Question

In Exercises 9-12, 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. $$ \vec{x}=\left[\begin{array}{c} 1 \\ -2 \\ 5 \end{array}\right], a=2 $$

Short answer

The length of \( \vec{x} \) is \( \sqrt{30} \) and the length of \( 2\vec{x} \) is \( \sqrt{120} \).

Step-by-step solution

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

To calculate the length (norm) of the vector \( \vec{x} \), we use the formula \( \| \vec{x} \| = \sqrt{x_1^2 + x_2^2 + x_3^2} \). Given \( \vec{x} = \begin{bmatrix} 1 \ -2 \ 5 \end{bmatrix} \), substitute the values: \( \| \vec{x} \| = \sqrt{1^2 + (-2)^2 + 5^2} = \sqrt{1 + 4 + 25} = \sqrt{30} \).

Calculate the Scaled Vector \( a \vec{x} \)

First, find the new vector \( a \vec{x} \) by multiplying each component of \( \vec{x} \) by \( a = 2 \). So, \( a \vec{x} = 2 \begin{bmatrix} 1 \ -2 \ 5 \end{bmatrix} = \begin{bmatrix} 2 \ -4 \ 10 \end{bmatrix} \).

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

Now calculate the length of \( a \vec{x} \) using the same formula: \( \| a \vec{x} \| = \sqrt{(2)^2 + (-4)^2 + (10)^2} = \sqrt{4 + 16 + 100} = \sqrt{120} \).

Compare the Lengths

According to Definition 32, the length \( \| a \vec{x} \| \) should be \(|a| \) times the length of \( \| \vec{x} \| \). Therefore, \( |a| \| \vec{x} \| = 2 \cdot \sqrt{30} = \sqrt{120} \). This confirms that the length \( \| a \vec{x} \| = \sqrt{120} \) is indeed correct and the comparison holds.

Key concepts

Scalar Multiplication

Scalar multiplication refers to the process of multiplying each component of a vector by a scalar, which is just a real number or... even sometimes a complex number. Imagine you have a vector \(\vec{x} = \begin{bmatrix}1 \ -2 \ 5\end{bmatrix}\) and a scalar \(a = 2\). By multiplying each component of \(\vec{x}\) by \(a\), you get a new vector \(a\vec{x} = \begin{bmatrix}2 \ -4 \ 10\end{bmatrix}\).
This operation stretches or shrinks the vector. If \(a > 1\), the vector stretches. If \(0 < a < 1\), it shrinks. If \(a < 0\), the vector not only changes size but also reverses direction.
Scalar multiplication is a basic yet crucial operation in vector algebra, as it directly influences the length of a vector.

Vector Norm

The vector norm, often referred to as the length or magnitude, measures how long a vector is. For any vector \(\vec{x} = \begin{bmatrix}x_1 \ x_2 \ x_3\end{bmatrix}\), the length is calculated using the formula \(\|\vec{x}\| = \sqrt{x_1^2 + x_2^2 + x_3^2}\). This creates a straight-line distance from the origin to the point in space defined by the vector, analogous to finding the hypotenuse of a right triangle.
For example, the length of \(\vec{x}\) in \(\begin{bmatrix}1 \ -2 \ 5\end{bmatrix}\) is \(\sqrt{30}\)... because \(\|\vec{x}\| = \sqrt{1^2 + (-2)^2 + 5^2} = \sqrt{1 + 4 + 25}\). The length provides a method for comparing vectors, checking vector equality, and understanding vector scaling in scalar multiplication.

Vector Comparison

When comparing vectors, we often consider their magnitudes and directions. Two vectors are considered equal if they have the same magnitude and direction. This means that even scaled versions of a vector... using scalar multiplication... can relate to each other through their lengths.
In exercises involving vector comparison, if a scalar multiple maintains the vector's direction but alters its length, the comparison boils down to confirming that \(\|a\vec{x}\| = |a|\cdot \|\vec{x}\|\).
For instance, with \(a\vec{x} = \begin{bmatrix}2 \ -4 \ 10\end{bmatrix}\) and \(\vec{x}\), their lengths are \(\sqrt{120}\) and \(\sqrt{30}\) respectively, fulfilling \(2\sqrt{30} = \sqrt{120}\). This demonstrates that multiplying by a scalar scales the vector's length, making it indispensable for clear vector comparisons.