Vector (Cross) Product of Two Vectors

Definition of Vector (Cross) Product

The vector product (or cross product) of two non-zero vectors $\vec{a}$ and $\vec{b}$, denoted by $\vec{a} \times \vec{b}$, is defined as a vector whose magnitude is $|\vec{a}||\vec{b}|\sin\theta$ and whose direction is perpendicular to the plane containing both $\vec{a}$ and $\vec{b}$.

Mathematically, $$\vec{a} \times \vec{b} = |\vec{a}| |\vec{b}| \sin \theta \, \hat{n}$$

where:

• $\theta$ is the angle between $\vec{a}$ and $\vec{b}$, with $0 \le \theta \le \pi$.
• $\hat{n}$ is a unit vector perpendicular to both $\vec{a}$ and $\vec{b}$.

The direction of $\hat{n}$ is determined by the Right-Hand Rule: if the fingers of the right hand are curled from $\vec{a}$ towards $\vec{b}$ through the smaller angle, then the thumb points in the direction of $\vec{a} \times \vec{b}$.

Important observations

• The result of a cross product is always a vector quantity.
• If either vector is the zero vector, then the cross product is defined as the zero vector.
• The magnitude is $$|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta.$$
• If $\theta = 0^\circ$ or $180^\circ$, then $\sin\theta = 0$, so the cross product is zero. Hence parallel vectors have zero cross product.
• If $\theta = 90^\circ$, then $\sin\theta = 1$, so $$|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}|.$$

Properties of the Vector Product

1. Non-Commutative (Anti-Commutative): $$\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})$$ Reversing the order reverses the direction.

2. Condition for Collinearity: Two non-zero vectors are parallel if and only if their cross product is zero. $$\vec{a} \times \vec{b} = \vec{0} \iff \vec{a} \parallel \vec{b}$$ In particular, $$\vec{a} \times \vec{a} = \vec{0}.$$

3. Distributivity over Addition: $$\vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}$$ and similarly, $$(\vec{a} + \vec{b}) \times \vec{c} = \vec{a} \times \vec{c} + \vec{b} \times \vec{c}.$$

4. Scalar Multiplication: For any scalar $\lambda$, $$\lambda(\vec{a} \times \vec{b}) = (\lambda\vec{a}) \times \vec{b} = \vec{a} \times (\lambda\vec{b}).$$

5. Not Associative: In general, $$(\vec{a} \times \vec{b}) \times \vec{c} \ne \vec{a} \times (\vec{b} \times \vec{c}).$$

These properties are fundamental in vector algebra and applications.

Cross Product of Standard Unit Vectors

For the mutually perpendicular unit vectors $\hat{i}$, $\hat{j}$, and $\hat{k}$:

• Since the angle between a unit vector and itself is $0^\circ$, $$\hat{i} \times \hat{i} = \vec{0}, \quad \hat{j} \times \hat{j} = \vec{0}, \quad \hat{k} \times \hat{k} = \vec{0}.$$

• Using the cyclic order $\hat{i} \to \hat{j} \to \hat{k} \to \hat{i}$ and the right-hand rule, $$\hat{i} \times \hat{j} = \hat{k}, \quad \hat{j} \times \hat{k} = \hat{i}, \quad \hat{k} \times \hat{i} = \hat{j}.$$

• Reversing the order introduces a minus sign: $$\hat{j} \times \hat{i} = -\hat{k}, \quad \hat{k} \times \hat{j} = -\hat{i}, \quad \hat{i} \times \hat{k} = -\hat{j}.$$

These identities are used to compute cross products in component form.

Cross Product in Component Form

If $$\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}, \qquad \vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k},$$ then their cross product is given by the determinant

$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}.$$

Expanding along the first row, $$\vec{a} \times \vec{b} = (a_2 b_3 - a_3 b_2)\hat{i} - (a_1 b_3 - a_3 b_1)\hat{j} + (a_1 b_2 - a_2 b_1)\hat{k}.$$

This form is especially useful in coordinate geometry, mechanics, and geometry problems involving area and perpendicular directions.

Geometric Interpretation: Areas

The magnitude of the cross product has an important geometric meaning.

• Area of a Parallelogram: If $\vec{a}$ and $\vec{b}$ are adjacent sides of a parallelogram, then $$\text{Area of Parallelogram} = |\vec{a} \times \vec{b}|.$$

• Area of a Triangle: If $\vec{a}$ and $\vec{b}$ are adjacent sides of a triangle, then $$\text{Area of Triangle} = \frac{1}{2}|\vec{a} \times \vec{b}|.$$

• Area of a Parallelogram using Diagonals: If $\vec{d}_1$ and $\vec{d}_2$ are the diagonals of a parallelogram, then $$\text{Area} = \frac{1}{2}|\vec{d}_1 \times \vec{d}_2|.$$

Thus the cross product is closely tied to area and perpendicularity.

Example 1: Basic Cross Product

Find $\vec{a} \times \vec{b}$ if $\vec{a} = 2\hat{i} + \hat{j} + 3\hat{k}$ and $\vec{b} = 3\hat{i} + 5\hat{j} - 2\hat{k}$.

Solution: Step 1: Set up the determinant. $$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 3 \\ 3 & 5 & -2 \end{vmatrix}$$

Step 2: Expand: $$= \hat{i}[(1)(-2) - (3)(5)] - \hat{j}[(2)(-2) - (3)(3)] + \hat{k}[(2)(5) - (1)(3)]$$

Step 3: Simplify: $$= \hat{i}[-2 - 15] - \hat{j}[-4 - 9] + \hat{k}[10 - 3]$$ $$= -17\hat{i} - (-13)\hat{j} + 7\hat{k}$$ $$= -17\hat{i} + 13\hat{j} + 7\hat{k}.$$

Answer: $-17\hat{i} + 13\hat{j} + 7\hat{k}$

Example 2: Finding a Perpendicular Unit Vector

Find a unit vector perpendicular to each of the vectors $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$, where $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$.

Solution: Step 1: Compute $$\vec{u} = \vec{a} + \vec{b} = 2\hat{i} + 3\hat{j} + 4\hat{k},$$ $$\vec{v} = \vec{a} - \vec{b} = 0\hat{i} - \hat{j} - 2\hat{k}.$$

Step 2: A vector perpendicular to both is $$\vec{u} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 0 & -1 & -2 \end{vmatrix}.$$

Expanding, $$= \hat{i}[3(-2) - 4(-1)] - \hat{j}[2(-2) - 4(0)] + \hat{k}[2(-1) - 3(0)]$$ $$= \hat{i}(-6 + 4) - \hat{j}(-4) + \hat{k}(-2)$$ $$= -2\hat{i} + 4\hat{j} - 2\hat{k}.$$

Step 3: Its magnitude is $$\sqrt{(-2)^2 + 4^2 + (-2)^2} = \sqrt{24} = 2\sqrt{6}.$$

Step 4: The required unit vector is $$\pm \frac{-2\hat{i} + 4\hat{j} - 2\hat{k}}{2\sqrt{6}} = \pm\left(-\frac{1}{\sqrt{6}}\hat{i} + \frac{2}{\sqrt{6}}\hat{j} - \frac{1}{\sqrt{6}}\hat{k}\right).$$

Answer: $\pm\left(-\frac{1}{\sqrt{6}}\hat{i} + \frac{2}{\sqrt{6}}\hat{j} - \frac{1}{\sqrt{6}}\hat{k}\right)$

Example 3: Area of a Parallelogram

Find the area of a parallelogram whose adjacent sides are given by the vectors $\vec{a} = 3\hat{i} + \hat{j} + 4\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$.

Solution: Step 1: The area is $$|\vec{a} \times \vec{b}|.$$

Step 2: Compute the cross product: $$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 4 \\ 1 & -1 & 1 \end{vmatrix}.$$

Expanding, $$= \hat{i}[1(1) - 4(-1)] - \hat{j}[3(1) - 4(1)] + \hat{k}[3(-1) - 1(1)]$$ $$= \hat{i}(1+4) - \hat{j}(3-4) + \hat{k}(-3-1)$$ $$= 5\hat{i} + \hat{j} - 4\hat{k}.$$

Step 3: Find its magnitude: $$|\vec{a} \times \vec{b}| = \sqrt{5^2 + 1^2 + (-4)^2} = \sqrt{25+1+16} = \sqrt{42}.$$

Answer: $\sqrt{42}$ square units.

Example 4: Area of a Triangle

Find the area of a triangle having the points $A(1, 1, 1)$, $B(1, 2, 3)$, and $C(2, 3, 1)$ as its vertices.

Solution: Step 1: Find two side vectors: $$\vec{AB} = (1-1)\hat{i} + (2-1)\hat{j} + (3-1)\hat{k} = \hat{j} + 2\hat{k},$$ $$\vec{AC} = (2-1)\hat{i} + (3-1)\hat{j} + (1-1)\hat{k} = \hat{i} + 2\hat{j}.$$

Step 2: Area of triangle is $$\frac{1}{2}|\vec{AB} \times \vec{AC}|.$$

Step 3: Compute the cross product: $$\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 2 \\ 1 & 2 & 0 \end{vmatrix}.$$

Expanding, $$= \hat{i}[1(0) - 2(2)] - \hat{j}[0(0) - 2(1)] + \hat{k}[0(2) - 1(1)]$$ $$= -4\hat{i} + 2\hat{j} - \hat{k}.$$

Step 4: Magnitude: $$|\vec{AB} \times \vec{AC}| = \sqrt{(-4)^2 + 2^2 + (-1)^2} = \sqrt{21}.$$

Step 5: Therefore, $$\text{Area} = \frac{\sqrt{21}}{2}.$$

Answer: $\frac{\sqrt{21}}{2}$ square units.

Example 5: Collinear Vectors Condition

Find $\lambda$ and $\mu$ if $(2\hat{i} + 6\hat{j} + 27\hat{k}) \times (\hat{i} + \lambda\hat{j} + \mu\hat{k}) = \vec{0}$.

Solution: Step 1: Since the cross product is zero, the vectors are parallel.

Step 2: Compute the determinant: $$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 6 & 27 \\ 1 & \lambda & \mu \end{vmatrix} = \vec{0}.$$

Expanding, $$\hat{i}(6\mu - 27\lambda) - \hat{j}(2\mu - 27) + \hat{k}(2\lambda - 6) = \vec{0}.$$

Step 3: Equate components to zero: $$2\lambda - 6 = 0 \implies \lambda = 3,$$ $$-(2\mu - 27) = 0 \implies 2\mu = 27 \implies \mu = \frac{27}{2}.$$

Check the $\hat{i}$-component: $$6\left(\frac{27}{2}\right) - 27(3) = 81 - 81 = 0.$$ So the values are correct.

Answer: $\lambda = 3$, $\mu = 27/2$

Example 6: Sine of the Angle Between Vectors

Find the sine of the angle between the vectors $\vec{a} = 3\hat{i} + \hat{j} + 2\hat{k}$ and $\vec{b} = 2\hat{i} - 2\hat{j} + 4\hat{k}$.

Solution: Use $$\sin\theta = \frac{|\vec{a} \times \vec{b}|}{|\vec{a}| |\vec{b}|}.$$

Step 1: Compute the cross product: $$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 2 \\ 2 & -2 & 4 \end{vmatrix}.$$

Expanding, $$= \hat{i}[1(4) - 2(-2)] - \hat{j}[3(4) - 2(2)] + \hat{k}[3(-2) - 1(2)]$$ $$= 8\hat{i} - 8\hat{j} - 8\hat{k}.$$

Step 2: Magnitude of cross product: $$|\vec{a} \times \vec{b}| = \sqrt{8^2 + (-8)^2 + (-8)^2} = \sqrt{192} = 8\sqrt{3}.$$

Step 3: Magnitudes of vectors: $$|\vec{a}| = \sqrt{3^2 + 1^2 + 2^2} = \sqrt{14},$$ $$|\vec{b}| = \sqrt{2^2 + (-2)^2 + 4^2} = \sqrt{24} = 2\sqrt{6}.$$

Step 4: Therefore, $$\sin\theta = \frac{8\sqrt{3}}{\sqrt{14} \cdot 2\sqrt{6}} = \frac{8\sqrt{3}}{2\sqrt{84}} = \frac{4\sqrt{3}}{2\sqrt{21}} = \frac{2\sqrt{3}}{\sqrt{3}\sqrt{7}} = \frac{2}{\sqrt{7}}.$$

Answer: $2/\sqrt{7}$

Example 7: Vector Algebra Identity

Prove that $(\vec{a} - \vec{b}) \times (\vec{a} + \vec{b}) = 2(\vec{a} \times \vec{b})$.

Solution: Start with the left-hand side: $$(\vec{a} - \vec{b}) \times (\vec{a} + \vec{b}).$$

Using distributivity, $$= \vec{a}\times\vec{a} + \vec{a}\times\vec{b} - \vec{b}\times\vec{a} - \vec{b}\times\vec{b}.$$

Now, $$\vec{a}\times\vec{a} = \vec{0}, \qquad \vec{b}\times\vec{b} = \vec{0},$$ and $$\vec{b}\times\vec{a} = -(\vec{a}\times\vec{b}).$$

So, $$= \vec{0} + \vec{a}\times\vec{b} - [-(\vec{a}\times\vec{b})] - \vec{0}$$ $$= \vec{a}\times\vec{b} + \vec{a}\times\vec{b}$$ $$= 2(\vec{a}\times\vec{b}).$$

Hence proved.

Answer: Proved.

Example 8: Area of Parallelogram via Diagonals

Find the area of the parallelogram whose diagonals are represented by the vectors $\vec{d}_1 = 3\hat{i} + \hat{j} - 2\hat{k}$ and $\vec{d}_2 = \hat{i} - 3\hat{j} + 4\hat{k}$.

Solution: Step 1: Use the formula $$\text{Area} = \frac{1}{2}|\vec{d}_1 \times \vec{d}_2|.$$

Step 2: Compute the cross product: $$\vec{d}_1 \times \vec{d}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -2 \\ 1 & -3 & 4 \end{vmatrix}.$$

Expanding, $$= \hat{i}[1(4) - (-2)(-3)] - \hat{j}[3(4) - (-2)(1)] + \hat{k}[3(-3) - 1(1)]$$ $$= -2\hat{i} - 14\hat{j} - 10\hat{k}.$$

Step 3: Magnitude: $$|\vec{d}_1 \times \vec{d}_2| = \sqrt{(-2)^2 + (-14)^2 + (-10)^2} = \sqrt{300} = 10\sqrt{3}.$$

Step 4: Therefore, $$\text{Area} = \frac{1}{2}(10\sqrt{3}) = 5\sqrt{3}.$$

Answer: $5\sqrt{3}$ square units.

Example 9: Lagrange's Identity Application

If $|\vec{a}| = 2$, $|\vec{b}| = 5$, and $|\vec{a} \times \vec{b}| = 8$, find the value of $\vec{a} \cdot \vec{b}$.

Solution: Use Lagrange's identity: $$|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2.$$

Substitute the given values: $$8^2 + (\vec{a} \cdot \vec{b})^2 = 2^2 \cdot 5^2$$ $$64 + (\vec{a} \cdot \vec{b})^2 = 100$$ $$(\vec{a} \cdot \vec{b})^2 = 36.$$

Hence, $$\vec{a} \cdot \vec{b} = \pm 6.$$

The sign cannot be uniquely determined from the given information alone because the angle could be acute or obtuse.

Answer: $\pm 6$

Example 10: Angle Yielding a Unit Vector

Let the vectors $\vec{a}$ and $\vec{b}$ be such that $|\vec{a}| = 3$ and $|\vec{b}| = \frac{\sqrt{2}}{3}$. If $\vec{a} \times \vec{b}$ is a unit vector, find the angle between $\vec{a}$ and $\vec{b}$.

Solution: Since $\vec{a} \times \vec{b}$ is a unit vector, $$|\vec{a} \times \vec{b}| = 1.$$

Using the cross product magnitude formula, $$|\vec{a}| |\vec{b}| \sin\theta = 1.$$

Substitute the given values: $$3\left(\frac{\sqrt{2}}{3}\right)\sin\theta = 1$$ $$\sqrt{2}\sin\theta = 1$$ $$\sin\theta = \frac{1}{\sqrt{2}}.$$

Therefore, $$\theta = 45^\circ \quad \text{or} \quad 135^\circ.$$

If the angle between vectors is taken to be the acute angle in context, then the answer is $45^\circ$.

Answer: $45^\circ$ or $135^\circ$