This tutorial explains how to find the magnitude and direction of vectors using their components or given angles. Several worked examples are provided with clear explanations.
If a vector v is given in component form
\[ \mathbf{v} = \langle a , b \rangle \]its magnitude is calculated using the Pythagorean theorem:
\[ \|\mathbf{v}\| = \sqrt{a^2 + b^2} \]The direction of the vector is defined as the angle \( \theta \) (in standard position) measured from the positive x-axis to the vector. It satisfies:
\[ \tan(\theta) = \frac{b}{a}, \quad 0 \le \theta < 2\pi \]When the magnitude and direction are known, the vector components can be recovered using:
\[ a = \|\mathbf{v}\|\cos(\theta), \qquad b = \|\mathbf{v}\|\sin(\theta) \]Find the magnitude and direction of the vector
\[ \mathbf{v} = \langle 2, 2 \rangle \]Magnitude
\[ \|\mathbf{v}\| = \sqrt{2^2 + 2^2} = 2\sqrt{2} \]Direction
\[ \tan(\theta) = \frac{2}{2} = 1 \]Both components are positive, so the vector lies in Quadrant I.
\[ \theta = \arctan(1) = 45^\circ \]Calculate the magnitude and direction of
\[ \mathbf{u} = \langle -7\sqrt{3}, 7 \rangle \]Magnitude
\[ \|\mathbf{u}\| = \sqrt{(-7\sqrt{3})^2 + 7^2} = 14 \]Direction
\[ \tan(\theta) = \frac{7}{-7\sqrt{3}} = -\frac{1}{\sqrt{3}} \]Since the x-component is negative and the y-component is positive, the vector lies in Quadrant II.
\[ \theta = 180^\circ - \arctan\left(\frac{1}{\sqrt{3}}\right) = 150^\circ \]Calculate the magnitude and direction of
\[ \mathbf{v} = \langle -5, -5\sqrt{3} \rangle \]Magnitude
\[ \|\mathbf{v}\| = \sqrt{(-5)^2 + (-5\sqrt{3})^2} = 10 \]Direction
\[ \tan(\theta) = \frac{-5\sqrt{3}}{-5} = \sqrt{3} \]Both components are negative, so the vector lies in Quadrant III.
\[ \theta = 180^\circ + \arctan(\sqrt{3}) = 240^\circ \]Compare the magnitude and direction of vector \( \mathbf{u} \) and \( 3\mathbf{u} \), where
\[ \mathbf{u} = \langle 4, 1 \rangle \]Using scalar multiplication:
\[ 3\mathbf{u} = \langle 12, 3 \rangle \]Magnitudes
\[ \|\mathbf{u}\| = \sqrt{17}, \qquad \|3\mathbf{u}\| = 3\sqrt{17} \]Directions
\[ \tan(\theta) = \frac{1}{4} \]Both vectors lie in Quadrant I and have the same direction:
\[ \theta \approx 14.04^\circ \]Conclusion: Scalar multiplication by a positive number changes the magnitude but not the direction.
Compare vector \( \mathbf{u} \) and \( -6\mathbf{u} \), where
\[ \mathbf{u} = \langle 1, 1 \rangle \]Magnitudes
\[ \|\mathbf{u}\| = \sqrt{2}, \qquad \|-6\mathbf{u}\| = 6\sqrt{2} \]Directions
\[ \theta_{\mathbf{u}} = 45^\circ, \qquad \theta_{-6\mathbf{u}} = 225^\circ \]Conclusion: Multiplying by a negative scalar reverses the direction by \(180^\circ\).
Find the components of a vector with magnitude 5 and direction \(270^\circ\).
Vectors \( \mathbf{u} \) and \( \mathbf{v} \) have magnitudes 2 and 4 and directions \(90^\circ\) and \(180^\circ\). Find the magnitude and direction of \(2\mathbf{u} + 3\mathbf{v}\).
Magnitude
\[ \|\mathbf{w}\| = \sqrt{(-12)^2 + 4^2} = 4\sqrt{10} \]Direction
\[ \tan(\theta) = -\frac{1}{3} \]The vector lies in Quadrant II.
\[ \theta = 180^\circ - \arctan\left(\frac{1}{3}\right) \approx 161.57^\circ \]
Magnitude and Direction Calculator
Vector Tutorials
Vector Calculators