Problem on scalar multiplication and vector direction
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Let $c$ be a nonzero scalar. Does $c \ \mathbf{x}$ have the same direction as $\mathbf{x}$?
Solution
Direction of a vector
We are seeking to determine if $\text{dir }( c \ \mathbf{x})$ and the direction $\text{dir } \mathbf{x}$.
We will write the expression of $\text{dir } c \ \mathbf{x}$ and manipulate it to involve $\text{dir } \mathbf{x}$ explicitly.
Recall that:
Definition of the direction of a vector
The direction of $\mathbf{x}$ is defined by
$$\text{dir } \mathbf{x} = \frac{\mathbf{x}}{| \mathbf{x} | }.$$
We are thus seeking the relationship of $\frac{\mathbf{x}}{|\mathbf{x}|}$ and $\frac{c \mathbf{x}}{|c \mathbf{x}|}$. Are they equal?
In order to make the two terms more alike, we try to factor out $c$ from the denominator of $\text{dir } (c \ \mathbf{x})$. Recall that:
Length of a scalar vector multiplication
For any scalar $c$ and vector $\mathbf{x}$, $|c \mathbf{x} | = |c| |\mathbf{x} |$.
Thus, $\text{dir } (c \mathbf{x}) = \frac{c}{|c|} \frac{\mathbf{x}}{|\mathbf{x}|} = \frac{c}{|c|} \text{dir } \mathbf{x}$.
Whether or not $\mathbf{x}$ and $c \ \mathbf{x}$ have the same direction is determined by whether or not $\frac{c}{|c|}=1 $.
If $c> 0$: $\frac{c}{|c|} = 1$ and $\text{dir } (c \mathbf{x}) = \text{dir } \mathbf{x}$.
If $c< 0$: $\frac{c}{|c|} = 1$ and $\text{dir } (c \mathbf{x}) = - \text{dir } \mathbf{x}$.
If $c=0$: $\text{dir } (c \mathbf{x})$ is undefined.
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