Vector Products

Introduction to Vector Products

Beyond simple addition and subtraction, there are two important ways to multiply vectors. One method results in a scalar, while the other results in a new vector. These operations are fundamental in physics for defining quantities like work, power, torque, and angular momentum.

Multiplication of a Vector by a Scalar

Definition: When a vector is multiplied by a scalar (a real number), the result is a vector whose magnitude is scaled by that scalar, and direction is preserved or reversed based on the sign of the scalar.

Mathematical Representation: If $\vec{A}$ is a vector and $k$ is a scalar, then: $$\vec{B} = k\vec{A}$$

• $|\vec{B}| = |k||\vec{A}|$
• Direction of $\vec{B}$ is:
• Same as $\vec{A}$ if $k > 0$
• Opposite to $\vec{A}$ if $k < 0$
• Zero vector if $k = 0$

Properties:

• Scalar multiplication follows distributive law: $$k(\vec{A} + \vec{B}) = k\vec{A} + k\vec{B}$$
• Associative with respect to scalars: $$(ab)\vec{A} = a(b\vec{A})$$
• Unit vector scaling: $$\vec{A} = |\vec{A}| \cdot \hat{A}$$ where $\hat{A}$ is the unit vector in the direction of $\vec{A}$.

The Scalar Product (Dot Product)

The scalar product of two vectors $\vec{A}$ and $\vec{B}$, written as $\vec{A} \cdot \vec{B}$, is a scalar quantity. It is defined as: $$\vec{A} \cdot \vec{B} = AB\cos(\theta)$$ where A and B are the magnitudes of the vectors and $\theta$ is the angle between them. The dot product can be interpreted as the product of the magnitude of one vector with the magnitude of the component of the other vector in its direction.

Properties of the Dot Product:

• It is commutative: $\vec{A} \cdot \vec{B} = \vec{B} \cdot \vec{A}$.
• It is distributive: $\vec{A} \cdot (\vec{B} + \vec{C}) = \vec{A} \cdot \vec{B} + \vec{A} \cdot \vec{C}$.
• For perpendicular vectors, $\theta = 90^\circ$, so $\vec{A} \cdot \vec{B} = 0$.
• In terms of components: $\vec{A} \cdot \vec{B} = (A_x\hat{i} + A_y\hat{j} + A_z\hat{k}) \cdot (B_x\hat{i} + B_y\hat{j} + B_z\hat{k}) = A_x B_x + A_y B_y + A_z B_z$.
• Application: Work done by a constant force is $W = \vec{F} \cdot \vec{d}$.

Cross Product (Vector Product) of Vectors

Definition: The cross product of two vectors $\vec{A}$ and $\vec{B}$ is a vector given by: $$\vec{A} \times \vec{B} = AB \sin\theta \, \hat{n}$$ where $\theta$ is the angle between the vectors and $\hat{n}$ is a unit vector perpendicular to the plane containing $\vec{A}$ and $\vec{B}$ (using right-hand rule).

Direction:

• Given by Right Hand Rule: Curl your fingers from $\vec{A}$ to $\vec{B}$, the thumb points in direction of $\vec{A} \times \vec{B}$.

Properties:

• $\vec{A} \times \vec{B} = -\vec{B} \times \vec{A}$ (Anti-commutative)
• $\vec{A} \times \vec{A} = \vec{0}$
• $\vec{i} \times \vec{j} = \vec{k}$, $\vec{j} \times \vec{k} = \vec{i}$, $\vec{k} \times \vec{i} = \vec{j}$

Algebraic Expression: If $\vec{A} = A_x\hat{i} + A_y\hat{j} + A_z\hat{k}$ and $\vec{B} = B_x\hat{i} + B_y\hat{j} + B_z\hat{k}$: $$\vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix}$$

Example 1:

Question: Let $\vec{A} = 2\hat{i} + 3\hat{j}$. Find the result of $-2\vec{A}$.

Solution: $$-2\vec{A} = -2(2\hat{i} + 3\hat{j}) = -4\hat{i} -6\hat{j}$$ This vector has magnitude $\sqrt{(-4)^2 + (-6)^2} = \sqrt{52}$ and is in opposite direction to $\vec{A}$.

Example 2 :

Question: Find the dot product of $\vec{A} = 2\hat{i} + 3\hat{j}$ and $\vec{B} = -\hat{i} + 4\hat{j}$.

Solution: $$\vec{A} \cdot \vec{B} = (2)(-1) + (3)(4) = -2 + 12 = 10$$

Example 3 :

Find the angle between force $\vec{F} = (3\hat{i} + 4\hat{j} - 5\hat{k})$ unit and displacement $\vec{d} = (5\hat{i} + 4\hat{j} + 3\hat{k})$ unit. Also, find the projection of F on d.

Solution:

1. Find the dot product: $$\vec{F} \cdot \vec{d} = (3)(5) + (4)(4) + (-5)(3) = 15 + 16 - 15 = 16\ \text{units}$$
2. Find the magnitudes: $$|\vec{F}| = \sqrt{3^2 + 4^2 + (-5)^2} = \sqrt{9+16+25} = \sqrt{50} = 5\sqrt{2}$$ $$|\vec{d}| = \sqrt{5^2 + 4^2 + 3^2} = \sqrt{25+16+9} = \sqrt{50} = 5\sqrt{2}$$
3. Find the angle: $$\cos(\theta) = \frac{\vec{F} \cdot \vec{d}}{|\vec{F}||\vec{d}|} = \frac{16}{(5\sqrt{2})(5\sqrt{2})} = \frac{16}{50} = 0.32$$ $$\theta = \cos^{-1}(0.32) \approx 71.3^\circ$$
4. Find the projection of F on d: This is given by $F \cos(\theta) = \frac{\vec{F} \cdot \vec{d}}{|\vec{d}|} = \frac{16}{5\sqrt{2}} \approx 2.26\ \text{units}$.

Example 1:

Question: Find the cross product of $\vec{A} = 2\hat{i} + 3\hat{j}$ and $\vec{B} = \hat{i} + 4\hat{j}$.

Solution: Use determinant method: $$\vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 0 \\ 1 & 4 & 0 \end{vmatrix} = (0)\hat{i} - (0)\hat{j} + (8 - 3)\hat{k} = 5\hat{k}$$

Example 2:

Find the torque about the origin when a force of $3\sqrt{3}\hat{j}\ N$ acts on a particle whose position vector is $2\hat{k}\ m$.

Solution: Torque is given by $\vec{\tau} = \vec{r} \times \vec{F}$. $$\vec{\tau} = (2\hat{k}) \times (3\sqrt{3}\hat{j})$$ $$\vec{\tau} = (2)(3\sqrt{3}) (\hat{k} \times \hat{j})$$ Using the cyclic property of unit vectors, we know that $\hat{j} \times \hat{k} = \hat{i}$, so $\hat{k} \times \hat{j} = -\hat{i}$. $$\vec{\tau} = 6\sqrt{3} (- \hat{i}) = -6\sqrt{3}\hat{i}\ N \cdot m$$