How to Find the Angle Between Two Vectors

If you are a mathematician or a graphics programmer, you will probably have to find the angle between two given vectors. In this article, the wikiHow will show you how to do just that.

Table of Contents

Find the angle between two vectors

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Define vector. Write down all the information related to the two vectors you have. Assume you only have certain parameters about their dimensional coordinates (also called components). ^{[1] X Research Source} If you already know the length (magnitude) of a vector, you can skip some of the steps below.

Example: Two-dimensional vector $u$ $\to$ ${displaystyle {overrightarrow {u}}}$ ${overrightarrow {u}}$ = (2,2) and two-dimensional vector $v$ $\to$ ${displaystyle {overrightarrow {v}}}$ ${overrightarrow {v}}$ = (0,3). They can also be written as $u$ $\to$ ${displaystyle {overrightarrow {u}}}$ ${overrightarrow {u}}$ = 2 i + 2 j and $v$ $\to$ ${displaystyle {overrightarrow {v}}}$ ${overrightarrow {v}}$ = 0 i + 3 j = 3 j .
Although two-dimensional vectors are used in this article’s example, the following instructions can be applied to vectors of any number of dimensions.

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Write the cosine formula. To find the angle θ between two vectors, we start with the formula for finding the cosine for that angle. You can learn about this formula below, or just write it down like this: ^{[2] X Research Source}

cosθ = ( $u$ $\to$ ${displaystyle {overrightarrow {u}}}$ ${overrightarrow {u}}$ • $v$ $\to$ ${displaystyle {overrightarrow {v}}}$ ${overrightarrow {v}}$ ) / ( || $u$ $\to$ ${displaystyle {overrightarrow {u}}}$ ${overrightarrow {u}}$ || || $v$ $\to$ ${displaystyle {overrightarrow {v}}}$ ${overrightarrow {v}}$ || )
|| $u$ $\to$ ${displaystyle {overrightarrow {u}}}$ ${overrightarrow {u}}$ || means “the length of the vector $u$ $\to$ ${displaystyle {overrightarrow {u}}}$ ${overrightarrow {u}}$ “.
$u$ $\to$ ${displaystyle {overrightarrow {u}}}$ ${overrightarrow {u}}$ • $v$ $\to$ ${displaystyle {overrightarrow {v}}}$ ${overrightarrow {v}}$ is the dot product of two vectors – this will be explained below.

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Calculate the length of each vector. Imagine a right triangle made up of the x component, the y component of the vector, and the vector itself. The vector forms the hypotenuse of the triangle, so to find its length we use the Pythagorean theorem. In fact, this formula can be easily extended to vectors of any dimension. ^{[3] X Research Sources}

|| u || ² = u ₁² + u ₂² . If the vector has more than two components, we just keep adding +u ₃² + u ₄² + …
Therefore, for a two-dimensional vector, || u || = √(u ₁² + u ₂² ) .
In this example, || $u$ $\to$ ${displaystyle {overrightarrow {u}}}$ ${overrightarrow {u}}$ || = √(2 ² + 2 ² ) = √(8) = 2√2 . || $v$ $\to$ ${displaystyle {overrightarrow {v}}}$ ${overrightarrow {v}}$ || = √(0 ² + 3 ² ) = √(9) = 3 .

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Calculate the dot product of two vectors. You’ve probably learned this method of vector multiplication, also known as dot product . ^{[4] X Research Source} To calculate dot products relative to their components, multiply the components in each direction, then add up the entire result. ^{[5] X Research Sources}

With a graphics program, refer to Tips before reading on.
In math $u$ $\to$ ${displaystyle {overrightarrow {u}}}$ ${overrightarrow {u}}$ • $v$ $\to$ ${displaystyle {overrightarrow {v}}}$ ${overrightarrow {v}}$ = u ₁ v ₁ + u ₂ v ₂ , where, u = (u ₁ , u ₂ ). If the vector has more than two components, you just add + u ₃ v ₃ + u ₄ v ₄ …
In this example, $u$ $\to$ ${displaystyle {overrightarrow {u}}}$ ${overrightarrow {u}}$ • $v$ $\to$ ${displaystyle {overrightarrow {v}}}$ ${overrightarrow {v}}$ = u ₁ v ₁ + u ₂ v ₂ = (2)(0) + (2)(3) = 0 + 6 = 6 . This is the dot product of the vector $u$ $\to$ ${displaystyle {overrightarrow {u}}}$ ${overrightarrow {u}}$ and vector $v$ $\to$ ${displaystyle {overrightarrow {v}}}$ ${overrightarrow {v}}$ .

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Enter the result obtained into the formula. Remember that cosθ = (

u

\to

{displaystyle {overrightarrow {u}}}

{overrightarrow {u}}

•

v

\to

{displaystyle {overrightarrow {v}}}

{overrightarrow {v}}

) / ( ||

u

\to

{displaystyle {overrightarrow {u}}}

{overrightarrow {u}}

|| ||

v

\to

{displaystyle {overrightarrow {v}}}

{overrightarrow {v}}

|| ). At this point, we know both the dot product and the length of each vector. Enter these values into the formula to calculate the cosine of the angle.

In our example, cosθ = 6 / ( 2√2 * 3 ) = 1 / √2 = √2 / 2.

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Find the angle based on its cosine value. You can use the arccos or cos ^-1 function in your calculator to find θ from a known cos θ value. With some results, you can probably find the angle based on the unit circle.

In the example, cosθ = √2 / 2. Type “arccos(√2 / 2)” into the calculator to find the angle. Or, you can find the angle θ on the unit circle, where cosθ = √2 / 2. It is true for θ = ^π / ₄ or 45º .
Combining everything, the final formula is: angle θ = arccosine(( $u$ $\to$ ${displaystyle {overrightarrow {u}}}$ ${overrightarrow {u}}$ • $v$ $\to$ ${displaystyle {overrightarrow {v}}}$ ${overrightarrow {v}}$ ) / ( || $u$ $\to$ ${displaystyle {overrightarrow {u}}}$ ${overrightarrow {u}}$ || || $v$ $\to$ ${displaystyle {overrightarrow {v}}}$ ${overrightarrow {v}}$ || ))

Determine the angle formula

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Understand the purpose of the formula. This formula is not derived from existing rules. Instead, it is formulated as the definition of the dot product and the angle between two vectors. ^{[6] X Source of Research} However, it is not an arbitrary decision. Going back to basic geometry, we can see why this formula provides useful and intuitive definitions.

The examples below use two-dimensional vectors because they are easiest to understand and simple. Vectors of three dimensions or more have properties defined by an almost similar general formula.

READ More: Tracee Ellis Ross

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Review the Cosine theorem. Consider an ordinary triangle with angle θ between sides a and b, opposite side c. The Cosine theorem states that c ² = a ² + b ² -2ab cos (θ). This result is derived quite simply from basic geometry.

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Connect two vectors, forming a triangle. Draw a pair of two-dimensional vectors on paper, vector

a

\to

{displaystyle {overrightarrow {a}}}

{overrightarrow {a}}

and vector

b

\to

{displaystyle {overrightarrow {b}}}

{overrightarrow {b}}

, where θ is the angle between them. Draw a third vector between these two vectors to create a triangle. In other words, draw the vector

c

\to

{displaystyle {overrightarrow {c}}}

{overrightarrow {c}}

so that

b

\to

{displaystyle {overrightarrow {b}}}

{overrightarrow {b}}

+

c

\to

{displaystyle {overrightarrow {c}}}

{overrightarrow {c}}

=

a

\to

{displaystyle {overrightarrow {a}}}

{overrightarrow {a}}

. Vector

c

\to

{displaystyle {overrightarrow {c}}}

{overrightarrow {c}}

=

a

\to

{displaystyle {overrightarrow {a}}}

{overrightarrow {a}}

–

b

\to

{displaystyle {overrightarrow {b}}}

{overrightarrow {b}}

. ^{[7] X Research Sources}

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Write the Cosine theorem for this triangle. Substitute the length of our “vector triangle” into the Cosine theorem:

|| (a – b) || ² = || a || ² + || b || ² – 2 || a || || b || cos (θ)

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Rewrite with dot product. Remember, the dot product is the image of one vector on the other. The dot product of a vector with itself does not need any projection, because here, there is no dimensional difference. ^{[8] X Research Source} Means

a

\to

{displaystyle {overrightarrow {a}}}

{overrightarrow {a}}

•

a

\to

{displaystyle {overrightarrow {a}}}

{overrightarrow {a}}

= || a || ² . Using this, we rewrite the equation:

( $a$ $\to$ ${displaystyle {overrightarrow {a}}}$ ${overrightarrow {a}}$ – $b$ $\to$ ${displaystyle {overrightarrow {b}}}$ ${overrightarrow {b}}$ ) • ( $a$ $\to$ ${displaystyle {overrightarrow {a}}}$ ${overrightarrow {a}}$ – $b$ $\to$ ${displaystyle {overrightarrow {b}}}$ ${overrightarrow {b}}$ ) = $a$ $\to$ ${displaystyle {overrightarrow {a}}}$ ${overrightarrow {a}}$ • $a$ $\to$ ${displaystyle {overrightarrow {a}}}$ ${overrightarrow {a}}$ + $b$ $\to$ ${displaystyle {overrightarrow {b}}}$ ${overrightarrow {b}}$ • $b$ $\to$ ${displaystyle {overrightarrow {b}}}$ ${overrightarrow {b}}$ – 2 || a || || b || cos (θ)

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Rewrite the same formula. Expand the left side of the formula, then reduce it to get the formula used to find the measures of angles.

$a$ $\to$ ${displaystyle {overrightarrow {a}}}$ ${overrightarrow {a}}$ • $a$ $\to$ ${displaystyle {overrightarrow {a}}}$ ${overrightarrow {a}}$ – $a$ $\to$ ${displaystyle {overrightarrow {a}}}$ ${overrightarrow {a}}$ • $b$ $\to$ ${displaystyle {overrightarrow {b}}}$ ${overrightarrow {b}}$ – $b$ $\to$ ${displaystyle {overrightarrow {b}}}$ ${overrightarrow {b}}$ • $a$ $\to$ ${displaystyle {overrightarrow {a}}}$ ${overrightarrow {a}}$ + $b$ $\to$ ${displaystyle {overrightarrow {b}}}$ ${overrightarrow {b}}$ • $b$ $\to$ ${displaystyle {overrightarrow {b}}}$ ${overrightarrow {b}}$ = $a$ $\to$ ${displaystyle {overrightarrow {a}}}$ ${overrightarrow {a}}$ • $a$ $\to$ ${displaystyle {overrightarrow {a}}}$ ${overrightarrow {a}}$ + $b$ $\to$ ${displaystyle {overrightarrow {b}}}$ ${overrightarrow {b}}$ • $b$ $\to$ ${displaystyle {overrightarrow {b}}}$ ${overrightarrow {b}}$ – 2 || a || || b || cos (θ)
– $a$ $\to$ ${displaystyle {overrightarrow {a}}}$ ${overrightarrow {a}}$ • $b$ $\to$ ${displaystyle {overrightarrow {b}}}$ ${overrightarrow {b}}$ – $b$ $\to$ ${displaystyle {overrightarrow {b}}}$ ${overrightarrow {b}}$ • $a$ $\to$ ${displaystyle {overrightarrow {a}}}$ ${overrightarrow {a}}$ = -2 || a || || b || cos (θ)
-2( $a$ $\to$ ${displaystyle {overrightarrow {a}}}$ ${overrightarrow {a}}$ • $b$ $\to$ ${displaystyle {overrightarrow {b}}}$ ${overrightarrow {b}}$ ) = -2 || a || || b || cos (θ)
$a$ $\to$ ${displaystyle {overrightarrow {a}}}$ ${overrightarrow {a}}$ • $b$ $\to$ ${displaystyle {overrightarrow {b}}}$ ${overrightarrow {b}}$ = || a || || b || cos (θ)

Advice

To swap values and solve problems quickly, use this formula for any pair of two-dimensional vectors: cosθ = (u ₁ • v ₁ + u ₂ • v ₂ ) / (√(u ₁² •) u ₂² ) • √(v ₁² • v ₂² )).
If you’re working with computer graphics software, you’ll most likely only have to care about the dimensions of the vectors and not their length. Use these steps to shorten the equation and speed up your program: ^{[9] X Research Source}^{[10] X Research Source}
- Normalize each vector so that they have a length of 1. To do so, divide each component of the vector by its length.
- Take the dot product of the normalized vector instead of the original vector.
- Since the vector has a length of 1, we can remove the length element from the equation. Finally, the angle equation we get is arccos( $u$ $\to$ ${displaystyle {overrightarrow {u}}}$ ${overrightarrow {u}}$ • $v$ $\to$ ${displaystyle {overrightarrow {v}}}$ ${overrightarrow {v}}$ ).
Based on the cosine formula, we can quickly determine whether an angle is an acute or obtuse angle. Start with cosθ = ( u→{displaystyle {overrightarrow {u}}} ${overrightarrow {u}}$ • v→{displaystyle {overrightarrow {v}}} ${overrightarrow {v}}$ ) / ( || u→{displaystyle {overrightarrow {u}}} ${overrightarrow {u}}$ || || v→{displaystyle {overrightarrow {v}}} ${overrightarrow {v}}$ || ):
- The left and right sides of the equation must have the same sign (positive or negative).
- Since the length is always positive, cosθ must have the same sign as the dot product.
- Therefore, if the dot product is positive, cosθ is also positive. We’re in the first quadrant of the unit circle, with θ < π / 2 or 90º. The required angle is an acute angle.
- If the dot product is negative, cosθ is negative. We’re in the second quadrant of the unit circle, with π / 2 < θ ≤ π or 90º < θ ≤ 180º. It’s an obtuse angle.

Steps

Find the angle between two vectors

Determine the angle formula

Advice

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