How to Find the Angle Between Two Vectors: Formula & Examples (2024)

Co-authored byDevin McSweenReviewed byGrace Imson, MA

Last Updated: April 7, 2023Fact Checked

Things You Should Know

1. 1

   Use the formula: ( • ) / (|||| ||||). The angle between 2 vectors is where the tails of 2 vectors, or line segments, meet. Each vector has a magnitude, or length, and a direction that it’s heading. So, to find the angle between 2 vectors, you use the above formula where:^[1]

   • is the angle between the vectors.
   • is the inverse of cosine, or the arc cos.
   • • is the dot product of vector and .
   • |||| |||| is the magnitude of vector and .
2. 2

   Identify the vectors’ coordinates in your math problem. Most math problems give you the dimensional coordinates of each vector, which are sometimes also called components. You use each vector’s coordinates to find their magnitudes and combined dot product. If your math problem already gives the vectors’ magnitudes, skip the magnitude step below.^[2]

   • For example, find the angle between vector and vector . Vector has coordinates at (2, 2) and vector has coordinates at (0, 3).
   • Sometimes, vectors are written as = 2i + 2j and = 0i + 3j.
   • While our example uses two-dimensional vectors, finding the angle between 3-dimensional vectors follows the same steps.

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3. 3

   Calculate the magnitude of each vector. Picture a right triangle drawn from the vector's x-component, its y-component, and the vector itself. The vector forms the hypotenuse of the triangle, so to find its magnitude, simply use the Pythagorean theorem. Just plug each vector’s coordinates into the theorem.^[3]

   • In the Pythagorean theorem of a² + b² + c², the vector’s magnitude is denoted by c. So, just rewrite the equation to isolate the magnitude on one side: ||u|| = √(u₁² + u₂²) with u₁² + u₂² being the vector’s x and y coordinates.
   • Using our example, find the magnitudes for vector at (2, 2) and vector at (0, 3).
     • Insert coordinates into the theorem: √2² + 2² = √8 = 2√2. So, ||u|| = 2√2.
       
     • Find magnitude: √0² + 3² = √9. So, |||| = 3.
   • If a vector is 3-dimensional or has more than 2 components, simply continue adding +u₃² + u₄² + … to the Pythagorean Theorem.
4. 4

   Calculate the dot product of the 2 vectors. The dot product is a way to multiply vectors, which is also commonly called the scalar product.

   To calculate the dot product, multiply the same direction coordinates of the vectors, then add the results together.

   For computer graphics programs, see Tips before you continue.^[4]

   • Using our example, = (2, 2) and = (0, 3). Find • .
     • Multiply the x-coordinates of and and the y-coordinates. u_xv_x + u_yv_y = (2)(0) + (2)(3) = 0 + 6 = 6.
     • 6 is the dot product of vector and .

   Defining Dot Product
   In mathematical terms, • = u₁v₁ + u₂v₂, where (u₁, u₂) are the coordinates for vector u. If your vector has more than 2 components, simply continue to add + u₃v₃ + u₄v₄...

5. 5

   Plug the dot product and each vector’s magnitude into the formula. Remember, the formula is ( • ) / (|||| ||||) Now that you know both the dot product and the magnitudes of each vector, simply enter them into this formula.

   Finding Cosine with Dot Product and Magnitude
   In our example, θ = cos^-16 / (2√2•3). Simplify to get θ = cos^-1(√2 / 2).

6. 6

   Use a scientific calculator to find the angle based on the cosine. On most calculators, use either the arccos or cos^-1 function on your calculator to find the angle θ. Simply enter “arccos” and the dot product divided by the vectors’ magnitudes. For some results, use the unit circle to work out the angle.

   Finding an Angle with Cosine
   In our example, θ = cos^-1(√2 / 2). Enter "arccos(√2 / 2)" in your calculator to get θ = 45º. Alternatively, find the angle θ on the unit circle where cosθ = √2 / 2.

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1. 1

   Use the formula: ( ) / (|||| ||||). This formula uses sine and the cross product of vectors to find the angle between them. Unlike the dot product formula, which gives you a scalar answer, the cross product formula gives you an answer as a vector. In this formula:^[5]

   • is the angle between the vectors.
   • is the inverse of sine, or the arc sin.
   • is the cross product of vector and .
   • |||| |||| is the magnitude of vector and .
2. 2

   Find the cross product using the vectors’ coordinates. In most math problems, you have the dimensional coordinates, or components, of each vector written as . To find the cross product, make a matrix with the first vector’s coordinates in the first row and the second vector’s coordinates in the second row. Calculate the i, j, and k values for each matrix section.^[6]

   • For example, find the angle between 2 vectors where is 1i - 2j + 3k and is 10i + 1j - 3k.
     • Draw a matrix for and : 1 2 3 is on the top row, and 10 1 -3 is on the bottom row.
     • Solve the matrix for i: i = (u_j * v_k) - (v_j * u_k)
       i = (6 - 3) = 3
       
     • Solve the matrix for j: j = (u_i * v_k) - (v_i * u_k)
       j = (-3 - 30) = -33
       
     • Solve the matrix for k: k = (u_i * v_j) - (v_i * u_j)
       k = (1 - -20) = 21
       
     • Find the coordinates for i - j + k: 3i - -33j + 21k = 3i + 33j + 21k or <3, 33, 21>
3. 3

   Calculate the magnitude of the cross product. The final step in finding the cross product of vectors is finding the magnitude of their coordinates. Remember, use the Pythagorean Theorem to find a vector’s magnitude. Just plug in the i, k, j coordinates of the cross product of the vectors to get their magnitude.^[7]

   • The cross product of x is 3i + 33j + 21k or <3, 33, 21>.
     • Use the Pythagorean Theorem to find the magnitude: ||u x v|| = √(i² + j² + k²)
     • Plug in 3i + 33j + 21k into the theorem: √((3)² + (33)² + (21)²)
     • Solve: √9 + 1089 + 441 = √1539
     • The cross product of vector x = √1539
4. 4

   Find the magnitude of each vector. Now, calculate the magnitude of each vector using their dimensional coordinates. Just plug the coordinates into the Pythagorean Theorem like in the step above.^[8]

   • In the example, is 1i - 2j + 3k and is 10i + 1j - 3k.
     • Find the magnitude of : ||u|| = √i² + j² + k² = √((1)² + (-2)² + (3)²) = √1 + 4 + 9 = √14
     • Find the magnitude of : ||v|| = √((10)² + (1)² + (-3)²) = √100 + 1 + 9 = √110
5. 5

   Plug the cross product and the vectors’ magnitude into the formula. Now that you have the vectors’ cross product and magnitudes, simply enter them into the formula ( ) / (|||| ||||).^[9]

   See Also

   Vector Calculator - with all steps2.4 The Cross Product - Calculus Volume 3 | OpenStax6.E: Triangles and Vectors (Exercises)

   • In our example, θ = sin^-1(√1539 / √14 * √110)
6. 6

   Find the angle using a calculator. Simply take the inverse sine of the cross product and magnitudes to find the angle between the vectors. Using your calculator, find the arcsin or sin^-1 function. Then, enter in the cross product and magnitude.^[10]

   • In our example, enter “arcsin(√1539 / √14 * √110) into your calculator to get θ = 88.5º.

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1. 1

   Understand the purpose of the angle formula. This formula was not derived from existing rules. Instead, it was created as a definition of 2 vectors' dot product and the angle between them. However, this decision was not arbitrary. With a look back to basic geometry, you see why this formula results in intuitive and useful definitions.

   • The examples below use 2-dimensional vectors because these are the most intuitive to use. Vectors with 3 or more components use the same formula.
2. 2

   Review the Law of Cosines used in the formula. Take an ordinary triangle, with angle θ between sides a and b, and opposite side c. The Law of Cosines states that c² = a² + b² -2abcos(θ). This is derived fairly easily from basic geometry.^[11]

3. 3

   Connect 2 vectors to form a triangle. Sketch a pair of 2D vectors on paper, vectors and , with angle θ between them. Draw a third vector between them to make a triangle. In other words, draw vector such that + = . This vector = - .^[12]

4. 4

   Write the Law of Cosines for the triangle. Insert the length of the "vector triangle" sides in our example into the Law of Cosines:^[13]

   • ||(a - b)||² = ||a||² + ||b||² - 2||a|| ||b||cos(θ)
5. 5

   Write the Law of Cosines using the dot product of vector a and b. Remember, the dot product is the magnification of 1 vector projected onto another. A vector's dot product with itself doesn't require any projection, since there is no difference in direction. This means that • = ||a||². Use this fact to rewrite the equation:^[14]

   • ( - ) • ( - ) = • + • - 2||a|| ||b||cos(θ)
6. 6

   Rewrite the dot product into the angle formula. Expand the left side of the formula, then simplify to reach the formula used to find angles.^[15]

   • • - • - • + • = • + • - 2||a|| ||b||cos(θ)
   • - • - • = -2||a|| ||b||cos(θ)
   • -2( • ) = -2||a|| ||b||cos(θ)
   • • = ||a|| ||b||cos(θ)

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