Vector (Cross) Product of Vectors

Vector (or Cross) Product of Two Vectors: Definition and Formula

In addition to the scalar product, there is another method for "multiplying" two vectors, which yields a vector quantity. This operation is called the vector product or cross product. Unlike the dot product, the cross product is defined only for vectors in three-dimensional space.

Definition of the Vector (Cross) Product

The vector product or cross product of two non-zero vectors $\vec{a}$ and $\vec{b}$ is a vector, denoted by $\vec{a} \times \vec{b}$ (read as "a cross b"). This resulting vector is defined by its magnitude and direction as follows:

1. Magnitude: The magnitude of the vector $\vec{a} \times \vec{b}$ is equal to the product of the magnitudes of $\vec{a}$ and $\vec{b}$ and the sine of the angle $\theta$ between them. The angle $\theta$ is the smaller angle between $\vec{a}$ and $\vec{b}$ when they are placed tail-to-tail, such that $0 \le \theta \le \pi$ radians (or $0^\circ \le \theta \le 180^\circ$).

   $$ |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta $$

2. Direction: The direction of the vector $\vec{a} \times \vec{b}$ is perpendicular (orthogonal) to the plane containing both vectors $\vec{a}$ and $\vec{b}$. There are two directions perpendicular to any given plane (pointing out of one side or the other). The specific direction of $\vec{a} \times \vec{b}$ is determined by the Right-Hand Rule.
   • Right-Hand Rule: If you align your right hand along the direction of the first vector ($\vec{a}$) and curl your fingers towards the direction of the second vector ($\vec{b}$) through the smaller angle $\theta$, your extended thumb will point in the direction of $\vec{a} \times \vec{b}$.

   This directional property implies that the cross product $\vec{a} \times \vec{b}$ is orthogonal (perpendicular) to both the vector $\vec{a}$ and the vector $\vec{b}$. We can verify this using the dot product:

   $(\vec{a} \times \vec{b}) \cdot \vec{a} = 0$

   (The dot product of orthogonal vectors is zero)

   $(\vec{a} \times \vec{b}) \cdot \vec{b} = 0$

   (The dot product of orthogonal vectors is zero)

Combining the magnitude and direction, the vector product can be written as:

$$ \vec{a} \times \vec{b} = (|\vec{a}| |\vec{b}| \sin \theta) \hat{n} $$

where $\hat{n}$ is a unit vector perpendicular to the plane containing $\vec{a}$ and $\vec{b}$, and its direction is given by the Right-Hand Rule when going from $\vec{a}$ to $\vec{b}$.

Result of the Vector Product

The result of the cross product $\vec{a} \times \vec{b}$ is a vector quantity. This vector is always perpendicular to the plane spanned by the two vectors $\vec{a}$ and $\vec{b}$ (unless they are collinear).

Special Cases of the Cross Product

• If either vector is the zero vector:

  If $\vec{a} = \vec{0}$ or $\vec{b} = \vec{0}$ (or both), the cross product is defined to be the zero vector, $\vec{0}$.

  $\vec{a} \times \vec{0} = \vec{0}$

  $\vec{0} \times \vec{b} = \vec{0}$

  This aligns with the magnitude formula, as $|\vec{0}| = 0$, so $|\vec{a} \times \vec{0}| = |\vec{a}| \cdot 0 \cdot \sin \theta = 0$.
• If $\vec{a}$ and $\vec{b}$ are non-zero and collinear:

  If $\vec{a}$ and $\vec{b}$ are parallel or anti-parallel, the angle $\theta$ between them is $0^\circ$ or $180^\circ$ ($\pi$ radians). In both these cases, $\sin \theta = \sin 0^\circ = 0$ and $\sin \pi = 0$.

  Using the magnitude formula:

  $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta = |\vec{a}| |\vec{b}| (0) = 0$

  Since the magnitude of the resulting vector is zero, the vector itself must be the zero vector.

  Thus, the cross product of two non-zero vectors is the zero vector if and only if the vectors are collinear (parallel or anti-parallel).

  $\vec{a} \times \vec{b} = \vec{0} \iff \vec{a} \text{ and } \vec{b} \text{ are collinear}$

  (for non-zero vectors)

  A specific case of this is the cross product of any non-zero vector with itself, or with its negative:

  $\vec{a} \times \vec{a} = \vec{0}$

  (Angle is $0^\circ$)

  $\vec{a} \times (-\vec{a}) = \vec{0}$

  (Angle is $180^\circ$)

• If $\vec{a}$ and $\vec{b}$ are perpendicular:

  If $\vec{a}$ and $\vec{b}$ are perpendicular, the angle $\theta$ between them is $90^\circ$ ($\pi/2$ radians). $\sin 90^\circ = 1$.

  The magnitude of the cross product is:

  $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin 90^\circ = |\vec{a}| |\vec{b}| (1) = |\vec{a}| |\vec{b}|$

  The magnitude is simply the product of the magnitudes. The direction is perpendicular to the plane containing $\vec{a}$ and $\vec{b}$, as given by the right-hand rule.

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Properties of Vector Product (Non-commutativity, Distributivity, etc.)

The vector product, being a different operation from the scalar product, has its own set of properties. Some are similar to scalar multiplication, while others are quite different, particularly concerning commutativity and associativity.

Let $\vec{a}, \vec{b}, \vec{c}$ be any three vectors in 3D space, and let $k$ be any scalar (real number).

1. Non-Commutativity (Anti-Commutativity):

   Unlike scalar multiplication or the dot product, the order of vectors in a cross product matters. Reversing the order results in a vector with the same magnitude but the opposite direction.

   $\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})$

   Reason: The magnitude $|\vec{a}||\vec{b}|\sin\theta$ is the same for both $\vec{a} \times \vec{b}$ and $\vec{b} \times \vec{a}$. However, applying the Right-Hand Rule to $\vec{a} \times \vec{b}$ gives a direction (e.g., upwards from the plane), while applying it to $\vec{b} \times \vec{a}$ requires curling fingers from $\vec{b}$ to $\vec{a}$, resulting in the exact opposite direction (e.g., downwards from the plane).

2. Non-Associativity:

   The cross product is generally not associative. The grouping of vectors in multiple cross products affects the result.

   $(\vec{a} \times \vec{b}) \times \vec{c} \neq \vec{a} \times (\vec{b} \times \vec{c})$

   (in general)

   Example: Let's use the standard unit vectors $\hat{i}, \hat{j}, \hat{k}$.

   Calculate $(\hat{i} \times \hat{j}) \times \hat{j}$:

   $(\hat{i} \times \hat{j}) \times \hat{j} = \hat{k} \times \hat{j}$

   Using the unit vector cross product rules ($\hat{k} \times \hat{j} = -\hat{i}$):

   $\phantom{(\hat{i} \times \hat{j}) \times \hat{j}} = -\hat{i}$

   Now, calculate $\hat{i} \times (\hat{j} \times \hat{j})$:

   $\hat{i} \times (\hat{j} \times \hat{j}) = \hat{i} \times \vec{0}$

   (Since $\hat{j} \times \hat{j} = \vec{0}$)

   The cross product with the zero vector is the zero vector:

   $\phantom{\hat{i} \times (\hat{j} \times \hat{j})} = \vec{0}$

   Since $-\hat{i} \neq \vec{0}$, $(\hat{i} \times \hat{j}) \times \hat{j} \neq \hat{i} \times (\hat{j} \times \hat{j})$. This confirms that the cross product is not associative.

3. Distributivity over Vector Addition:

   The cross product distributes over vector addition. This property holds from both the left and the right, but due to non-commutativity, the order of the vectors must be maintained.

   • Left Distributive Property:

     $\vec{a} \times (\vec{b} + \vec{c}) = (\vec{a} \times \vec{b}) + (\vec{a} \times \vec{c})$

   • Right Distributive Property:

     $(\vec{b} + \vec{c}) \times \vec{a} = (\vec{b} \times \vec{a}) + (\vec{c} \times \vec{a})$

   Note that $(\vec{b} \times \vec{a}) + (\vec{c} \times \vec{a})$ is not equal to $(\vec{a} \times \vec{b}) + (\vec{a} \times \vec{c})$ unless $(\vec{b} \times \vec{a}) = (\vec{a} \times \vec{b}) + (\vec{a} \times \vec{c}) - (\vec{c} \times \vec{a})$, which involves anti-commutativity. So, the left and right distributive laws are distinct statements.

4. Association with Scalar Multiplication:

   A scalar factor in a cross product can be moved freely or factored out, similar to other multiplication operations.

   $(k\vec{a}) \times \vec{b} = k(\vec{a} \times \vec{b}) = \vec{a} \times (k\vec{b})$

   Reason: The magnitude and direction both scale appropriately. For example, $|(k\vec{a}) \times \vec{b}| = |k\vec{a}||\vec{b}|\sin\theta = |k||\vec{a}||\vec{b}|\sin\theta = |k||\vec{a} \times \vec{b}|$. The direction depends on the sign of $k$, but the formula handles this correctly. If $k>0$, the direction is the same. If $k<0$, the direction is opposite, and the $-1$ factor is handled by $k$ being negative.

5. Condition for Collinearity:

   For any two non-zero vectors $\vec{a}$ and $\vec{b}$, their cross product is the zero vector if and only if they are collinear.

   $\vec{a} \times \vec{b} = \vec{0} \iff \vec{a} \text{ is parallel to } \vec{b}$

   (for $\vec{a} \ne \vec{0}, \vec{b} \ne \vec{0}$)

   This is a very useful criterion for checking if two vectors are parallel or lie on the same line.

6. Cross Product with Zero Vector:

   The cross product of any vector with the zero vector is the zero vector.

   $\vec{a} \times \vec{0} = \vec{0}$

   $\vec{0} \times \vec{a} = \vec{0}$

7. Lagrange's Identity:

   This identity relates the magnitude of the cross product to the magnitudes and dot product of the vectors. It is a consequence of the trigonometric identity $\sin^2\theta + \cos^2\theta = 1$.

   $|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2$

   Proof: Start with the square of the magnitude of the cross product definition:

   $|\vec{a} \times \vec{b}|^2 = (|\vec{a}| |\vec{b}| \sin \theta)^2$

   $\phantom{|\vec{a} \times \vec{b}|^2} = |\vec{a}|^2 |\vec{b}|^2 \sin^2 \theta$

   Use the identity $\sin^2 \theta = 1 - \cos^2 \theta$:

   $\phantom{|\vec{a} \times \vec{b}|^2} = |\vec{a}|^2 |\vec{b}|^2 (1 - \cos^2 \theta)$

   $\phantom{|\vec{a} \times \vec{b}|^2} = |\vec{a}|^2 |\vec{b}|^2 - |\vec{a}|^2 |\vec{b}|^2 \cos^2 \theta$

   Recall the dot product definition: $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$. Squaring this gives $(\vec{a} \cdot \vec{b})^2 = (|\vec{a}| |\vec{b}| \cos \theta)^2 = |\vec{a}|^2 |\vec{b}|^2 \cos^2 \theta$. Substitute this into the expression:

   $|\vec{a}|^2 |\vec{b}|^2 - (|\vec{a}| |\vec{b}| \cos \theta)^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2$

   Thus, we have shown that $|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2$. This identity is useful in various vector calculations.

Vector Product in terms of Components (using Determinant)

Similar to the scalar product, the vector product also has a convenient algebraic formula when vectors are expressed in terms of their components with respect to the standard orthonormal basis $\{\hat{i}, \hat{j}, \hat{k}\}$. This formula is most easily derived and remembered using the properties of determinants.

Cross Products of Standard Unit Vectors

Based on the definition of the cross product (magnitude, direction, and the Right-Hand Rule) and the fact that $\hat{i}, \hat{j}, \hat{k}$ are mutually orthogonal unit vectors:

• Cross product of a unit vector with itself:

  The angle is $\theta = 0^\circ$, $\sin 0^\circ = 0$. The magnitude is $|\hat{i} \times \hat{i}| = |\hat{i}||\hat{i}|\sin 0^\circ = (1)(1)(0) = 0$. A vector with zero magnitude is the zero vector.

  $\hat{i} \times \hat{i} = \vec{0}$

  Similarly,

  $\hat{j} \times \hat{j} = \vec{0}$

  $\hat{k} \times \hat{k} = \vec{0}$

• Cross product of two different unit vectors:

  The angle between any two different standard unit vectors is $\theta = 90^\circ$, $\sin 90^\circ = 1$. The magnitude of their cross product is $|\hat{i} \times \hat{j}| = |\hat{i}||\hat{j}|\sin 90^\circ = (1)(1)(1) = 1$. The direction is perpendicular to the plane of the two vectors (e.g., for $\hat{i}$ and $\hat{j}$, the XY-plane) and given by the Right-Hand Rule.

  Applying the Right-Hand Rule (curl fingers from $\hat{i}$ to $\hat{j}$):

  $\hat{i} \times \hat{j} = \hat{k}$

  Applying the Right-Hand Rule (curl fingers from $\hat{j}$ to $\hat{k}$):

  $\hat{j} \times \hat{k} = \hat{i}$

  Applying the Right-Hand Rule (curl fingers from $\hat{k}$ to $\hat{i}$):

  $\hat{k} \times \hat{i} = \hat{j}$

• Using Anti-Commutativity:

  Using the property $\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})$, we get the reverse products:

  $\hat{j} \times \hat{i} = -(\hat{i} \times \hat{j}) = -\hat{k}$

  $\hat{k} \times \hat{j} = -(\hat{j} \times \hat{k}) = -\hat{i}$

  $\hat{i} \times \hat{k} = -(\hat{k} \times \hat{i}) = -\hat{j}$

A cyclic diagram can help remember these relationships for different unit vectors:

Multiplying two adjacent vectors in the clockwise direction gives the next vector in the cycle (e.g., $\hat{i} \times \hat{j} = \hat{k}$). Multiplying in the counter-clockwise direction gives the negative of the next vector (e.g., $\hat{j} \times \hat{i} = -\hat{k}$).

Derivation of the Component Formula

Let $\vec{a}$ and $\vec{b}$ be two vectors in 3D space given in component form:

We can compute their cross product using the distributive property and the cross products of the unit vectors derived above:

$\vec{a} \times \vec{b} = (a_1\hat{i} + a_2\hat{j} + a_3\hat{k}) \times (b_1\hat{i} + b_2\hat{j} + b_3\hat{k})$

Expanding this product using distributivity and the property $(k\vec{u}) \times (m\vec{v}) = km (\vec{u} \times \vec{v})$, we get nine terms:

$\phantom{\vec{a} \times \vec{b}} = a_1 b_1 (\hat{i} \times \hat{i}) + a_1 b_2 (\hat{i} \times \hat{j}) + a_1 b_3 (\hat{i} \times \hat{k})$

$\phantom{\vec{a} \times \vec{b} =} + a_2 b_1 (\hat{j} \times \hat{i}) + a_2 b_2 (\hat{j} \times \hat{j}) + a_2 b_3 (\hat{j} \times \hat{k})$

$\phantom{\vec{a} \times \vec{b} =} + a_3 b_1 (\hat{k} \times \hat{i}) + a_3 b_2 (\hat{k} \times \hat{j}) + a_3 b_3 (\hat{k} \times \hat{k})$

Substitute the cross products of the unit vectors ($\hat{i} \times \hat{i} = \vec{0}$, $\hat{i} \times \hat{j} = \hat{k}$, etc.):

$\phantom{\vec{a} \times \vec{b}} = a_1 b_1 (\vec{0}) + a_1 b_2 (\hat{k}) + a_1 b_3 (-\hat{j})$

$\phantom{\vec{a} \times \vec{b} =} + a_2 b_1 (-\hat{k}) + a_2 b_2 (\vec{0}) + a_2 b_3 (\hat{i})$

$\phantom{\vec{a} \times \vec{b} =} + a_3 b_1 (\hat{j}) + a_3 b_2 (-\hat{i}) + a_3 b_3 (\vec{0})$

The terms with $\hat{i} \times \hat{i}$, $\hat{j} \times \hat{j}$, $\hat{k} \times \hat{k}$ are zero vectors and disappear. Group the remaining terms by the unit vectors $\hat{i}, \hat{j}, \hat{k}$:

Terms with $\hat{i}$: $a_2 b_3 \hat{i} - a_3 b_2 \hat{i} = (a_2 b_3 - a_3 b_2)\hat{i}$

Terms with $\hat{j}$: $- a_1 b_3 \hat{j} + a_3 b_1 \hat{j} = (a_3 b_1 - a_1 b_3)\hat{j}$

Terms with $\hat{k}$: $a_1 b_2 \hat{k} - a_2 b_1 \hat{k} = (a_1 b_2 - a_2 b_1)\hat{k}$

Combining these, we get the component form of the cross product:

$$ \vec{a} \times \vec{b} = (a_2 b_3 - a_3 b_2)\hat{i} + (a_3 b_1 - a_1 b_3)\hat{j} + (a_1 b_2 - a_2 b_1)\hat{k} $$

Determinant Formula for Vector Product

The component form derived above can be elegantly expressed and easily remembered as the expansion of a $3 \times 3$ determinant. We construct a determinant where the first row consists of the standard unit vectors $\hat{i}, \hat{j}, \hat{k}$, the second row consists of the scalar components of the first vector $\vec{a}$, and the third row consists of the scalar components of the second vector $\vec{b}$.

Let $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$.

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

Expanding this determinant along the first row using cofactor expansion:

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

Calculate the $2 \times 2$ determinants:

• $\begin{vmatrix} a_2 & a_3 \\ b_2 & b_3 \end{vmatrix} = a_2 b_3 - a_3 b_2$
• $\begin{vmatrix} a_1 & a_3 \\ b_1 & b_3 \end{vmatrix} = a_1 b_3 - a_3 b_1$
• $\begin{vmatrix} a_1 & a_2 \\ b_1 & b_2 \end{vmatrix} = a_1 b_2 - a_2 b_1$

Substitute these back into the expansion:

$$ \vec{a} \times \vec{b} = (a_2 b_3 - a_3 b_2)\hat{i} - (a_1 b_3 - a_3 b_1)\hat{j} + (a_1 b_2 - a_2 b_1)\hat{k} $$

This is the same component formula derived earlier. The determinant form is the standard way to calculate the cross product algebraically.

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Geometric Interpretation and Area of Parallelogram/Triangle

The magnitude of the vector product $|\vec{a} \times \vec{b}|$ has a powerful and intuitive geometric interpretation. It directly relates to the area of the parallelogram formed by the two vectors.

Geometric Interpretation: Area of Parallelogram

Recall the definition of the magnitude of the cross product of two non-zero vectors $\vec{a}$ and $\vec{b}$: $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta$, where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$ ($0 \le \theta \le \pi$).

Consider a parallelogram formed by taking vectors $\vec{a}$ and $\vec{b}$ as its adjacent sides, originating from a common vertex.

Let's calculate the area of this parallelogram using the standard formula: Area = Base $\times$ Height.

Choose the side represented by vector $\vec{a}$ as the base. The length of the base is the magnitude of $\vec{a}$, which is $|\vec{a}|$.

The height $h$ of the parallelogram corresponding to this base is the perpendicular distance from the terminal point of $\vec{b}$ to the line containing $\vec{a}$. From trigonometry in the right triangle formed by $\vec{b}$ and the height, this height is given by $h = |\vec{b}| \sin \theta$.

Now, substitute these into the area formula:

We know that the magnitude of the cross product is defined as $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta$.

Therefore, the area of the parallelogram formed by adjacent vectors $\vec{a}$ and $\vec{b}$ is equal to the magnitude of their cross product:

$$ \mathbf{\text{Area of Parallelogram} = |\vec{a} \times \vec{b}|} $$

If $\vec{a}$ and $\vec{b}$ are collinear, the angle $\theta$ is $0$ or $\pi$, $\sin \theta = 0$, and $|\vec{a} \times \vec{b}| = 0$. This makes sense, as collinear vectors would form a degenerate parallelogram with zero area.

Area of Triangle

Consider a triangle formed by two vectors $\vec{a}$ and $\vec{b}$ as two of its adjacent sides, sharing a common vertex. This triangle is exactly half the area of the parallelogram formed by the same two vectors.

Therefore, the area of the triangle with adjacent sides $\vec{a}$ and $\vec{b}$ is:

$$ \mathbf{\text{Area of Triangle} = \frac{1}{2} |\vec{a} \times \vec{b}|} $$

If the vertices of a triangle are given by points A, B, and C, we can form vectors representing two sides of the triangle originating from a common vertex. For example, we can use vectors $\vec{AB}$ and $\vec{AC}$. These vectors are adjacent sides of the triangle ABC. The area of triangle ABC is then given by:

$$ \text{Area of } \triangle ABC = \frac{1}{2} |\vec{AB} \times \vec{AC}| $$

Alternatively, we could use other pairs of vectors originating from a common vertex, such as $\vec{BA}$ and $\vec{BC}$ (Area = $\frac{1}{2} |\vec{BA} \times \vec{BC}|$), or $\vec{CA}$ and $\vec{CB}$ (Area = $\frac{1}{2} |\vec{CA} \times \vec{CB}|$). All these will give the same area, as $|\vec{AB} \times \vec{AC}| = |\vec{BA} \times \vec{BC}| = |\vec{CA} \times \vec{CB}|$. For instance, $\vec{BA} = -\vec{AB}$ and $\vec{BC} = \vec{AC} - \vec{AB}$. $\vec{BA} \times \vec{BC} = (-\vec{AB}) \times (\vec{AC} - \vec{AB}) = -\vec{AB} \times \vec{AC} + (-\vec{AB}) \times (-\vec{AB}) = -(\vec{AB} \times \vec{AC}) + \vec{AB} \times \vec{AB} = -(\vec{AB} \times \vec{AC}) + \vec{0} = -(\vec{AB} \times \vec{AC})$. The magnitude is $|-(\vec{AB} \times \vec{AC})| = |\vec{AB} \times \vec{AC}|$.

Answer:

Answer:

Applications of Cross Product (Finding Normal Vector, Torque)

The vector product is a powerful mathematical tool used extensively in various fields, particularly in physics and geometry, where quantities involving perpendicular directions or rotations are involved.

1. Finding a Normal Vector to a Plane

A key property of the cross product $\vec{a} \times \vec{b}$ is that the resulting vector is always perpendicular to both $\vec{a}$ and $\vec{b}$. If $\vec{a}$ and $\vec{b}$ are two non-collinear vectors lying in a plane, their cross product $\vec{a} \times \vec{b}$ provides a vector that is perpendicular to that plane.

This perpendicular vector is called a normal vector to the plane. If $\vec{n} = \vec{a} \times \vec{b}$, then $\vec{n}$ is normal to the plane containing $\vec{a}$ and $\vec{b}$. Any non-zero scalar multiple of $\vec{n}$, $k\vec{n}$ (where $k \neq 0$), is also a normal vector to the same plane.

This application is fundamental in:

• Defining the orientation of a plane in 3D space.
• Finding the equation of a plane (as the normal vector is a key component of the plane equation).
• Determining the angle between two planes (which is related to the angle between their normal vectors).
• Calculating the distance from a point to a plane.

Answer:

2. Torque (Moment of a Force)

In rotational mechanics, torque is the rotational equivalent of linear force. It measures how effectively a force causes an object to rotate about a pivot point or axis. The torque ($\vec{\tau}$) produced by a force ($\vec{F}$) acting at a point P relative to a pivot point O is defined using the cross product of the position vector of P relative to O ($\vec{r} = \vec{OP}$) and the force vector $\vec{F}$.

$$ \vec{\tau} = \vec{r} \times \vec{F} $$

where $\vec{r}$ is the vector from the pivot point O to the point where the force $\vec{F}$ is applied.

• Magnitude of Torque: The magnitude of the torque is $|\vec{\tau}| = |\vec{r} \times \vec{F}| = |\vec{r}| |\vec{F}| \sin \theta$, where $\theta$ is the angle between the position vector $\vec{r}$ and the force vector $\vec{F}$. This magnitude is often interpreted as the product of the lever arm (the perpendicular distance from the pivot to the line of action of the force, which is $|\vec{r}|\sin\theta$) and the magnitude of the force, or the product of the distance from the pivot ($|\vec{r}|$) and the component of the force perpendicular to the position vector ($|\vec{F}|\sin\theta$).
• Direction of Torque: The direction of the torque vector $\vec{\tau}$ is perpendicular to the plane containing $\vec{r}$ and $\vec{F}$, and it points along the axis about which the force tends to cause rotation. The specific sense of rotation is determined by the Right-Hand Rule applied to $\vec{r} \times \vec{F}$. If the force tends to cause counter-clockwise rotation about the axis, the torque vector points towards the viewer (assuming $\vec{r}$ and $\vec{F}$ are in the plane of the page); if it tends to cause clockwise rotation, the torque vector points away.

3. Other Important Applications in Physics

• Magnetic Force on a Moving Charge: The force $\vec{F}$ experienced by a point charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ is given by the magnetic part of the Lorentz force law: $\vec{F} = q(\vec{v} \times \vec{B})$. The force is proportional to the charge, speed, magnitude of the magnetic field, and the sine of the angle between the velocity and the magnetic field, and its direction is perpendicular to both $\vec{v}$ and $\vec{B}$.
• Angular Momentum: For a particle with linear momentum $\vec{p} = m\vec{v}$ located at position $\vec{r}$ relative to an origin, its angular momentum $\vec{L}$ about that origin is defined as the cross product: $\vec{L} = \vec{r} \times \vec{p}$. Angular momentum is a crucial quantity in the description of rotational motion.