Cross Product (3D)
Vectors
Intro: We expand the determinant and simplify components; we also return $\|\vec{a}\times\vec{b}\|$ and the parallelogram area.
Worked example
- a=(1,2,3), b=(4,−5,6)
- Write component form: $$\vec{a}\times\vec{b} = (a_2 b_3 - a_3 b_2,\; a_3 b_1 - a_1 b_3,\; a_1 b_2 - a_2 b_1).$$
- Substitute $\vec{a}=(1,2,3)$ and $\vec{b}=(4,-5,6)$:
- • $x$-component: $a_2 b_3 - a_3 b_2 = 2\cdot 6 - 3\cdot(-5) = 12 + 15 = 27$.
- • $y$-component: $a_3 b_1 - a_1 b_3 = 3\cdot 4 - 1\cdot 6 = 12 - 6 = 6$.
- • $z$-component: $a_1 b_2 - a_2 b_1 = 1\cdot(-5) - 2\cdot 4 = -5 - 8 = -13$.
- Therefore: $$\boxed{\vec{a}\times\vec{b}=(27,\,6,\,-13)}.$$
- Magnitude (parallelogram area): $$\|\vec{a}\times\vec{b}\| = \sqrt{27^2 + 6^2 + (-13)^2} = \sqrt{729 + 36 + 169} = \sqrt{934} \approx 30.56.$$
- Orthogonality checks (dot with inputs should be 0):
- • $(27,6,-13)\cdot(1,2,3)=27+12-39=0$ ✔️
- • $(27,6,-13)\cdot(4,-5,6)=108-30-78=0$ ✔️
- Angle relation (optional): $$\|\vec{a}\times\vec{b}\|=\|\vec{a}\|\,\|\vec{b}\|\sin\theta.$$ With $\|\vec{a}\|=\sqrt{14}$ and $\|\vec{b}\|=\sqrt{77}$, $$\sin\theta=\frac{\sqrt{934}}{\sqrt{14\cdot 77}}=\frac{\sqrt{934}}{\sqrt{1078}}\approx 0.9180\Rightarrow \theta\approx 66.9^\circ.$$
FAQs
2D vectors?
Embed as 3D with zero z-components: (x,y) → (x,y,0). The result points along ±k and its magnitude equals the signed area magnitude |x1 y1; x2 y2|.
Geometric meaning?
Direction is perpendicular to both a and b (right-hand rule). Magnitude equals the area of the parallelogram spanned by a and b.
Zero cross product?
If a and b are parallel or one is the zero vector, then a×b = (0,0,0).
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How this calculator works
- Type or paste your function (LaTeX like
\sin,\lnworks too). - Press Generate a practice question button to generate the derivative and the full reasoning.
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