circ\)).
This conversion helps us follow the conventional direction, ensuring accurate representation in contexts like physics and navigation.
To find this angle for the vector \( \mathbf{x} \), use the arctangent of the ratio of the y-component to the x-component:
\[ \theta = \tan^{-1}\left(\frac{x_2}{x_1}\right) \] In our example,
\( \theta = \tan^{-1}\left(\frac{-\sqrt{3}}{1}\right) = \tan^{-1}(-\sqrt{3}) \).
The principal value here is \(-\frac{\pi}{3}\) radians (or \(-60^\circ\)). Since this places the vector in the fourth quadrant, where angles are measured clockwise from the positive x-axis, we convert this to a positive counterclockwise standard angle:
\( 2\pi - \frac{\pi}{3} = \frac{5\pi}{3} \) radians (or \(300^\circ\)).
This conversion helps us follow the conventional direction, ensuring accurate representation in contexts like physics and navigation.
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