Scalar Triple Product Calculator

Enter the vector components to calculate their scalar triple product (a · (b × c)) and see the step-by-step resolution.

Vector a
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Vector b
⟨ , , ⟩
Vector c
⟨ , , ⟩

Quick Examples

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How to Use the Calculator

This online scalar triple product solver is an analytical tool designed to solve combined operations involving three vectors in three-dimensional space. The algorithm evaluates the expression a · (b × c) and, in addition to providing the algebraic value, uses geometric properties to deduce the volume of the solid formed by the vectors.

Setting up the data:

  • Entering components in R3: The interface features three independent blocks for you to input the coordinates (x, y, z) corresponding to vectors a, b, and c. The order is important, as the calculator will always take the first vector for the dot product and the following two for the cross product.
  • Supported number types: The text boxes are designed to interpret integers, decimals, exact fractions, and irrational numbers (such as square roots). The mathematical system will preserve the symbolic format during the operation to ensure exact results, without rounding errors.

Reading the results:

Once you start the calculation, the tool will provide a report focused on the algebraic steps in two sections:

  1. Quick answer: The main highlighted box will display the scalar value (a single number) resulting from the operation. Additionally, this box will indicate the volume of the parallelepiped formed by the three vectors in space; to obtain this geometric value, the tool automatically applies the absolute value to the scalar triple product result.
  2. Step-by-step analytical method: In the section below, you can follow the algebraic step-by-step process. You will see how the system assembles the determinant of a 3x3 matrix by placing the components of vector a in the first row, followed by those of vector b in the second, and those of vector c in the third. Next, it will show the cofactor expansion (solving the 2x2 subdeterminants for each term in the top row), adding and subtracting the components until reaching the final scalar number.

Solved Exercises

The following are examples of problems solved by the calculator.

Calculate the scalar triple product of the three vectors \(\vec{a}=\langle 1, -2, 3\rangle\), \(\vec{b}=\langle 0, 4, -1\rangle\), and \(\vec{c}=\langle 2, 1, 5\rangle.\)

Result

The scalar triple product of the given vectors is:

$$ \displaystyle \vec{a} \cdot (\vec{b} \times \vec{c}) = 1 $$

Volume of the parallelepiped: \( \displaystyle |\vec{a} \cdot (\vec{b} \times \vec{c})| = 1 \text{ u}^3 \)

Step-by-step resolution

1. Identify the vectors.

The vectors to work with are:

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

2. Formulate the general determinant.

We express the scalar triple product using a 3x3 determinant, where the rows correspond to the components of the ordered vectors a, b, and c.

$$ \displaystyle \vec{a} \cdot (\vec{b} \times \vec{c}) = \begin{vmatrix} a_x & a_y & a_z \\[0.3em] b_x & b_y & b_z \\[0.3em] c_x & c_y & c_z \end{vmatrix} $$

3. Substitute the vector components.

We replace the entered vector values into the determinant.

$$ \displaystyle \vec{a} \cdot (\vec{b} \times \vec{c}) = \begin{vmatrix} 1 & -2 & 3 \\[0.5em] 0 & 4 & -1 \\[0.5em] 2 & 1 & 5 \end{vmatrix} $$

4. Expand the determinant by minors.

We expand the determinant along the first row. We multiply each element of the row by its respective 2x2 minor determinant. Remember to alternate the signs of the terms (positive, negative, positive).

$$ \displaystyle \vec{a} \cdot (\vec{b} \times \vec{c}) = (1) \begin{vmatrix} 4 & -1 \\[0.5em] 1 & 5 \end{vmatrix} - (-2) \begin{vmatrix} 0 & -1 \\[0.5em] 2 & 5 \end{vmatrix} + (3) \begin{vmatrix} 0 & 4 \\[0.5em] 2 & 1 \end{vmatrix} $$

5. Solve and simplify.

We calculate the 2x2 determinants by cross-multiplying and subtracting their products, and finally, we multiply by the outer component.

First term:

$$ \displaystyle (1) \begin{vmatrix} 4 & -1 \\[0.5em] 1 & 5 \end{vmatrix} = (1) [(4)(5) - (-1)(1)] \\[1em] = (1) [20 - (-1)] = 21 $$

Second term:

It is important not to forget the negative sign that affects this term due to the expansion rule.

$$ \displaystyle - (-2) \begin{vmatrix} 0 & -1 \\[0.5em] 2 & 5 \end{vmatrix} = - (-2) [(0)(5) - (-1)(2)] \\[1em] = - (-2) [0 - (-2)] = 4 $$

Third term:

$$ \displaystyle (3) \begin{vmatrix} 0 & 4 \\[0.5em] 2 & 1 \end{vmatrix} = (3) [(0)(1) - (4)(2)] \\[1em] = (3) [0 - 8] = -24 $$

6. Add the terms

Now that we have the three terms, we add them together to get the final result of the scalar triple product:

$$ \displaystyle \vec{a} \cdot (\vec{b} \times \vec{c}) = 21 + 4 - 24 = 1 $$
Find the result of the scalar triple product \(\langle 2, 0, 0\rangle \cdot (\langle 0, 3, 0\rangle \times \langle 0, 0, 4\rangle).\)

Result

The scalar triple product of the given vectors is:

$$ \displaystyle \vec{a} \cdot (\vec{b} \times \vec{c}) = 24 $$

Volume of the parallelepiped: \( \displaystyle |\vec{a} \cdot (\vec{b} \times \vec{c})| = 24 \text{ u}^3 \)

Step-by-step resolution

1. Identify the vectors.

The vectors to work with are:

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

2. Formulate the general determinant.

We express the scalar triple product using a 3x3 determinant, where the rows correspond to the components of the ordered vectors a, b, and c.

$$ \displaystyle \vec{a} \cdot (\vec{b} \times \vec{c}) = \begin{vmatrix} a_x & a_y & a_z \\[0.3em] b_x & b_y & b_z \\[0.3em] c_x & c_y & c_z \end{vmatrix} $$

3. Substitute the vector components.

We replace the entered vector values into the determinant.

$$ \displaystyle \vec{a} \cdot (\vec{b} \times \vec{c}) = \begin{vmatrix} 2 & 0 & 0 \\[0.5em] 0 & 3 & 0 \\[0.5em] 0 & 0 & 4 \end{vmatrix} $$

4. Expand the determinant by minors.

We expand the determinant along the first row. We multiply each element of the row by its respective 2x2 minor determinant. Remember to alternate the signs of the terms (positive, negative, positive).

$$ \displaystyle \vec{a} \cdot (\vec{b} \times \vec{c}) = (2) \begin{vmatrix} 3 & 0 \\[0.5em] 0 & 4 \end{vmatrix} - (0) \begin{vmatrix} 0 & 0 \\[0.5em] 0 & 4 \end{vmatrix} + (0) \begin{vmatrix} 0 & 3 \\[0.5em] 0 & 0 \end{vmatrix} $$

5. Solve and simplify.

We calculate the 2x2 determinants by cross-multiplying and subtracting their products, and finally, we multiply by the outer component.

First term:

$$ \displaystyle (2) \begin{vmatrix} 3 & 0 \\[0.5em] 0 & 4 \end{vmatrix} = (2) [(3)(4) - (0)(0)] \\[1em] = (2) [12 - 0] = 24 $$

Second term:

It is important not to forget the negative sign that affects this term due to the expansion rule.

$$ \displaystyle - (0) \begin{vmatrix} 0 & 0 \\[0.5em] 0 & 4 \end{vmatrix} = - (0) [(0)(4) - (0)(0)] \\[1em] = - (0) [0 - 0] = 0 $$

Third term:

$$ \displaystyle (0) \begin{vmatrix} 0 & 3 \\[0.5em] 0 & 0 \end{vmatrix} = (0) [(0)(0) - (3)(0)] \\[1em] = (0) [0 - 0] = 0 $$

6. Add the terms

Now that we have the three terms, we add them together to get the final result of the scalar triple product:

$$ \displaystyle \vec{a} \cdot (\vec{b} \times \vec{c}) = 24 + 0 + 0 = 24 $$
Find the product of the three vectors \(\langle 1/2, 1, -3/4\rangle \cdot (\langle 2, 5/2, 1\rangle \times \langle 3,-1/3, 2/5\rangle).\)

Result

The scalar triple product of the given vectors is:

$$ \displaystyle \vec{a} \cdot (\vec{b} \times \vec{c}) = \frac{1151}{120} \approx 9.59 $$

Volume of the parallelepiped: \( \displaystyle |\vec{a} \cdot (\vec{b} \times \vec{c})| = \frac{1151}{120} \approx 9.59 \text{ u}^3 \)

Step-by-step resolution

1. Identify the vectors.

The vectors to work with are:

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

2. Formulate the general determinant.

We express the scalar triple product using a 3x3 determinant, where the rows correspond to the components of the ordered vectors a, b, and c.

$$ \displaystyle \vec{a} \cdot (\vec{b} \times \vec{c}) = \begin{vmatrix} a_x & a_y & a_z \\[0.3em] b_x & b_y & b_z \\[0.3em] c_x & c_y & c_z \end{vmatrix} $$

3. Substitute the vector components.

We replace the entered vector values into the determinant.

$$ \displaystyle \vec{a} \cdot (\vec{b} \times \vec{c}) = \begin{vmatrix} \frac{1}{2} & 1 & -\frac{3}{4} \\[0.5em] -2 & \frac{5}{2} & 1 \\[0.5em] 3 & -\frac{1}{3} & \frac{2}{5} \end{vmatrix} $$

4. Expand the determinant by minors.

We expand the determinant along the first row. We multiply each element of the row by its respective 2x2 minor determinant. Remember to alternate the signs of the terms (positive, negative, positive).

$$ \displaystyle \vec{a} \cdot (\vec{b} \times \vec{c}) = \left(\frac{1}{2}\right) \begin{vmatrix} \frac{5}{2} & 1 \\[0.5em] -\frac{1}{3} & \frac{2}{5} \end{vmatrix} - (1) \begin{vmatrix} -2 & 1 \\[0.5em] 3 & \frac{2}{5} \end{vmatrix} + \left(-\frac{3}{4}\right) \begin{vmatrix} -2 & \frac{5}{2} \\[0.5em] 3 & -\frac{1}{3} \end{vmatrix} $$

5. Solve and simplify.

We calculate the 2x2 determinants by cross-multiplying and subtracting their products, and finally, we multiply by the outer component.

First term:

$$ \displaystyle \left(\frac{1}{2}\right) \begin{vmatrix} \frac{5}{2} & 1 \\[0.5em] -\frac{1}{3} & \frac{2}{5} \end{vmatrix} = \left(\frac{1}{2}\right) \left[\left(\frac{5}{2}\right)\left(\frac{2}{5}\right) - (1)\left(-\frac{1}{3}\right)\right] \\[1em] = \left(\frac{1}{2}\right) \left[1 - \left(-\frac{1}{3}\right)\right] = \frac{2}{3} $$

Second term:

It is important not to forget the negative sign that affects this term due to the expansion rule.

$$ \displaystyle - (1) \begin{vmatrix} -2 & 1 \\[0.5em] 3 & \frac{2}{5} \end{vmatrix} = - (1) \left[(-2)\left(\frac{2}{5}\right) - (1)(3)\right] \\[1em] = - (1) \left[-\frac{4}{5} - 3\right] = \frac{19}{5} $$

Third term:

$$ \displaystyle \left(-\frac{3}{4}\right) \begin{vmatrix} -2 & \frac{5}{2} \\[0.5em] 3 & -\frac{1}{3} \end{vmatrix} = \left(-\frac{3}{4}\right) \left[(-2)\left(-\frac{1}{3}\right) - \left(\frac{5}{2}\right)(3)\right] \\[1em] = \left(-\frac{3}{4}\right) \left[\frac{2}{3} - \frac{15}{2}\right] = \frac{41}{8} $$

6. Add the terms

Now that we have the three terms, we add them together to get the final result of the scalar triple product:

$$ \displaystyle \vec{a} \cdot (\vec{b} \times \vec{c}) = \frac{2}{3} + \frac{19}{5} + \frac{41}{8} = \frac{1151}{120} \approx 9.59 $$
Find the result of the scalar triple product a · (b × c) where \(\vec{a}=\langle -5, -3, -2\rangle,\) \(\vec{b}=\langle -1, -4, -6\rangle\), and \(\vec{c}=\langle -7, -2, -1\rangle.\)

Result

The scalar triple product of the given vectors is:

$$ \displaystyle \vec{a} \cdot (\vec{b} \times \vec{c}) = -31 $$

Volume of the parallelepiped: \( \displaystyle |\vec{a} \cdot (\vec{b} \times \vec{c})| = 31 \text{ u}^3 \)

Step-by-step resolution

1. Identify the vectors.

The vectors to work with are:

$$ \displaystyle \vec{a} = \langle -5, -3, -2 \rangle \\[1em] \vec{b} = \langle -1, -4, -6 \rangle \\[1em] \vec{c} = \langle -7, -2, -1 \rangle $$

2. Formulate the general determinant.

We express the scalar triple product using a 3x3 determinant, where the rows correspond to the components of the ordered vectors a, b, and c.

$$ \displaystyle \vec{a} \cdot (\vec{b} \times \vec{c}) = \begin{vmatrix} a_x & a_y & a_z \\[0.3em] b_x & b_y & b_z \\[0.3em] c_x & c_y & c_z \end{vmatrix} $$

3. Substitute the vector components.

We replace the entered vector values into the determinant.

$$ \displaystyle \vec{a} \cdot (\vec{b} \times \vec{c}) = \begin{vmatrix} -5 & -3 & -2 \\[0.5em] -1 & -4 & -6 \\[0.5em] -7 & -2 & -1 \end{vmatrix} $$

4. Expand the determinant by minors.

We expand the determinant along the first row. We multiply each element of the row by its respective 2x2 minor determinant. Remember to alternate the signs of the terms (positive, negative, positive).

$$ \displaystyle \vec{a} \cdot (\vec{b} \times \vec{c}) = (-5) \begin{vmatrix} -4 & -6 \\[0.5em] -2 & -1 \end{vmatrix} - (-3) \begin{vmatrix} -1 & -6 \\[0.5em] -7 & -1 \end{vmatrix} + (-2) \begin{vmatrix} -1 & -4 \\[0.5em] -7 & -2 \end{vmatrix} $$

5. Solve and simplify.

We calculate the 2x2 determinants by cross-multiplying and subtracting their products, and finally, we multiply by the outer component.

First term:

$$ \displaystyle (-5) \begin{vmatrix} -4 & -6 \\[0.5em] -2 & -1 \end{vmatrix} = (-5) [(-4)(-1) - (-6)(-2)] \\[1em] = (-5) [4 - 12] = 40 $$

Second term:

It is important not to forget the negative sign that affects this term due to the expansion rule.

$$ \displaystyle - (-3) \begin{vmatrix} -1 & -6 \\[0.5em] -7 & -1 \end{vmatrix} = - (-3) [(-1)(-1) - (-6)(-7)] \\[1em] = - (-3) [1 - 42] = -123 $$

Third term:

$$ \displaystyle (-2) \begin{vmatrix} -1 & -4 \\[0.5em] -7 & -2 \end{vmatrix} = (-2) [(-1)(-2) - (-4)(-7)] \\[1em] = (-2) [2 - 28] = 52 $$

6. Add the terms

Now that we have the three terms, we add them together to get the final result of the scalar triple product:

$$ \displaystyle \vec{a} \cdot (\vec{b} \times \vec{c}) = 40 - 123 + 52 = -31 $$

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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: (1 week ago)