Cross Product
Right-Hand Rule
A triple of non-coplanar vectors a , b and c are said to form a right-handed coordinate system if, being reduced to a common origin, these vectors are located, respectively, as the index finger , middle finger and the thumb of the right hand.
Figure 1. Definition of the Cross Product
The cross product of vector a and vector b is called vector c , denoted by c = a × b and satisfying the following requirements:
The length of vector c is equal to the product of the lengths of vectors a and b and the sine of the angle φ between them, that is
\[\Vert{\mathbf{c}}\Vert = \Vert{\mathbf{a} \times \mathbf{b}}\Vert = \Vert{\mathbf{a}}\Vert \Vert{\mathbf{b}}\Vert \sin\varphi\]
Vector c is orthogonal to each of the vectors a and b ;
Vector c is directed so that the three vectors a , b and c is right-handed .
Figure 2. Some Properties of the Cross Product
The absolute value of the cross product of a and b is equal to the area of the parallelogram spanned by these vectors:
\[A = \Vert{\mathbf{a} \times \mathbf{b}}\Vert = \Vert{\mathbf{a}}\Vert \Vert{\mathbf{b}}\Vert \sin\varphi\]
The angle between two vectors can be expressed in terms of their cross product:
\[\sin\varphi = \frac{\Vert{\mathbf{a} \times \mathbf{b}}\Vert}{\Vert{\mathbf{a}}\Vert \Vert{\mathbf{b}}\Vert}\]
The cross product is anticommutative :
\[\mathbf{a} \times \mathbf{b} = - \left({\mathbf{a} \times \mathbf{b}}\right)\]
The cross product is associative only with respect to scalar multiplication :
\[\left({\lambda\mathbf{a}}\right) \times \left({\mu\mathbf{b}}\right) = \lambda\mu\left({\mathbf{a} \times \mathbf{b}}\right),\]
where \(\lambda, \mu\) are real numbers.
Distributive property of the cross product over addition of vectors :
\[\mathbf{a} \times \left({\mathbf{b} + \mathbf{c}}\right) = \mathbf{a} \times \mathbf{b} + \mathbf{a} \times \mathbf{c}\]
The cross product of vectors a and b is the zero vector if a and b are parallel (collinear ):
\[\mathbf{a} \times \mathbf{b} = \mathbf{0} \text{ if } \mathbf{a} \parallel \mathbf{b} \left({\varphi = 0}\right)\]
Cross product of the unit vectors:
\[\mathbf{i} \times \mathbf{i} = \mathbf{j} \times \mathbf{j} = \mathbf{k} \times \mathbf{k} = \mathbf{0}\]
Cross product of distinct unit vectors:
\[\mathbf{i} \times \mathbf{j} = \mathbf{k}, \;\mathbf{j} \times \mathbf{k} = \mathbf{i},\; \mathbf{k} \times \mathbf{i} = \mathbf{j}\]
Cross Product in Coordinate Form
If \(\mathbf{a} = \left({X_1, Y_1, Z_1}\right)\) and \(\mathbf{b} = \left({X_2, Y_2, Z_2}\right),\) then the cross product of these vectors is given by
\[\mathbf{c} = \mathbf{a} \times \mathbf{b} = \left| {\begin{array}{*{20}{c}}
\mathbf{i} & \mathbf{j} & \mathbf{k}\\
{{X_1}} & {{Y_1}} & {{Z_1}}\\
{{X_2}} & {{Y_2}} & {{Z_2}}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
{{Y_1}} & {{Z_1}}\\
{{Y_2}} & {{Z_2}}
\end{array}} \right|\mathbf{i} - \left| {\begin{array}{*{20}{c}}
{{X_1}} & {{Z_1}}\\
{{X_2}} & {{Z_2}}
\end{array}} \right|\mathbf{j} + \left| {\begin{array}{*{20}{c}}
{{X_1}} & {{Y_1}}\\
{{X_2}} & {{Y_2}}
\end{array}} \right|\mathbf{k}\]
Moment of a Force (Torque)
If vector \(\mathbf{F_B}\) represents the force applied at some point B and vector \(\mathbf{r_{AB}}\) goes from some point A to point B , then the cross product \(\mathbf{M_A} = \mathbf{r_{AB}} \times \mathbf{F_B}\) represents the moment of force \(\mathbf{F_B}\) relative to point A .
\[\mathbf{M_A} = \mathbf{r_{AB}} \times \mathbf{F_B}\]
Figure 3. Solved Problems
Example 1.
Given vectors \(\mathbf{a}\left({1,2,3}\right)\) and \(\mathbf{b}\left({2,4,7}\right).\) Find the coordinates of the cross product \(\mathbf{a} \times \mathbf{b}.\)
Solution.
We determine the cross product by the formula
\[\mathbf{c} = \mathbf{a} \times \mathbf{b} = \left| {\begin{array}{*{20}{c}}
\mathbf{i} & \mathbf{j} & \mathbf{k}\\
{{X_1}} & {{Y_1}} & {{Z_1}}\\
{{X_2}} & {{Y_2}} & {{Z_2}}
\end{array}} \right|.\]
Substituting the coordinates of vectors \(\mathbf{a}\) and \(\mathbf{b},\) we get:
\[\mathbf{c} = \mathbf{a} \times \mathbf{b} = \left| {\begin{array}{*{20}{c}}
\mathbf{i} & \mathbf{j} & \mathbf{k}\\
1 & 2 & 3\\
2 & 4 & 7
\end{array}} \right| = \mathbf{i}\left| {\begin{array}{*{20}{c}}
2 & 3\\
4 & 7
\end{array}} \right| - \mathbf{j}\left| {\begin{array}{*{20}{c}}
1 & 3\\
2 & 7
\end{array}} \right| + \mathbf{k}\left| {\begin{array}{*{20}{c}}
1 & 2\\
2 & 4
\end{array}} \right| = \left({14-12}\right)\mathbf{i} - \left({7-6}\right)\mathbf{j} + \left({4-4}\right)\mathbf{k} = 2\mathbf{i} - \mathbf{j}.\]
Therefore, the coordinates of the vector \(\mathbf{c} = \mathbf{a} \times \mathbf{b}\) are \(\left({2,-1,0}\right).\)
Example 2.
Force \(\mathbf{F}\left({2,-3,4}\right)\) is applied to point \(B\left({3,-1,5}\right).\) Determine the moment of this force relative to point \(A\left({1,2,-2}\right).\)
Solution.
The moment of force \(\mathbf{F}\) about a point \(A\) is defined as the cross product:
\[\mathbf{M_A} = \mathbf{r_{AB}} \times \mathbf{F}.\]
Calculate the position vector \(\mathbf{r_{AB}}:\)
\[\mathbf{r_{AB}} = \underbrace{\left({3,-1,5}\right)}_{B} - \underbrace{\left({1,2,-2}\right)}_{A} = \left({2,-3,7}\right).\]
The moment of the force is
\[\mathbf{M_A} = \mathbf{r_{AB}} \times \mathbf{F} = \left| {\begin{array}{*{20}{c}}
\mathbf{i} & \mathbf{j} & \mathbf{k}\\
2 & -3 & 7\\
2 & -3 & 4
\end{array}} \right| = \mathbf{i}\left| {\begin{array}{*{20}{c}}
-3 & 7\\
-3 & 4
\end{array}} \right| - \mathbf{j}\left| {\begin{array}{*{20}{c}}
2 & 7\\
2 & 4
\end{array}} \right| + \mathbf{k}\left| {\begin{array}{*{20}{c}}
2 & -3\\
2 & -3
\end{array}} \right| = \left({-12+21}\right)\mathbf{i} - \left({8-14}\right)\mathbf{j} + 0\cdot\mathbf{k} = 9\mathbf{i} + 6\mathbf{j}.\]
Hence, \(\mathbf{M_A} = \left({9,6,0}\right).\)
Example 3.
Calculate the area of a triangle with vertices \(A\left({1,-2,0}\right),\) \(B\left({3,1,-1}\right)\) and \(C\left({2,-1,3}\right).\)
Solution.
The area of the triangle is equal to half the area of the parallelogram spanned by the vectors AB and AC , that is
\[A_{\triangle} = \frac{1}{2}\Vert{\mathbf{AB} \times \mathbf{AC}}\Vert.\]
Calculate the coordinates of vectors AB and AC :
\[\mathbf{AB} = \underbrace{\left({3,1,-1}\right)}_B - \underbrace{\left({1,-2,0}\right)}_A = \left({2,3,-1}\right);\]
\[\mathbf{AC} = \underbrace{\left({2,-1,3}\right)}_C - \underbrace{\left({1,-2,0}\right)}_A = \left({1,1,3}\right);\]
Find the cross product:
\[\mathbf{AB} \times \mathbf{AC} = \left| {\begin{array}{*{20}{c}}
\mathbf{i} & \mathbf{j} & \mathbf{k}\\
2 & 3 & -1\\
1 & 1 & 3
\end{array}} \right| = \mathbf{i}\left| {\begin{array}{*{20}{c}}
3 & -1\\
1 & 3
\end{array}} \right| - \mathbf{j}\left| {\begin{array}{*{20}{c}}
2 & -1\\
1 & 3
\end{array}} \right| + \mathbf{k}\left| {\begin{array}{*{20}{c}}
2 & 3\\
1 & 1
\end{array}} \right| = \left({9+1}\right)\mathbf{i} - \left({6+1}\right)\mathbf{j} + \left({2-3}\right)\mathbf{k} = 10\mathbf{i} - 7\mathbf{j} - \mathbf{k}.\]
The absolute value of this vector is
\[\Vert{\mathbf{AB} \times \mathbf{AC}}\Vert = \sqrt{10^2 + \left({-7}\right)^2 + \left({-1}\right)^2} = \sqrt{150} =5\sqrt{6}.\]
Then the area of the triangle is
\[A_{\triangle} = \frac{1}{2}\Vert{\mathbf{AB} \times \mathbf{AC}}\Vert = \frac{5\sqrt{6}}{2}.\]
Example 4.
Calculate the area of a parallelogram built on the vectors \(\mathbf{a} = \mathbf{i} + \mathbf{j} + \mathbf{k}\) and \(\mathbf{b} = \mathbf{j} - \mathbf{i}.\)
Solution.
Determine the cross product of vectors \(\mathbf{a}\) and \(\mathbf{b}:\)
\[\mathbf{a} \times \mathbf{b} = \left({\mathbf{i} + \mathbf{j} + \mathbf{k}}\right) \times \left({\mathbf{j} - \mathbf{i}}\right) = \left({\mathbf{i} + \mathbf{j} + \mathbf{k}}\right) \times \mathbf{j} - \left({\mathbf{i} + \mathbf{j} + \mathbf{k}}\right) \times \mathbf{i}.\]
Since cross product is anticommutative, we have
\[\mathbf{a} \times \mathbf{b} = \mathbf{i} \times \left({\mathbf{i} + \mathbf{j} + \mathbf{k}}\right) - \mathbf{j} \times \left({\mathbf{i} + \mathbf{j} + \mathbf{k}}\right).\]
By the distributive property,
\[\mathbf{a} \times \mathbf{b} = \mathbf{i} \times \mathbf{i} + \mathbf{i} \times \mathbf{j} + \mathbf{i} \times \mathbf{k} - \mathbf{j} \times \mathbf{i} - \mathbf{j} \times \mathbf{j} - \mathbf{j} \times \mathbf{k}.\]
Substitute the results of multiplication of the unit vectors:
\[\mathbf{a} \times \mathbf{b} = \mathbf{0} + \mathbf{k} + \left({-\mathbf{j}}\right) - \left({-\mathbf{k}}\right) - \mathbf{0} - \mathbf{i} = -\mathbf{i} - \mathbf{j} + 2\mathbf{k}.\]
Now we can calculate the area of the parallelogram:
\[A = \Vert{\mathbf{a} \times \mathbf{b}}\Vert = \Vert{-\mathbf{i} - \mathbf{j} + 2\mathbf{k}}\Vert = \sqrt{\left({-1}\right)^2 + \left({-1}\right)^2 + 2^2} = \sqrt{6}.\]
Example 5.
Vectors \(\mathbf{a}\) and \(\mathbf{b}\) are not collinear. Find the values of \(\lambda\) for which the vectors \(\lambda\mathbf{a} + \mathbf{b}\) and \(2\mathbf{a} + \lambda\mathbf{b}\) are collinear.
Solution.
If two vectors are collinear then their cross product (denote it by c ) is equal to the zero vector. Hence
\[\mathbf{c} = \left({\lambda\mathbf{a} + \mathbf{b}}\right) \times \left({2\mathbf{a} + \lambda\mathbf{b}}\right) = \mathbf{0}.\]
Using the distributive property, we can write:
\[\mathbf{c} = \left({\lambda\mathbf{a} + \mathbf{b}}\right) \times 2\mathbf{a} + \left({\lambda\mathbf{a} + \mathbf{b}}\right) \times \lambda\mathbf{b} = \mathbf{0}.\]
The cross product is anticommutative, so
\[\mathbf{c} = - 2\mathbf{a} \times \left({\lambda\mathbf{a} + \mathbf{b}}\right) - \lambda\mathbf{b} \times \left({\lambda\mathbf{a} + \mathbf{b}}\right) = \mathbf{0}.\]
Apply again the distributive property:
\[\mathbf{c} = - \underbrace{2\mathbf{a} \times \lambda\mathbf{a}}_1 - \underbrace{2\mathbf{a} \times \mathbf{b}}_2 - \underbrace{\lambda\mathbf{b} \times \lambda\mathbf{a}}_3 - \underbrace{\lambda\mathbf{b} \times \mathbf{b}}_4 = \mathbf{0}.\]
Consider the terms \(1\) and \(4:\)
\[2\mathbf{a} \times \lambda\mathbf{a} = 2\lambda\left({\mathbf{a} \times \mathbf{a}}\right) = 2\lambda \times \mathbf{0} = \mathbf{0};\]
\[\lambda\mathbf{b} \times \mathbf{b} = \lambda\left({\mathbf{b} \times \mathbf{b}}\right) = \lambda \times \mathbf{0} = \mathbf{0};\]
Therefore
\[\mathbf{c} = - 2\mathbf{a} \times \mathbf{b} - \lambda\mathbf{b} \times \lambda\mathbf{a} = \mathbf{0},\]
\[\Rightarrow \mathbf{c} = - 2\left({\mathbf{a} \times \mathbf{b}}\right) + \lambda\left({\mathbf{a} \times \mathbf{b}}\right) = \mathbf{0},\]
or
\[\left({\lambda - 2}\right)\left({\mathbf{a} \times \mathbf{b}}\right) = \mathbf{0}.\]
From the description of the problem it is known that \({\mathbf{a} \times \mathbf{b} \ne \mathbf{0}}.\) Therefore
\[\lambda - 2 = 0, \Rightarrow \lambda = 0.\]
Example 6.
Prove the identity
\[{\Vert{\mathbf{a} \times \mathbf{b}}\Vert}^2 = \left| {\begin{array}{*{20}{c}}
\left({\mathbf{a}\cdot\mathbf{a}}\right) & \left({\mathbf{a}\cdot\mathbf{b}}\right)\\
\left({\mathbf{a}\cdot\mathbf{b}}\right) & \left({\mathbf{b}\cdot\mathbf{b}}\right)
\end{array}} \right|\]
Solution.
Expand the determinant on the right side:
\[RHS = \left| {\begin{array}{*{20}{c}}
\left({\mathbf{a}\cdot\mathbf{a}}\right) & \left({\mathbf{a}\cdot\mathbf{b}}\right)\\
\left({\mathbf{a}\cdot\mathbf{b}}\right) & \left({\mathbf{b}\cdot\mathbf{b}}\right)
\end{array}} \right| = \left({\mathbf{a}\cdot\mathbf{a}}\right)\left({\mathbf{b}\cdot\mathbf{b}}\right) - \left({\mathbf{a}\cdot\mathbf{b}}\right)\left({\mathbf{a}\cdot\mathbf{b}}\right).\]
Remember that
\[\mathbf{a}\cdot\mathbf{a} = {\Vert{\mathbf{a}}\Vert}^2,\;\mathbf{b}\cdot\mathbf{b} = {\Vert{\mathbf{b}}\Vert}^2,\;\mathbf{a}\cdot\mathbf{b} = {\Vert{\mathbf{a}}\Vert}{\Vert{\mathbf{b}}\Vert}\cos\varphi,\;\]
where \(\varphi\) is the angle between vectors \(\mathbf{a}\) and \(\mathbf{b}.\) Then
\[RHS = {\Vert{\mathbf{a}}\Vert}^2 {\Vert{\mathbf{b}}\Vert}^2 - {\Vert{\mathbf{a}}\Vert}^2{\Vert{\mathbf{b}}\Vert}^2\cos^2\varphi = {\Vert{\mathbf{a}}\Vert}^2 {\Vert{\mathbf{b}}\Vert}^2 \left({1 - \cos^2\varphi}\right) = {\Vert{\mathbf{a}}\Vert}^2{\Vert{\mathbf{b}}\Vert}^2\sin^2\varphi\]
It's clear that
\[RHS = {\Vert{\mathbf{a}}\Vert}^2{\Vert{\mathbf{b}}\Vert}^2\sin^2\varphi = \left({{\Vert{\mathbf{a}}\Vert}{\Vert{\mathbf{b}}\Vert}\sin\varphi}\right)^2 = \left({\mathbf{a} \times \mathbf{b}}\right)^2 = {\Vert{\mathbf{a} \times \mathbf{b}}\Vert}^2 = LHS. \blacksquare\]
