# Angle between Two Vectors

*. A vector is said to be in standard position if its initial point is the origin (0, 0).*

__standard position__Figure 1 shows two vectors in standard position.

The angle between two vectors in standard position can be calculated as follows:

ANGLE BETWEEN TWO VECTORS:

*If θ is the angle between two non-zero vectors in standard position u and v:*

Where $0\le \theta \le 2\pi $ and $\parallel \text{v}\parallel =\sqrt{{v}_{1}{}^{2}+{v}_{2}{}^{2}}$

Let's look at some examples.

*To work these examples requires the use of various vector rules. If you are not familiar with a rule go to the associated topic for a review.*

**Example 1: Find the angle θ between $u=\langle 6,3\rangle $ and $v=\langle 5,13\rangle $.**

Remember the result will be a scalar. $u\xb7v={u}_{1}{v}_{1}+{u}_{2}{v}_{2}$ |
$u\xb7v$ $\left(6\xb75\right)+\left(3\xb713\right)$ $30+39={69}$ |

$\parallel v\parallel =\sqrt{{v}_{1}{}^{2}+{v}_{2}{}^{2}}$ |
||u|| = $\sqrt{{u}_{1}{}^{2}+{u}_{2}{}^{2}}$ ||u|| = $\sqrt{{6}^{2}+{3}^{2}}$ ||u|| = $\sqrt{45}={3}\sqrt{5}$ ______________________________ ||v|| = $\sqrt{{v}_{1}{}^{2}+{v}_{2}{}^{2}}$ ||v|| = $\sqrt{{5}^{2}+{13}^{2}}$ ||v|| = $\sqrt{194}$ |

$cos\theta =\frac{u\xb7v}{\parallel u\parallel \xb7\parallel v\parallel}$ |
$cos\theta =\frac{u\xb7v}{\parallel u\parallel \xb7\parallel v\parallel}$ $\text{cos}\theta =\frac{69}{3\sqrt{5}\xb7\sqrt{194}}=\frac{23}{\sqrt{970}}$ $\theta ={\mathrm{cos}}^{-1}\frac{23}{\sqrt{970}}$ $\theta \approx {42}{\xb0}$ |

**Example 2: Find the angle θ between $u=\langle 3,-6\rangle $ and $v=\langle 8,4\rangle $.**

Remember the result will be a scalar. $u\xb7v={u}_{1}{v}_{1}+{u}_{2}{v}_{2}$ |
$u\xb7v$ $\left(3\xb78\right)+\left(-6\xb74\right)$ $24-24={0}$ |

$\parallel v\parallel =\sqrt{{v}_{1}{}^{2}+{v}_{2}{}^{2}}$ |
||u|| = $\sqrt{{u}_{1}{}^{2}+{u}_{2}{}^{2}}$ ||u|| = $\sqrt{{3}^{2}+{\left(-6\right)}^{2}}$ ||u|| = $\sqrt{45}={3}\sqrt{5}$ ______________________________ ||v|| = $\sqrt{{v}_{1}{}^{2}+{v}_{2}{}^{2}}$ ||v|| = $\sqrt{{8}^{2}+{4}^{2}}$ ||v|| = $\sqrt{80}={2}\sqrt{20}$ |

$cos\theta =\frac{u\xb7v}{\parallel u\parallel \xb7\parallel v\parallel}$ As soon as you determine that the dot product is 0 you do not need to calculate the magnitudes. They are completed here for your benefit. Note that when two vectors in standard position have a dot product of 0 the angle between them is 90°. |
$cos\theta =\frac{u\xb7v}{\parallel u\parallel \xb7\parallel v\parallel}$ $\text{cos}\theta =\frac{0}{3\sqrt{5}\xb72\sqrt{20}}=\frac{0}{6\sqrt{100}}=\frac{0}{60}$ $\theta ={\mathrm{cos}}^{-1}0$ $\theta ={90}{\xb0}$ |

Related Links:Math algebra Decomposing a Vector into Components The First Derivative Rule Pre Calculus |

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