Vector Dot Product Calculator

Enter the vector components to calculate their dot or scalar product (a · b) and see the step-by-step solution.

Vector a
⟨ , ⟩
Vector b
⟨ , ⟩

Quick Examples

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How to Use the Calculator

This online dot product solver is a mathematical tool designed to quickly evaluate the scalar product (also known as the inner product) of two vectors. It works for solving this form of vector multiplication in both the two-dimensional plane (R2) and three-dimensional space (R3). Since it is a direct operation, the system will provide the exact numerical value along with its step-by-step algebraic solution.

How to enter your data:

  • Select the dimension: use the main drop-down menu to indicate whether you are working with two-dimensional (R2) or three-dimensional (R3) vectors.
  • Enter the components: for this type of calculation, data entry is done using the rectangular components of each vector (x, y for the plane; x, y, z for space).
  • Supported formats: the input fields support integers, decimals, exact fractions, and irrational numbers.

How to read the results:

Once the values are processed, the calculator structures the answer into two main sections:

  1. Quick answer (scalar value): the final result of your operation will be displayed in the highlighted top box. Keep in mind that, by mathematical definition, the inner product of two vectors does not result in a new vector, but rather a single numerical magnitude (a scalar).
  2. Step-by-step solution: the detailed procedure will be displayed at the bottom. You will see how the calculator sequentially multiplies the corresponding components of both vectors. Finally, it will show the algebraic sum of these sub-products to reach the answer, applying the formula a · b = axbx + ayby + azbz.

Solved Exercises

The following are examples of problems solved by the calculator.

Calculate the dot product of vectors \(\vec{a}=\langle 4, 3 \rangle\) and \(\vec{b}=\langle 2, 5 \rangle.\)

Result

The dot product of the given vectors is:

$$ \displaystyle \vec{a} \cdot \vec{b} = 23 $$

Step-by-step solution

1. Identify the vectors.

The vectors to be used are:

$$ \displaystyle \vec{a} = \langle 4, 3 \rangle \\[1em] \vec{b} = \langle 2, 5 \rangle $$

2. Calculate the dot product.

The dot product of two 2D vectors is obtained by multiplying their corresponding components and adding the results:

$$ \displaystyle \vec{a} \cdot \vec{b} = a_x b_x + a_y b_y $$

In our case:

$$ \displaystyle \vec{a} \cdot \vec{b} = \langle 4, 3 \rangle \cdot \langle 2, 5 \rangle \\[1em] \vec{a} \cdot \vec{b} = (4)(2) + (3)(5) \\[1em] \vec{a} \cdot \vec{b} = 8 + 15 \\[1em] \vec{a} \cdot \vec{b} = 23 $$
Determine the dot product of \(\langle -2, -6 \rangle\) and \(\langle -3, 4 \rangle.\)

Result

The dot product of the given vectors is:

$$ \displaystyle \vec{a} \cdot \vec{b} = -18 $$

Step-by-step solution

1. Identify the vectors.

The vectors to be used are:

$$ \displaystyle \vec{a} = \langle -2, -6 \rangle \\[1em] \vec{b} = \langle -3, 4 \rangle $$

2. Calculate the dot product.

The dot product of two 2D vectors is obtained by multiplying their corresponding components and adding the results:

$$ \displaystyle \vec{a} \cdot \vec{b} = a_x b_x + a_y b_y $$

In our case:

$$ \displaystyle \vec{a} \cdot \vec{b} = \langle -2, -6 \rangle \cdot \langle -3, 4 \rangle \\[1em] \vec{a} \cdot \vec{b} = (-2)(-3) + (-6)(4) \\[1em] \vec{a} \cdot \vec{b} = 6 - 24 \\[1em] \vec{a} \cdot \vec{b} = -18 $$
Find the result of the vector multiplication \(\langle \frac{1}{2}, \frac{3}{4}\rangle \cdot \langle 4, -2\rangle.\)

Result

The scalar product of the given vectors is:

$$ \displaystyle \vec{a} \cdot \vec{b} = \frac{1}{2} = 0.5 $$

Step-by-step solution

1. Identify the vectors.

The vectors to be used are:

$$ \displaystyle \vec{a} = \left\langle \frac{1}{2}, \frac{3}{4} \right\rangle \\[1em] \vec{b} = \langle 4, -2 \rangle $$

2. Calculate the dot product.

The dot product of two 2D vectors is obtained by multiplying their corresponding components and adding the results:

$$ \displaystyle \vec{a} \cdot \vec{b} = a_x b_x + a_y b_y $$

In our case:

$$ \displaystyle \vec{a} \cdot \vec{b} = \left\langle \frac{1}{2}, \frac{3}{4} \right\rangle \cdot \langle 4, -2 \rangle \\[1em] \vec{a} \cdot \vec{b} = \left(\frac{1}{2}\right)(4) + \left(\frac{3}{4}\right)(-2) \\[1em] \vec{a} \cdot \vec{b} = 2 - \frac{3}{2} \\[1em] \vec{a} \cdot \vec{b} = \frac{1}{2} = 0.5 $$
Find the dot product of the three-dimensional vectors \(\vec{a}=\langle -4, 0, 2\rangle\) and \(\vec{b}=\langle 3, -1, -5\rangle.\)

Result

The dot product of the given vectors is:

$$ \displaystyle \vec{a} \cdot \vec{b} = -22 $$

Step-by-step solution

1. Identify the vectors.

The vectors to be used are:

$$ \displaystyle \vec{a} = \langle -4, 0, 2 \rangle \\[1em] \vec{b} = \langle 3, -1, -5 \rangle $$

2. Calculate the dot product.

The dot product of two 3D vectors is obtained by multiplying their corresponding components and adding the results:

$$ \displaystyle \vec{a} \cdot \vec{b} = a_x b_x + a_y b_y + a_z b_z $$

In our case:

$$ \displaystyle \vec{a} \cdot \vec{b} = \langle -4, 0, 2 \rangle \cdot \langle 3, -1, -5 \rangle \\[1em] \vec{a} \cdot \vec{b} = (-4)(3) + (0)(-1) + (2)(-5) \\[1em] \vec{a} \cdot \vec{b} = -12 + 0 - 10 \\[1em] \vec{a} \cdot \vec{b} = -22 $$
Calculate the result of the multiplication \(\langle 3, -2, 1 \rangle\cdot \langle 4, 5, -2 \rangle.\)

Result

The dot product of the given vectors is:

$$ \displaystyle \vec{a} \cdot \vec{b} = 0 $$

Step-by-step solution

1. Identify the vectors.

The vectors to be used are:

$$ \displaystyle \vec{a} = \langle 3, -2, 1 \rangle \\[1em] \vec{b} = \langle 4, 5, -2 \rangle $$

2. Calculate the dot product.

The dot product of two 3D vectors is obtained by multiplying their corresponding components and adding the results:

$$ \displaystyle \vec{a} \cdot \vec{b} = a_x b_x + a_y b_y + a_z b_z $$

In our case:

$$ \displaystyle \vec{a} \cdot \vec{b} = \langle 3, -2, 1 \rangle \cdot \langle 4, 5, -2 \rangle \\[1em] \vec{a} \cdot \vec{b} = (3)(4) + (-2)(5) + (1)(-2) \\[1em] \vec{a} \cdot \vec{b} = 12 - 10 - 2 \\[1em] \vec{a} \cdot \vec{b} = 0 $$

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Daniel Machado

Professor of Mathematics, graduated from the Faculty of Exact, Chemical and Natural Sciences of the National University of Misiones (UNAM). Developer and creator of RigelUp, dedicated to building tools for mathematical learning.

Last updated: (2 weeks ago)