Question

A vector \(\vec{x}\) and a scalar \(a\) are given. Using Definition 32, compute the lengths of \(\vec{x}\) and \(a \vec{x},\) then compare these lengths. $$ \vec{x}=\left[\begin{array}{c} 1 \\ 2 \\ -2 \end{array}\right], a=3 $$

Short answer

The length of \(\vec{x}\) is 3, and the length of \(3\vec{x}\) is 9, which is 3 times the length of \(\vec{x}\).

Step-by-step solution

Calculating the Length of Vector \(\vec{x}\)

The length (or norm) of a vector \(\vec{x}\) is given by \(\|\vec{x}\| = \sqrt{x_1^2 + x_2^2 + x_3^2}\). For \(\vec{x} = \begin{bmatrix} 1 \ 2 \ -2 \end{bmatrix}\), calculate \(\|\vec{x}\|\):\[\|\vec{x}\| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3.\]

Scaling the Vector \(\vec{x}\) by Scalar \(a\)

To compute \(a\vec{x}\), multiply each component of \(\vec{x}\) by \(a = 3\):\[a\vec{x} = 3\begin{bmatrix} 1 \ 2 \ -2 \end{bmatrix} = \begin{bmatrix} 3 \ 6 \ -6 \end{bmatrix}.\]

Calculating the Length of the Scaled Vector \(a\vec{x}\)

Now, calculate the norm of the vector \(a\vec{x}\), which is \(\|a\vec{x}\| = \sqrt{(3)^2 + (6)^2 + (-6)^2}\):\[\|a\vec{x}\| = \sqrt{9 + 36 + 36} = \sqrt{81} = 9.\]

Comparing the Lengths of \(\vec{x}\) and \(a\vec{x}\)

Notice that the length of \(a\vec{x}\) is \(9\), which is \(3\) times the length of \(\vec{x}\), which is \(3\). This confirms that \(\|a\vec{x}\| = |a|\cdot\|\vec{x}\|\). In this case, \(\|3\vec{x}\| = 3 \times 3 = 9\).

Key concepts

Scalar Multiplication

Scalar multiplication involves multiplying a vector by a scalar, which is a single number. If you have a vector \( \vec{x} = \begin{bmatrix} x_1 \ x_2 \ x_3 \end{bmatrix} \) and a scalar \( a \), the result of scalar multiplication \( a\vec{x} \) is a new vector where each component of \( \vec{x} \) is multiplied by \( a \). For example, given \( \vec{x} = \begin{bmatrix} 1 \ 2 \ -2 \end{bmatrix} \) and \( a = 3 \), the scalar multiplication is:

• \( a\vec{x} = 3 \begin{bmatrix} 1 \ 2 \ -2 \end{bmatrix} = \begin{bmatrix} 3 \ 6 \ -6 \end{bmatrix} \)

Scalar multiplication scales the vector's length but does not change its direction unless you multiply by a negative number, which reverses the direction. Each element is individually scaled by the scalar, affecting both the magnitude and the orientation in the case of negative scalars.

Vector Norm

The vector norm, often known as the vector length, is a measure of a vector's magnitude. Computing the norm of a vector \( \vec{x} \) consists of deriving the square root of the sum of the squares of its components. This gives a single, non-negative value representing the vector size. For the vector \( \vec{x} = \begin{bmatrix} 1 \ 2 \ -2 \end{bmatrix} \), the norm is calculated as:

• \( \|\vec{x}\| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{1 + 4 + 4} = 3 \)

This length is geometrically interpreted as the distance of the vector from the origin in three-dimensional space. A longer norm indicates a vector farther from the origin, while a shorter one means it is closer.

Algebraic Properties of Vectors

Vectors exhibit several algebraic properties that make them useful in mathematical computations and applications. These properties include operations like vector addition, scalar multiplication, and finding the norm, all of which demonstrate certain consistent behaviors:

• Distributive Property: For any scalars \( a \) and \( b \) and vector \( \vec{x} \), \((a + b)\vec{x} = a\vec{x} + b\vec{x}\).
• Associative Property: For any scalar \( a \) and vectors \( \vec{x} \) and \( \vec{y} \), \( a(\vec{x} + \vec{y}) = a\vec{x} + a\vec{y} \).
• Scalar Multiplication: If you scale a vector by a scalar \( a \), then by \( b \), the same result is achieved by multiplying the vector directly by the product \( ab \), i.e., \( b(a\vec{x}) = (ab)\vec{x} \).

These properties are fundamental for problem-solving in physics, engineering, and various fields requiring vector manipulation. Understanding them helps in tackling complex tasks systematically and ensures correct outcomes across various applications.