Vector algebra is the algebra of operations on vectors. You can add two vectors, or scale one by a number. Past that, there are two genuinely different ways to multiply two vectors together. Those two ways are this section's geometric products: the dot product and the cross product. Both reappear constantly in this course. The dot product becomes the model for every inner product this course meets. The cross product becomes the seed of the wedge product in chapter 06.
Vector algebra is just arithmetic for arrows instead of numbers. You can add two arrows tip to tail. You can stretch one arrow longer or shorter. Past that, there are two geometric products. Each one combines two arrows into something new — either a plain number or a third arrow.
The dot product is like a shadow. It measures how much of one vector points along another. A stick's shadow does the same job: it shows how much of the stick points along the ground.
The cross product is like a flagpole. It stands straight up out of the flat sheet the two vectors span. It points perpendicular to both. Its length equals how much of that sheet the two vectors actually enclose.
The geometric products are the dot product and the cross product. They are the two ways vector algebra multiplies two vectors together. The dot product \mathbf{u}\cdot\mathbf{v}=u_1v_1+u_2v_2+u_3v_3 answers one question: how much do these two vectors agree in direction? It hands back a plain number. The cross product \mathbf{u}\times\mathbf{v} answers a different question: what direction is perpendicular to both? It hands back another vector instead, computed component by component from \mathbf{u}=\langle u_1,u_2,u_3\rangle and \mathbf{v}=\langle v_1,v_2,v_3\rangle.
The two products fail in opposite ways. \mathbf{u}\cdot\mathbf{v}=0 signals perpendicularity. That is the richest special case of the dot product. \mathbf{u}\times\mathbf{v}=\mathbf{0} signals parallelism instead. That is the cross product's own special case: two parallel vectors span no single perpendicular direction, only a whole plane's worth of them. Confusing these two zero conditions is the most common vector-algebra error carried into this course.
Theorem
The Dot Product in 3D
also known as: the scalar product
Recalled from Calculus III — Multivariable Calculus · The dot product — first taught there.
For vectors \mathbf{u}=\langle u_1,u_2,u_3\rangle and \mathbf{v}=\langle v_1,v_2,v_3\rangle with angle \theta between them,
\mathbf{u}\cdot\mathbf{v} = u_1v_1+u_2v_2+u_3v_3 = |\mathbf{u}|\,|\mathbf{v}|\cos\theta.Draw \mathbf{u} and \mathbf{v} from a common point O, with tips at U and V. The triangle OUV has two sides |\mathbf{u}|, |\mathbf{v}| meeting at angle \theta, and third side |\mathbf{u}-\mathbf{v}|, so the Law of Cosines gives
|\mathbf{u}-\mathbf{v}|^2 = |\mathbf{u}|^2+|\mathbf{v}|^2-2|\mathbf{u}||\mathbf{v}|\cos\theta.Expanding the left side component-by-component, \mathbf{u}-\mathbf{v}=\langle u_1-v_1,\,u_2-v_2,\,u_3-v_3\rangle, so
|\mathbf{u}-\mathbf{v}|^2 = \sum_i (u_i-v_i)^2 = \sum_i u_i^2+\sum_i v_i^2 - 2\sum_i u_iv_i = |\mathbf{u}|^2+|\mathbf{v}|^2-2\sum_i u_iv_i.Both expressions equal |\mathbf{u}-\mathbf{v}|^2; setting them equal, canceling |\mathbf{u}|^2+|\mathbf{v}|^2 from both sides, and dividing by -2 leaves \sum_i u_iv_i = |\mathbf{u}||\mathbf{v}|\cos\theta, which is exactly \mathbf{u}\cdot\mathbf{v}=u_1v_1+u_2v_2+u_3v_3 on the left.
Formula
The Cross Product
also known as: the vector product
Recalled from Calculus III — Multivariable Calculus · The cross product — first taught there.
For \mathbf{u}=\langle u_1,u_2,u_3\rangle and \mathbf{v}=\langle v_1,v_2,v_3\rangle,
\mathbf{u}\times\mathbf{v} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\ u_1&u_2&u_3\\ v_1&v_2&v_3\end{vmatrix} = \langle u_2v_3-u_3v_2,\ u_3v_1-u_1v_3,\ u_1v_2-u_2v_1\rangle,a new vector perpendicular to both \mathbf{u} and \mathbf{v}, with magnitude |\mathbf{u}\times\mathbf{v}| = |\mathbf{u}||\mathbf{v}|\sin\theta, which is the area of the parallelogram \mathbf{u} and \mathbf{v} span.
Direct computation confirms perpendicularity: expanding the dot product \mathbf{u}\cdot(\mathbf{u}\times\mathbf{v}) component-by-component gives u_1(u_2v_3-u_3v_2)+u_2(u_3v_1-u_1v_3)+u_3(u_1v_2-u_2v_1), and every term cancels against another, leaving 0; the same cancellation shows \mathbf{v}\cdot(\mathbf{u}\times\mathbf{v})=0.
For the magnitude, Lagrange's identity states |\mathbf{u}\times\mathbf{v}|^2 = |\mathbf{u}|^2|\mathbf{v}|^2 - (\mathbf{u}\cdot\mathbf{v})^2 (both sides expand to the same sum of squared component products — a direct algebraic check). Substituting \mathbf{u}\cdot\mathbf{v}=|\mathbf{u}||\mathbf{v}|\cos\theta:
|\mathbf{u}\times\mathbf{v}|^2 = |\mathbf{u}|^2|\mathbf{v}|^2 - |\mathbf{u}|^2|\mathbf{v}|^2\cos^2\theta = |\mathbf{u}|^2|\mathbf{v}|^2(1-\cos^2\theta) = |\mathbf{u}|^2|\mathbf{v}|^2\sin^2\theta.Since 0\le\theta\le\pi makes \sin\theta\ge0, taking square roots gives |\mathbf{u}\times\mathbf{v}|=|\mathbf{u}||\mathbf{v}|\sin\theta.
Dot product goes to zero exactly at perpendicular
Rotate v until the dot-product readout crosses zero -- that's the instant u and v are perpendicular.
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Cross product goes to zero exactly at parallel
Set v to a scalar multiple of u -- the cross-product readout collapses to the zero vector.
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Worked example
Both products, one pair of vectors
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1
For \mathbf{u}=\langle 2,1\rangle and \mathbf{v}=\langle 1,-2\rangle in the plane, compute \mathbf{u}\cdot\mathbf{v}, then find \mathbf{u}\times\mathbf{v} by treating both as 3D vectors with a zero third component.
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2
\mathbf{u}\cdot\mathbf{v} = (2)(1)+(1)(-2) = 2-2=0 — the two vectors are perpendicular.
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3
With \mathbf{u}=\langle 2,1,0\rangle, \mathbf{v}=\langle 1,-2,0\rangle: the \mathbf{k} component is (2)(-2)-(1)(1)=-5, and the \mathbf{i}, \mathbf{j} components are both 0 (every z-component was 0).
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4
\boxed{\mathbf{u}\cdot\mathbf{v}=0,\quad \mathbf{u}\times\mathbf{v}=\langle 0,0,-5\rangle} — perpendicular vectors in a plane cross to a vector straight out of that plane, exactly as the geometry predicts.
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Where this goes next
Chapters 5 and 6 of this course
The dot product is the specific case n=3 of a much more general idea, the "inner product." This course never needs to name that general idea formally. But every symmetric-tensor calculation in chapter 05 relies on the same basic pattern: multiply matching pieces, then add. The cross product is different. It is special to three dimensions in a way the dot product isn't. Chapter 06 replaces it with the wedge product. A "cross product" doesn't exist at all in four or more dimensions.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Calculus III — Multivariable Calculus
Find \mathbf{u}\cdot\mathbf{v} for \mathbf{u}=\langle 1,2,3\rangle, \mathbf{v}=\langle 4,-1,2\rangle.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
(1)(4)+(2)(-1)+(3)(2)=4-2+6=8.
Correct answer
Also accepted:
Why
From Calculus III — Multivariable Calculus
Determine the angle (in degrees) between \mathbf{u}=\langle 1,1,1\rangle and \mathbf{v}=\langle -1,-1,-1\rangle.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
\mathbf{v}=-\mathbf{u}, so they point in exactly opposite directions: \theta=180°.
Correct answer
Also accepted:
Why
Notice that \mathbf{v} is a negative scalar multiple of \mathbf{u} before computing anything.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Calculus III — Multivariable Calculus
Compute \mathbf{u}\times\mathbf{v} for \mathbf{u}=\langle 1,2,0\rangle, \mathbf{v}=\langle 0,1,3\rangle.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Enter as components — <2, -1, 3>, (2, -1, 3), or 2i - j + 3k all count. Any algebraically equivalent component counts too.
Why
\langle (2)(3)-(0)(1),\,(0)(0)-(1)(3),\,(1)(1)-(2)(0)\rangle=\langle 6,-3,1\rangle.
Correct answer
Also accepted:
Why
Use the determinant formula, one component at a time.
From Calculus III — Multivariable Calculus
What kind of object does \mathbf{u}\times\mathbf{v} produce?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
That's the dot product's output -- the cross product returns another vector.
Why
The cross product returns a new vector perpendicular to both inputs, unlike the dot product's scalar output.