Problem on finding a vector with given length and direction
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Find the vector of length 2 in the direction of $\langle 1,-1 \rangle$.
Solution
Let $\mathbf{x}$ be the vector we seek. We recall the relationship of a vector to its length and direction:
Definition of the direction of a vector
$$\text{dir } \mathbf{x} = \frac{\mathbf{x}} {| \mathbf{x} | }$$
Building a vector from its length and direction
Because we art trying to find $\bfx$ from information on its length and direction, we rewrite this formula as, $$ \mathbf{x} = |\mathbf{x} |\ \text{dir } \mathbf{x}.$$
We are given that $| \mathbf{x} |= 2$. We need to compute $\text{dir } \mathbf{x}$ from the information that $\mathbf{x}$ is in the same direction as $\langle 1,-1 \rangle$:
Definition of the direction of a vector
$$\text{dir } \mathbf{x} = \text{dir } \langle 1,-1 \rangle = \frac{\langle 1, -1\rangle} { | \langle 1, -1 \rangle | }. $$
We need to compute the length $|\langle 1, -1 \rangle |$. To do this, recall that
Definition of vector length
The length of the vector $\bfy = \langle y_1, y_2\rangle$ is $
|\mathbf{y}| = \sqrt{y_1^2 + y_2^2}$.
Hence, $ | \langle 1, -1 \rangle | = \sqrt{2}$, and
$$\text{dir } \mathbf{x} = \left \langle \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}} \right \rangle.$$
Thus
\begin{align}
\mathbf{x} &= 2 \cdot \left \langle \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}} \right \rangle\\
&= \langle \sqrt{2}, -\sqrt{2} \rangle.
\end{align}
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