The Cross Product and the Triple Product: Quick Revision

A brief note on The Cross Product and the Triple Product. Read it before a test. Try each question at the end before you open its answer.

The cross product

• Definition: $\vec a\times\vec b = \lvert\vec a\rvert\lvert\vec b\rvert\sin\theta\ \hat n$, with $\hat n$ at right angles to both, by the right-hand rule.
• Components: the determinant with rows $\hat i\ \hat j\ \hat k$, $\vec a$ and $\vec b$. Mind the minus sign of the $\hat j$ term.
• Rules: $\vec b\times\vec a = -\vec a\times\vec b$, $\vec a\times\vec a = \vec 0$, and the product is zero for parallel vectors.
• Not associative: $(\vec a\times\vec b)\times\vec c \ne \vec a\times(\vec b\times\vec c)$.
• Notation: $\times$ in NCERT and IB. Some UK boards write $\wedge$.

Areas

• Parallelogram: $\lvert\vec a\times\vec b\rvert$. Triangle: $\dfrac{1}{2}\lvert\overrightarrow{AB}\times\overrightarrow{AC}\rvert$.
• Diagonals: $\dfrac{1}{2}\lvert\vec d_1\times\vec d_2\rvert$.
• Collinear points: $\overrightarrow{AB}\times\overrightarrow{AC} = \vec 0$.
• Lagrange: $\lvert\vec a\times\vec b\rvert^2 = \lvert\vec a\rvert^2\lvert\vec b\rvert^2 - (\vec a\cdot\vec b)^2$.

The triple product

• Definition: $[\vec a\ \vec b\ \vec c] = \vec a\cdot(\vec b\times\vec c)$, the determinant of the three rows.
• Volume: parallelepiped $\lvert[\vec a\ \vec b\ \vec c]\rvert$, tetrahedron one sixth of that.
• Rules: cyclic order keeps the value, swapping two vectors changes the sign.
• Coplanar: $[\vec a\ \vec b\ \vec c] = 0$. For four points, use $[\overrightarrow{AB}\ \overrightarrow{AC}\ \overrightarrow{AD}]$.
• Vector triple product: $\vec a\times(\vec b\times\vec c) = \vec b(\vec a\cdot\vec c) - \vec c(\vec a\cdot\vec b)$.

A worked example

Let $\vec a = 2\hat i - \hat k$, $\vec b = \hat i + 3\hat j + 2\hat k$ and $\vec c = \hat i + \hat j + \hat k$. Find the area of the triangle on $\vec a$ and $\vec b$, and $[\vec a\ \vec b\ \vec c]$.

1. The cross product. $\vec a\times\vec b = (0\cdot 2 - (-1)(3))\hat i - (2\cdot 2 - (-1)(1))\hat j + (2\cdot 3 - 0)\hat k = 3\hat i - 5\hat j + 6\hat k$.
2. The area. The size is $\sqrt{9 + 25 + 36} = \sqrt{70}$, so the triangle has area $\dfrac{\sqrt{70}}{2} \approx 4.18$.
3. The triple product. $\vec b\times\vec c = \hat i + \hat j - 2\hat k$, and $\vec a\cdot(\vec b\times\vec c) = 2 + 0 + 2 = 4$.

Practice with new numbers

A program to try

The program checks the worked example.

Mistakes to avoid

• Using the left hand.
• Forgetting the half for a triangle, or the sixth for a tetrahedron.
• Dropping the minus sign of the $\hat j$ term.
• Reading a negative triple product as a negative volume.

Check yourself

1. Find $(\hat i + \hat j)\times\hat j$.

2. Find the area of the triangle with the vertices $(0, 0, 0)$, $(1, 2, 2)$ and $(2, 1, -2)$.

3. Find $[\hat i\ \ \hat i + \hat j\ \ \hat i + \hat j + \hat k]$.

= 1$.

4. Are $(1, 0, 1)$, $(0, 1, 1)$ and $(1, 1, 2)$ coplanar?

5. If $\lvert\vec a\rvert = 3$, $\lvert\vec b\rvert = 4$ and $\vec a\cdot\vec b = 0$, find $\lvert\vec a\times\vec b\rvert$.

Sources

• NCERT, Mathematics Part II for Class XII, Chapter 10, section on the vector product of two vectors.
• OpenStax, Calculus Volume 3, Chapter 2, section on the cross product.