Dot, cross, or triple product — which one answers the question?
Three products, three completely different kinds of answer.
Ask what shape the answer needs to be. A single NUMBER (an angle, a work value, a length along a direction) is the dot product. Another VECTOR, perpendicular to both inputs (a normal to a plane, a torque) is the cross product. A signed VOLUME of the box three vectors span is the triple product — and its zero case is the fastest coplanarity test there is.
Scalar answers — the dot product
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.Example
Show proof
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.
Term
Dot product
For vectors \mathbf{u} and \mathbf{v} with \mathbf{u} \neq \mathbf{0}, the projection of \mathbf{v} onto \mathbf{u} is
\text{proj}_{\mathbf{u}}\mathbf{v} = \left(\frac{\mathbf{v}\cdot\mathbf{u}}{|\mathbf{u}|^2}\right)\mathbf{u}.Built directly on the dot product — the "how much of u points along v" question.
Show proof
Drop a perpendicular from the tip of \mathbf{v} down onto the line through \mathbf{u}; this forms a right triangle with hypotenuse |\mathbf{v}| and angle \theta between \mathbf{u} and \mathbf{v}. Exactly as in Chapter 8, the leg lying along \mathbf{u}'s line — the scalar projection — has length |\mathbf{v}|\cos\theta (adjacent equals hypotenuse times cosine). From the theorem above, \cos\theta = \frac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{u}|\,|\mathbf{v}|}, so the scalar projection simplifies to
|\mathbf{v}|\cos\theta = |\mathbf{v}|\cdot\frac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{u}|\,|\mathbf{v}|} = \frac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{u}|}.The vector projection points along \mathbf{u}'s own direction, so multiply this scalar length by the unit vector \frac{\mathbf{u}}{|\mathbf{u}|}:
\text{proj}_{\mathbf{u}}\mathbf{v} = \frac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{u}|}\cdot\frac{\mathbf{u}}{|\mathbf{u}|} = \frac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{u}|^2}\mathbf{u}.Vector answers — the cross product
Formula
The Cross Product
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.
Example
Show proof
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.
Volume and coplanarity — the triple product
For \mathbf{u},\mathbf{v},\mathbf{w} in \mathbb{R}^3,
\mathbf{u}\cdot(\mathbf{v}\times\mathbf{w}) = \begin{vmatrix}u_1&u_2&u_3\\ v_1&v_2&v_3\\ w_1&w_2&w_3\end{vmatrix},and |\mathbf{u}\cdot(\mathbf{v}\times\mathbf{w})| is the volume of the parallelepiped with edges \mathbf{u}, \mathbf{v}, \mathbf{w}.
Example
Show proof
\mathbf{v}\times\mathbf{w} is perpendicular to the base parallelogram spanned by \mathbf{v} and \mathbf{w}, with length equal to that base's area. Writing \mathbf{n}=\mathbf{v}\times\mathbf{w} and \hat{\mathbf{n}}=\mathbf{n}/|\mathbf{n}|, the parallelepiped's height — how far \mathbf{u} reaches in the direction perpendicular to the base — is the absolute value of the scalar projection of \mathbf{u} onto \hat{\mathbf{n}}, namely |\mathbf{u}\cdot\hat{\mathbf{n}}|. Volume is base area times height:
V = |\mathbf{n}|\cdot|\mathbf{u}\cdot\hat{\mathbf{n}}| = \left|\mathbf{u}\cdot\left(|\mathbf{n}|\hat{\mathbf{n}}\right)\right| = |\mathbf{u}\cdot(\mathbf{v}\times\mathbf{w})|.The determinant form follows from expanding \mathbf{v}\times\mathbf{w} into components and dotting with \mathbf{u} — exactly the cofactor expansion of the 3\times3 determinant along its first row.
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