# A vector has a magnitude of 16 and a direction of 120°. How do you write the vector in terms of unit vectors?

Apr 6, 2017

$- 8 \hat{i} + 8 \sqrt{3} \hat{j}$

#### Explanation:

The vector is of the form $a \hat{i} + b \hat{j} , a \in \mathbb{R} , b \in \mathbb{R}$. We need to find $a$ and $b$.

The magnitude of a vector is defined as $\sqrt{{a}^{2} + {b}^{2}}$.

The direction of a vector can be found using $\arctan \left(\frac{b}{a}\right)$.

From the question, we know that $\sqrt{{a}^{2} + {b}^{2}} = 16$ and $\arctan \left(\frac{b}{a}\right) = {120}^{\circ}$. Since ${90}^{\circ} < {120}^{\circ} < {180}^{\circ}$, the vector lies in the second quadrant. In other words, $a < 0$ and $b > 0$.

From $\arctan \left(\frac{b}{a}\right) = {120}^{\circ}$, we know that $\frac{b}{a} = - \sqrt{3}$, or $b = - a \sqrt{3}$.

Substitute this into $\sqrt{{a}^{2} + {b}^{2}} = 16$. We get $\sqrt{{a}^{2} + {\left(- a \sqrt{3}\right)}^{2}} = 16$. Simplifying this, we get $\sqrt{4 {a}^{2}} = 16$. Solving this gives $a = \pm 8$. However, we said before that $a < 0$. Thus, $a = - 8$.

Now, $b = - a \sqrt{3} = - \left(- 8\right) \sqrt{3} = 8 \sqrt{3}$.

The vector can be written as $- 8 \hat{i} + 8 \sqrt{3} \hat{j}$.