# Find a vector of magnitude 4 unit and parallel to vector I + j?

Jan 12, 2018

#### Answer:

$= 2 \sqrt{2} \cdot \vec{i} + 2 \sqrt{2} \vec{j}$

#### Explanation:

First we want to turn our given vector, $1 \cdot \vec{i} + 1 \cdot \vec{j}$, into a unit vector--which we do by dividing the vector by its own magnitude:

$| | 1 \cdot \vec{i} + 1 \cdot \vec{j} | | = \sqrt{{1}^{2} + {1}^{2}} = \sqrt{2}$,

so the unit vector we want is:

$\left(\frac{1}{\sqrt{2}}\right) \left(1 \cdot \vec{i} + 1 \cdot \vec{j}\right)$

$= \left(\frac{1}{\sqrt{2}}\right) \cdot \vec{i} + \left(\frac{1}{\sqrt{2}}\right) \cdot \vec{j}$

You might choose to rationalize this:

$= \left(\frac{\sqrt{2}}{2}\right) \cdot \vec{i} + \left(\frac{\sqrt{2}}{2}\right) \cdot \vec{j}$

Now to get the vector we're looking for in the problem we have to scale our unit vector by a factor of 4:

$4 \left(\left(\frac{\sqrt{2}}{2}\right) \cdot \vec{i} + \left(\frac{\sqrt{2}}{2}\right) \cdot \vec{j}\right)$

$= 2 \sqrt{2} \cdot \vec{i} + 2 \sqrt{2} \vec{j}$