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Vector product

Three vectors $$(\uvec,\vvec,\wvec)$$ in three-dimensional space are positively oriented if their directions obey the right-hand rule: if the index finger of the right hand points in the direction of $$\uvec$$, the middle finger in the direction of $$\vvec$$, then the thumb points in the direction of $$\wvec$$.

For instance, the basis vectors $$(\ivec,\jvec,\kvec)$$ are positively oriented, while $$(\ivec,\kvec,\jvec)$$ are negatively oriented.

The right-hand rule
The right-hand rule

The vector product, or cross-product, of two vectors $$\uvec$$ and $$\vvec$$ is only defined in three-dimensional space. It is a vector, denoted by $$\uvec\times\vvec$$, with the following properties.

  • $$\uvec\times\vvec$$ is normal to $$\uvec$$ and $$\vvec$$.
  • $$(\uvec,\vvec,\uvec\times\vvec)$$ is positively oriented.
  • Its length is determined by the angle $$\theta$$ between $$\uvec$$ and $$\vvec$$ and is given by $$$ \vert \uvec\times\vvec \vert = \sin \theta\vert\uvec\vert \vert\vvec\vert $$$
The cross-product
The cross-product

In a Cartesian coordinate system, the vector product is given by: \begin{align*} (x_1,y_1,z_1) & \times(x_2,y_2,z_2) =\Big(\begin{vmatrix} y_1& z_1\\ y_2& z_2\end{vmatrix}, \begin{vmatrix} z_1& x_1\\ z_2& x_2\end{vmatrix}, \begin{vmatrix} x_1& y_1\\ x_2& y_2\end{vmatrix}\Big)\\ & = (y_1z_2-z_1y_2, z_1x_2-x_1z_2,x_1y_2-y_1x_2) \end{align*}

For instance, the $$x$$ coordinate corresponds to the determinant of the $$y$$ and $$z$$ coordinates. The determinant is defined as the quantity $$$ \begin{vmatrix}a& b\\ c& d\end{vmatrix} = ad - bc.$$$

If $$\ivec$$, $$\jvec$$, $$\kvec$$ is a standard basis, we have $$$ \ivec\times\jvec = \kvec,\quad \jvec\times\kvec = \ivec ,\quad \kvec\times\ivec = \jvec $$$

Two non-zero vectors are collinear if and only if their vector product is the null vector.

Two vectors are orthogonal if and only if $$\vert \uvec\times\vvec \vert = \vert\uvec\vert \vert\vvec\vert$$.

The main properties of the vector product are:

  • Anti-symmetry: $$ \vvec\times\uvec = -\uvec\times \vvec $$
  • Bilinearity: \begin{gather*} (\lambda_1\uvec_1+\lambda_2\uvec_2) \times\vvec = \lambda_1\uvec_1 \times \vvec +\lambda_2\uvec_2 \times \vvec\\ \uvec\times(\lambda_1\vvec_1+\lambda_2\vvec_2)= \lambda_1\uvec\times \vvec_1+\lambda_2\uvec\times \vvec_2. \end{gather*}

If $$\uvec$$ and $$\vvec$$ are two non collinear vectors, then the area of the parallelogram with sides $$\uvec$$ and $$\vvec$$ is $$\vert\uvec\times\vvec\vert$$.

The area of the triangle with sides $$\uvec$$ and $$\vvec$$ is $$$\frac{1}{2}\vert\uvec\times\vvec\vert.$$$

Proof We assume that the vectors are unit vectors (otherwise we divide them by their length). The area of the triangle is the sum of two right angled triangles, with one common side of length $$\sin\theta$$.

As the area of a right angled triangle is half the area of the rectangle, we conclude that the area of our triangle is $$$ \frac{1}{2}\sin\theta \big(\cos\theta + (1-\cos\theta)\big) = \frac{1}{2}\sin\theta.$$$

The area of the parallelogram is twice the area of the triangle: $$\sin\theta$$.

Area of a parallelogram and of a triangle
Area of a parallelogram and of a triangle