Chapter 6: Problem 6
Compute the norm and the direction cosines for the vector \(x=\left[\begin{array}{l}4 \\ 2 \\ 6\end{array}\right]\).
Short Answer
Expert verified
The norm is \ 2\sqrt{14}\ and the direction cosines are \frac{\sqrt{14}}{7} , \frac{1}{\sqrt{14}}, \ and \frac{3}{\sqrt{14}}.\
Step by step solution
01
- Define the vector
The vector given is \(x = \begin{bmatrix} 4 \ 2 \ 6 \ \end{bmatrix}\). This is a vector in 3-dimensional space.
02
- Compute the norm of the vector
The norm (or magnitude) of a vector \ \begin{bmatrix} x \ y \ z \end{bmatrix}\ is calculated using the formula: \( \text{{norm}} = \sqrt{x^2 + y^2 + z^2} \).Substitute the given values: \( \text{{norm}} = \sqrt{4^2 + 2^2 + 6^2} = \sqrt{16 + 4 + 36} = \sqrt{56} = 2\sqrt{14} \). Thus, the norm is \2\sqrt{14}\.
03
- Compute the direction cosines
The direction cosines are the cosines of the angles that the vector makes with the coordinate axes. These are calculated by dividing each component of the vector by the norm.For the given vector \ x = \begin{bmatrix} 4 \ 2 \ 6 \ \end{bmatrix}\ and the norm \ 2\sqrt{14}:\[ \cos(\alpha) = \frac{4}{2\sqrt{14}} = \frac{2}{\sqrt{14}} = \frac{\sqrt{14}}{7}\ \]\[ \cos(\beta) = \frac{2}{2\sqrt{14}} = \frac{1}{\sqrt{14}}\ \]\[ \cos(\gamma) = \frac{6}{2\sqrt{14}} = \frac{3}{\sqrt{14}}\ \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vector Norm
In mathematics, the norm of a vector is a measure of its length (or magnitude) in a vector space. For a vector in 3-dimensional space, the norm is computed using the formula: \[ \text{norm} = \sqrt{x^2 + y^2 + z^2} \].
This formula originates from the Pythagorean theorem, which is used to determine the length of the hypotenuse in a right triangle. When applied to a vector, it sums the squares of its components and finds the square root of this sum.
If you have a vector such as \[\begin{bmatrix} 4 \ 2 \ 6 \end{bmatrix}\], you would calculate its norm as follows:
This formula originates from the Pythagorean theorem, which is used to determine the length of the hypotenuse in a right triangle. When applied to a vector, it sums the squares of its components and finds the square root of this sum.
If you have a vector such as \[\begin{bmatrix} 4 \ 2 \ 6 \end{bmatrix}\], you would calculate its norm as follows:
- Square each component: <\(4^2, 2^2, 6^2\)
- Add these squares: \(16 + 4 + 36 = 56\)
- Take the square root of the sum: \[ \sqrt{56} = 2\sqrt{14} \]
Direction Cosines
Direction cosines are the cosines of the angles that a vector makes with each of the coordinate axes in 3-dimensional space. They provide a way to understand the orientation of the vector relative to the standard coordinate system.
To find the direction cosines of a vector, you need to normalize the vector by dividing each of its components by its norm. The formulas for direction cosines are:
Let's use the given vector \[\begin{bmatrix} 4 \ 2 \ 6 \end{bmatrix} \] and its norm \[ 2\sqrt{14} \]. Substitute these values into the formulas:
To find the direction cosines of a vector, you need to normalize the vector by dividing each of its components by its norm. The formulas for direction cosines are:
- \[ \cos(\alpha) = \frac{x}{\text{norm}} \]
- \[ \cos(\beta) = \frac{y}{\text{norm}} \]
- \[ \cos(\gamma) = \frac{z}{\text{norm}} \]
Let's use the given vector \[\begin{bmatrix} 4 \ 2 \ 6 \end{bmatrix} \] and its norm \[ 2\sqrt{14} \]. Substitute these values into the formulas:
- \[ \cos(\alpha) = \frac{4}{2\sqrt{14}} = \frac{2}{\sqrt{14}} = \frac{\sqrt{14}}{7} \]
- \[ \cos(\beta) = \frac{2}{2\sqrt{14}} = \frac{1}{\sqrt{14}} \]
- \[ \cos(\gamma) = \frac{6}{2\sqrt{14}} = \frac{3}{\sqrt{14}} \]
Magnitude of a Vector
The magnitude of a vector, often referred to as its length or size, indicates how large the vector is. In a 3-dimensional space, the magnitude is also known as the vector norm.
The formula for the magnitude of a vector is: \[ \text{magnitude} = \sqrt{x^2 + y^2 + z^2} \]. This is the same as the formula for the norm.
For example, taking the vector \[ \begin{bmatrix} 4 \ 2 \ 6 \end{bmatrix} \], we can compute its magnitude similarly:
The magnitude of the vector \[ \begin{bmatrix} 4 \ 2 \ 6 \end{bmatrix} \] is \[ 2\sqrt{14} \].
The formula for the magnitude of a vector is: \[ \text{magnitude} = \sqrt{x^2 + y^2 + z^2} \]. This is the same as the formula for the norm.
For example, taking the vector \[ \begin{bmatrix} 4 \ 2 \ 6 \end{bmatrix} \], we can compute its magnitude similarly:
- Square each component: \[ 4^2 = 16, 2^2 = 4, 6^2 = 36 \]
- Add the squares: \[ 16 + 4 + 36 = 56 \]
- Take the square root of the sum: \[ \sqrt{56} = 2\sqrt{14} \]
The magnitude of the vector \[ \begin{bmatrix} 4 \ 2 \ 6 \end{bmatrix} \] is \[ 2\sqrt{14} \].
3-Dimensional Space
3-dimensional space is the setting in which we live and experience the physical world. It has three dimensions: length, width, and height. This space can be described using coordinate axes: x, y, and z, which extend infinitely in all directions.
Vectors in 3-dimensional space can represent a variety of quantities, including forces, velocities, and displacements. Each vector has three components corresponding to the three axes.
For instance, a vector \[\begin{bmatrix} 4 \ 2 \ 6 \end{bmatrix}\] means:
Visualize such vectors as arrows pointing from the origin \[ \begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix} \] to the point \[ \begin{bmatrix} 4 \ 2 \ 6 \end{bmatrix} \] in 3D space. This concept is essential for understanding spatial relationships and is fundamental in physics, engineering, and several other fields.
Vectors in 3-dimensional space can represent a variety of quantities, including forces, velocities, and displacements. Each vector has three components corresponding to the three axes.
For instance, a vector \[\begin{bmatrix} 4 \ 2 \ 6 \end{bmatrix}\] means:
- 4 units in the x-direction
- 2 units in the y-direction
- 6 units in the z-direction
Visualize such vectors as arrows pointing from the origin \[ \begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix} \] to the point \[ \begin{bmatrix} 4 \ 2 \ 6 \end{bmatrix} \] in 3D space. This concept is essential for understanding spatial relationships and is fundamental in physics, engineering, and several other fields.