\langle 1, 0, 4 \rangle \times \langle 2, -1, 3 \rangle = \langle (0)(3) - (4)(-1), -[(1)(3) - (4)(2)], (1)(-1) - (0)(2) \rangle = \langle 0 + 4, -(3 - 8), -1 - 0 \rangle = \langle 4, 5, -1 \rangle - RoadRUNNER Motorcycle Touring & Travel Magazine

Understanding Cross Products in 3D Space: A Step-by-Step Calculation of ⟨1, 0, 4⟩ × ⟨2, −1, 3⟩

The cross product of two vectors in three-dimensional space is a fundamental operation in linear algebra, physics, and engineering. Despite its seemingly abstract appearance, the cross product produces another vector perpendicular to the original two. This article explains how to compute the cross product of the vectors ⟨1, 0, 4⟩ and ⟨2, −1, 3⟩ using both algorithmic step-by-step methods and component-wise formulas—ultimately revealing why ⟨1, 0, 4⟩ × ⟨2, −1, 3⟩ = ⟨4, 5, −1⟩.

Understanding the Context

What Is a Cross Product?

Given two vectors a = ⟨a₁, a₂, a₃⟩ and b = ⟨b₁, b₂, b₃⟩ in ℝ³, their cross product a × b is defined as:

⟨a₂b₃ − a₃b₂,
−(a₁b₃ − a₃b₁),
a₁b₂ − a₂b₁⟩

This vector is always orthogonal to both a and b, and its magnitude equals the area of the parallelogram formed by a and b.

$\langle 1, 0, 4 \rangle \times \langle 2, -1, 3 \rangle = \langle (0)(3) - (4)(-1), -[(1)(3) - (4)(2)], (1)(-1) - (0)(2) \rangle = \langle 0 + 4, -(3 - 8), -1 - 0 \rangle = \langle 4, 5, -1 \rangle$

Key Insights

Applying the Formula to ⟨1, 0, 4⟩ × ⟨2, −1, 3⟩

Let a = ⟨1, 0, 4⟩ and b = ⟨2, −1, 3⟩.

Using the standard cross product formula:

Step 1: Compute the first component
(0)(3) − (4)(−1) = 0 + 4 = 4

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Final Thoughts

Step 2: Compute the second component
−[(1)(3) − (4)(2)] = −[3 − 8] = −[−5] = 5

Step 3: Compute the third component
(1)(−1) − (0)(2) = −1 − 0 = −1

Putting it all together:
⟨1, 0, 4⟩ × ⟨2, −1, 3⟩ = ⟨4, 5, −1⟩

Why Does This Work? Intuition Behind the Cross Product

The cross product’s components follow the determinant of a 3×3 matrix with unit vectors and the vector components:

⟨i, j, k⟩
| 1 0 4
|² −1 3

Expanding the determinant:

i-component: (0)(3) − (4)(−1) = 0 + 4 = 4
j-component: −[(1)(3) − (4)(2)] = −[3 − 8] = 5
k-component: (1)(−1) − (0)(2) = −1 − 0 = −1

This confirms that the formula used is equivalent to the cofactor expansion method, validating the result.