# Find the magnitude ||v|| and the direction angle \theta for the given vector v. v = -4i + 11j

## Question:

Find the magnitude {eq}\left\| \mathbf{v} \right\| {/eq} and the direction angle {eq}\,\theta\, {/eq} for the given vector {eq}\mathbf{v} {/eq}.

{eq}\displaystyle\;\mathbf{v} = -4\mathbf{i} + 11\mathbf{j} {/eq}

## Magnitude and Direction of a vector:

For a given vector, {eq}\vec{v}= x\vec{i} + y \vec{j} {/eq}:

(i) The magnitude is: {eq}||\vec{v}|| = \sqrt{x^2+y^2} {/eq}.

(ii) To find the direction, we will first find {eq}\alpha = \tan^{-1} \left|\dfrac{y}{x} \right| {/eq}. We then see which quadrant {eq}(x, y) {/eq} lies in and we will find the direction {eq}\theta {/eq} depending on the quadrant:

• {eq}1^{st} {/eq} Quadrant: {eq}\theta=\alpha {/eq}.
• {eq}2^{nd} {/eq} Quadrant: {eq}\theta= 180-\alpha {/eq}.
• {eq}3^{rd} {/eq} Quadrant: {eq}\theta = 180+ \alpha {/eq}.
• {eq}4^{th} {/eq} Quadrant: {eq}\theta = 360-\alpha {/eq}.

The given vector is:

$$\vec{v}= x\vec{i} + y \vec{j}=-4\vec{i} + 11 \vec{j}$$

Here, {eq}x=-4 {/eq} and {eq}y=11 {/eq}.

Finding magnitude:

The magnitude of the vector is, {eq}||\vec{v}|| = \sqrt{x^2+y^2}= \sqrt{(-4)^2+11^2} = \color{blue}{\boxed{\mathbf{\sqrt{137}}}} {/eq}.

Finding direction:

To find the direction, first, we will find:

\begin{align} \alpha &= \tan^{-1} \left|\dfrac{y}{x} \right|\\[0.4cm] &= \tan^{-1} \left|\dfrac{11}{-4} \right|\\[0.4cm] &=\tan^{-1} \dfrac{11}{4} \end{align}

We know that {eq}(x, y)= (-4, 11) {/eq} is in Quadrant II.

So the required direction is:

$$\theta = \color{blue}{\boxed{\mathbf{180^\circ - \tan^{-1} \dfrac{11}{4} }}} (OR) \color{blue}{\boxed{\mathbf{109.98311}}}$$

The final answer is rounded to five decimals. 