Scalar (Dot) Product of Vectors

Scalar (or Dot) Product of Two Vectors: Definition and Formula

Beyond the operations of vector addition, subtraction, and scalar multiplication, there are ways to combine two vectors through processes that can be thought of as "multiplication". One such operation is the scalar product, also known as the dot product. This operation takes two vectors and produces a scalar quantity.

Definition of the Scalar (Dot) Product

The scalar product or dot product of two non-zero vectors $\vec{a}$ and $\vec{b}$ is a scalar quantity defined as the product of their magnitudes and the cosine of the angle between them. The angle $\theta$ between the two vectors is always considered to be between $0$ and $\pi$ radians (or $0^\circ$ and $180^\circ$). When determining the angle, the vectors are imagined to originate from the same initial point.

• Notation: The scalar product of $\vec{a}$ and $\vec{b}$ is denoted by $\vec{a} \cdot \vec{b}$. The dot symbol $(\cdot)$ is used, which is why it is commonly called the dot product.
• Formula: The mathematical definition is:

  $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$

  Here, $|\vec{a}|$ is the magnitude of vector $\vec{a}$, $|\vec{b}|$ is the magnitude of vector $\vec{b}$, and $\theta$ is the angle between $\vec{a}$ and $\vec{b}$.

Result of the Scalar Product

The most important characteristic of the scalar product is its result. As the name suggests, $\vec{a} \cdot \vec{b}$ is always a scalar quantity (a real number), not a vector. This is distinct from the vector product (cross product), which results in a vector.

Special Cases of the Dot Product

• If either vector is the zero vector:

  If $\vec{a} = \vec{0}$ or $\vec{b} = \vec{0}$ (or both), the angle $\theta$ between the vectors is undefined. However, the scalar product is defined to be 0 in this case.

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

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

  This follows naturally from the formula since the magnitude of the zero vector is 0 ($|\vec{0}|=0$).
• If $\vec{a}$ and $\vec{b}$ are non-zero vectors:
  • Perpendicular Vectors: If $\vec{a}$ and $\vec{b}$ are perpendicular (or orthogonal), the angle between them is $\theta = 90^\circ = \pi/2$ radians.

    $\cos(\pi/2) = 0$.

    Therefore, the dot product is:

    $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos(\pi/2) = |\vec{a}| |\vec{b}| (0) = 0$

    This is a fundamental property: the dot product of two non-zero perpendicular vectors is zero. Conversely, if the dot product of two non-zero vectors is zero, they must be perpendicular.

  • Parallel Vectors (Same Direction): If $\vec{a}$ and $\vec{b}$ are parallel and point in the same direction, the angle between them is $\theta = 0^\circ = 0$ radians.

    $\cos(0) = 1$.

    The dot product is:

    $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos(0) = |\vec{a}| |\vec{b}| (1) = |\vec{a}| |\vec{b}|$.

    The dot product is the product of their magnitudes.

  • Parallel Vectors (Opposite Direction): If $\vec{a}$ and $\vec{b}$ are parallel but point in opposite directions (anti-parallel), the angle between them is $\theta = 180^\circ = \pi$ radians.

    $\cos(\pi) = -1$.

    The dot product is:

    $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos(\pi) = |\vec{a}| |\vec{b}| (-1) = -|\vec{a}| |\vec{b}|$.

    The dot product is the negative of the product of their magnitudes.

  • Dot Product of a Vector with Itself: The angle between a vector $\vec{a}$ and itself is $0^\circ$.

    $\vec{a} \cdot \vec{a} = |\vec{a}| |\vec{a}| \cos 0^\circ = |\vec{a}|^2 (1) = |\vec{a}|^2$

    The dot product of a vector with itself is equal to the square of its magnitude. This provides a way to calculate the magnitude using the dot product: $|\vec{a}| = \sqrt{\vec{a} \cdot \vec{a}}$.

Answer:

Properties of Scalar Product (Commutativity, Distributivity, etc.)

The scalar product of vectors possesses several algebraic properties that are useful in simplifying expressions and solving problems involving vectors. These properties are similar to some properties of multiplication of real numbers.

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

1. Commutativity:

   The order in which the dot product is taken does not affect the result.

   $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$

   Proof: Let $\theta$ be the angle between $\vec{a}$ and $\vec{b}$. By definition, $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$. The angle between $\vec{b}$ and $\vec{a}$ is also $\theta$. So, $\vec{b} \cdot \vec{a} = |\vec{b}| |\vec{a}| \cos \theta$. Since the multiplication of magnitudes (which are scalars) is commutative ($|\vec{a}| |\vec{b}| = |\vec{b}| |\vec{a}|$), we have $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$.

2. Distributivity over Vector Addition:

   The dot product distributes over vector addition. The dot product of a vector with the sum of two other vectors is equal to the sum of the dot products of the vector with each of the other two vectors separately.

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

   By commutativity, the right-hand distributive property also holds:

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

   While proving this geometrically requires the concept of projection, it is a fundamental property of the dot product.

3. Association with Scalar Multiplication:

   When a scalar is involved in a dot product, it can be associated with either vector or factored out of the dot product entirely.

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

   Proof Sketch: We can prove this by considering cases for $k$. Case 1: $k > 0$. The vector $k\vec{a}$ has magnitude $k|\vec{a}|$ and the same direction as $\vec{a}$. The angle between $k\vec{a}$ and $\vec{b}$ is the same as the angle between $\vec{a}$ and $\vec{b}$, say $\theta$. $(k\vec{a}) \cdot \vec{b} = |k\vec{a}||\vec{b}|\cos\theta = (k|\vec{a}|)|\vec{b}|\cos\theta = k(|\vec{a}||\vec{b}|\cos\theta) = k(\vec{a}\cdot\vec{b})$. Case 2: $k < 0$. The vector $k\vec{a}$ has magnitude $|k||\vec{a}| = (-k)|\vec{a}|$ and the opposite direction to $\vec{a}$. The angle between $k\vec{a}$ and $\vec{b}$ is $\pi - \theta$ (if $\theta$ is the angle between $\vec{a}$ and $\vec{b}$). $(k\vec{a}) \cdot \vec{b} = |k\vec{a}||\vec{b}|\cos(\pi-\theta) = (-k)|\vec{a}||\vec{b}|(-\cos\theta) = k|\vec{a}||\vec{b}|\cos\theta = k(\vec{a}\cdot\vec{b})$. Case 3: $k = 0$. $(0\vec{a}) \cdot \vec{b} = \vec{0} \cdot \vec{b} = 0$. Also, $0(\vec{a} \cdot \vec{b}) = 0$. And $\vec{a} \cdot (0\vec{b}) = \vec{a} \cdot \vec{0} = 0$. The property holds.

4. Non-negativity of Self Dot Product:

   The dot product of any vector with itself is always non-negative.

   $\vec{a} \cdot \vec{a} = |\vec{a}|^2 \ge 0$

   This is because the square of a real number (magnitude) is always non-negative.

5. Zero Self Dot Product:

   $\vec{a} \cdot \vec{a} = 0$ if and only if $\vec{a} = \vec{0}$.

   This follows directly from $\vec{a} \cdot \vec{a} = |\vec{a}|^2$. The square of the magnitude is zero if and only if the magnitude itself is zero, which occurs if and only if the vector is the zero vector.

6. Cauchy-Schwarz Inequality:

   For any two vectors $\vec{a}$ and $\vec{b}$, the absolute value of their dot product is less than or equal to the product of their magnitudes.

   $|\vec{a} \cdot \vec{b}| \le |\vec{a}| |\vec{b}|$

   Proof: By definition, $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$. Taking the absolute value of both sides:

   $|\vec{a} \cdot \vec{b}| = ||\vec{a}| |\vec{b}| \cos \theta| = |\vec{a}| |\vec{b}| |\cos \theta|$

   (Since magnitudes are non-negative, $|\vec{a}| |\vec{b}| = ||\vec{a}| |\vec{b}||$). We know that for any angle $\theta$, the value of $|\cos \theta|$ is between 0 and 1 (inclusive), i.e., $0 \le |\cos \theta| \le 1$.

   Multiplying the inequality by $|\vec{a}| |\vec{b}|$ (which is non-negative), we get:

   $|\vec{a}| |\vec{b}| |\cos \theta| \le |\vec{a}| |\vec{b}| \cdot 1$

   Substituting back $|\vec{a} \cdot \vec{b}| = |\vec{a}| |\vec{b}| |\cos \theta|$, we obtain:

   $|\vec{a} \cdot \vec{b}| \le |\vec{a}| |\vec{b}|$

   Equality holds if and only if $|\cos \theta| = 1$, which means $\theta = 0$ or $\theta = \pi$, i.e., the vectors are collinear.

7. Triangle Inequality:

   For any two vectors $\vec{a}$ and $\vec{b}$, the magnitude of their sum is less than or equal to the sum of their magnitudes. This is a fundamental geometric principle (the sum of two sides of a triangle is greater than or equal to the third side), which can be proven using the dot product and the Cauchy-Schwarz inequality.

   $|\vec{a} + \vec{b}| \le |\vec{a}| + |\vec{b}|$

   Proof using dot product: Consider the square of the magnitude of the sum vector:

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

   Using the distributive property of the dot product:

   $\phantom{|\vec{a} + \vec{b}|^2} = \vec{a}\cdot(\vec{a}+\vec{b}) + \vec{b}\cdot(\vec{a}+\vec{b})$

   $\phantom{|\vec{a} + \vec{b}|^2} = \vec{a}\cdot\vec{a} + \vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{a} + \vec{b}\cdot\vec{b}$

   Using $\vec{a} \cdot \vec{a} = |\vec{a}|^2$, $\vec{b} \cdot \vec{b} = |\vec{b}|^2$, and $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$ (commutativity):

   $\phantom{|\vec{a} + \vec{b}|^2} = |\vec{a}|^2 + 2(\vec{a}\cdot\vec{b}) + |\vec{b}|^2$

   Now, we know that $\vec{a} \cdot \vec{b} \le |\vec{a} \cdot \vec{b}|$. So, $2(\vec{a}\cdot\vec{b}) \le 2|\vec{a} \cdot \vec{b}|$.

   $|\vec{a} + \vec{b}|^2 \le |\vec{a}|^2 + 2|\vec{a}\cdot\vec{b}| + |\vec{b}|^2$

   Applying the Cauchy-Schwarz inequality, $|\vec{a} \cdot \vec{b}| \le |\vec{a}| |\vec{b}|$:

   $|\vec{a} + \vec{b}|^2 \le |\vec{a}|^2 + 2|\vec{a}||\vec{b}| + |\vec{b}|^2$

   The right side is a perfect square:

   $|\vec{a} + \vec{b}|^2 \le (|\vec{a}| + |\vec{b}|)^2$

   Taking the non-negative square root of both sides (since magnitudes are non-negative), we get the Triangle Inequality:

   $|\vec{a} + \vec{b}| \le |\vec{a}| + |\vec{b}|$

   Equality holds if and only if $\vec{a}$ and $\vec{b}$ are collinear and point in the same direction, i.e., $\theta=0$.

Scalar Product in terms of Components

While the geometric definition of the dot product ($\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$) is fundamental, it requires knowing the angle between the vectors, which might not always be readily available. If the vectors are given in terms of their components with respect to an orthonormal basis (like the standard Cartesian basis $\{\hat{i}, \hat{j}, \hat{k}\}$), we can compute the dot product using a simpler algebraic formula based on these components.

Dot Products of Standard Unit Vectors

The standard unit vectors $\hat{i}, \hat{j}, \hat{k}$ are unit vectors (magnitude 1) and are mutually perpendicular. Let's find their dot products based on the definition $\vec{u} \cdot \vec{v} = |\vec{u}| |\vec{v}| \cos \theta$.

• Dot product of a unit vector with itself:

  The angle is $\theta = 0^\circ$, $\cos 0^\circ = 1$. The magnitude of a unit vector is 1.

  $\hat{i} \cdot \hat{i} = |\hat{i}| |\hat{i}| \cos 0^\circ = (1)(1)(1) = 1$

  Similarly,

  $\hat{j} \cdot \hat{j} = |\hat{j}| |\hat{j}| \cos 0^\circ = (1)(1)(1) = 1$

  $\hat{k} \cdot \hat{k} = |\hat{k}| |\hat{k}| \cos 0^\circ = (1)(1)(1) = 1$

• Dot product of two different unit vectors:

  The angle between any two different standard unit vectors ($\hat{i}$ and $\hat{j}$, $\hat{j}$ and $\hat{k}$, $\hat{k}$ and $\hat{i}$) is $\theta = 90^\circ = \pi/2$ radians, since they are mutually perpendicular.

  $\cos 90^\circ = 0$.

  $\hat{i} \cdot \hat{j} = |\hat{i}| |\hat{j}| \cos 90^\circ = (1)(1)(0) = 0$

  Similarly,

  $\hat{j} \cdot \hat{i} = 0$ (by commutativity)

  $\hat{j} \cdot \hat{k} = |\hat{j}| |\hat{k}| \cos 90^\circ = (1)(1)(0) = 0$

  $\hat{k} \cdot \hat{j} = 0$

  $\hat{k} \cdot \hat{i} = |\hat{k}| |\hat{i}| \cos 90^\circ = (1)(1)(0) = 0$

  $\hat{i} \cdot \hat{k} = 0$

Summary of unit vector dot products:

(and their commutative pairs are also zero).

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 dot product using the distributive property of the dot product over vector addition, and the scalar multiplication property:

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

Expanding this using distributivity (like multiplying two trinomials):

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

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

Using the property $(k\vec{u}) \cdot (m\vec{v}) = km (\vec{u} \cdot \vec{v})$:

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

Now, substitute the dot products of the unit vectors (1 for same, 0 for different):

$\vec{a} \cdot \vec{b} = a_1 b_1 (1) + a_1 b_2 (0) + a_1 b_3 (0) + a_2 b_1 (0) + a_2 b_2 (1) + a_2 b_3 (0) + a_3 b_1 (0) + a_3 b_2 (0) + a_3 b_3 (1)$

Most terms vanish, leaving only the terms where the unit vectors are the same:

$$ \vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 $$

This derivation shows how the component formula arises directly from the definition and the properties of the standard basis vectors.

Component Formula for Scalar Product

If two vectors $\vec{a}$ and $\vec{b}$ are given in component form as $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$, their scalar product is the sum of the products of their corresponding scalar components:

This formula is extremely useful for calculating the dot product when vector components are known.

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Geometric Interpretation and Projection of a Vector on Another Vector

The scalar product $\vec{a} \cdot \vec{b}$ has a very insightful geometric interpretation that relates to the idea of 'shadows' or components of one vector along the direction of another. This interpretation leads directly to the concepts of scalar and vector projections.

Geometric Interpretation of the Dot Product

Recall the definition of the scalar product of two non-zero vectors $\vec{a}$ and $\vec{b}$: $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$, where $\theta$ is the angle between them.

We can rearrange this formula in two ways:

1. $\vec{a} \cdot \vec{b} = |\vec{a}| (|\vec{b}| \cos \theta)$

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

Consider the term $|\vec{b}| \cos \theta$. If we place vectors $\vec{a}$ and $\vec{b}$ with a common initial point, say O, and let P be the terminal point of $\vec{b}$, and draw a perpendicular from P to the line containing $\vec{a}$, meeting it at point M. The segment OM is the projection of $\vec{b}$ onto the line of $\vec{a}$. The length of this segment OM, taken with a sign depending on whether M is on the same side of O as the terminal point of $\vec{a}$ (positive) or the opposite side (negative), is given by $|\vec{b}| \cos \theta$.

The quantity $|\vec{b}| \cos \theta$ is called the scalar projection of $\vec{b}$ onto $\vec{a}$. It tells us "how much" of vector $\vec{b}$ lies in the direction of vector $\vec{a}$.

Thus, the dot product $\vec{a} \cdot \vec{b}$ can be interpreted as:

$$ \vec{a} \cdot \vec{b} = (\text{Magnitude of } \vec{a}) \times (\text{Scalar Projection of } \vec{b} \text{ onto } \vec{a}) $$

Similarly, the term $|\vec{a}| \cos \theta$ is the scalar projection of $\vec{a}$ onto $\vec{b}$.

$$ \vec{a} \cdot \vec{b} = (\text{Magnitude of } \vec{b}) \times (\text{Scalar Projection of } \vec{a} \text{ onto } \vec{b}) $$

Scalar Projection of a Vector onto Another Vector

The scalar projection of vector $\vec{b}$ onto vector $\vec{a}$ is the signed length of the projection of $\vec{b}$ onto the line containing $\vec{a}$. It is given by $|\vec{b}| \cos \theta$.

From the dot product formula, $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$, we can isolate the term $|\vec{b}| \cos \theta$ (assuming $|\vec{a}| \neq 0$):

$$ \text{Scalar projection of } \vec{b} \text{ onto } \vec{a} = |\vec{b}| \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|} $$

This formula allows us to calculate the scalar projection using the dot product and the magnitude of the vector we are projecting onto. Note that the scalar projection can be positive (if $0 \le \theta < \pi/2$), negative (if $\pi/2 < \theta \le \pi$), or zero (if $\theta = \pi/2$).

We can also express the scalar projection using the unit vector in the direction of $\vec{a}$. Let $\hat{a} = \frac{\vec{a}}{|\vec{a}|}$ be the unit vector in the direction of $\vec{a}$. Then the scalar projection of $\vec{b}$ onto $\vec{a}$ is:

$$ \text{Scalar projection of } \vec{b} \text{ onto } \vec{a} = |\vec{b}| \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|} = \left(\frac{\vec{a}}{|\vec{a}|}\right) \cdot \vec{b} = \hat{a} \cdot \vec{b} = \vec{b} \cdot \hat{a} $$

So, the scalar projection of $\vec{b}$ onto $\vec{a}$ is the dot product of $\vec{b}$ with the unit vector in the direction of $\vec{a}$.

Similarly, the scalar projection of $\vec{a}$ onto $\vec{b}$ is $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \vec{a} \cdot \hat{b}$.

Vector Projection of a Vector onto Another Vector

The vector projection of vector $\vec{b}$ onto vector $\vec{a}$ is a vector that lies along the direction of $\vec{a}$ and has a magnitude equal to the absolute value of the scalar projection of $\vec{b}$ onto $\vec{a}$. Its direction is the same as $\vec{a}$ if the scalar projection is positive, and opposite to $\vec{a}$ if the scalar projection is negative.

A vector is defined by its magnitude and direction. The magnitude of the vector projection of $\vec{b}$ onto $\vec{a}$ is $|\text{Scalar projection of } \vec{b} \text{ onto } \vec{a}| = ||\vec{b}| \cos \theta| = |\vec{b}| |\cos \theta|$. However, it is more common to use the directed projection. The vector projection is the scalar projection multiplied by the unit vector in the direction of $\vec{a}$.

$$ \text{Vector projection of } \vec{b} \text{ onto } \vec{a} = (\text{Scalar projection of } \vec{b} \text{ onto } \vec{a}) \times (\text{Unit vector in direction of } \vec{a}) $$

$$ \text{Vector projection of } \vec{b} \text{ onto } \vec{a} = \left( \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|} \right) \hat{a} $$

Substitute $\hat{a} = \frac{\vec{a}}{|\vec{a}|}$ into the formula:

$$ \text{Vector projection of } \vec{b} \text{ onto } \vec{a} = \left( \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|} \right) \frac{\vec{a}}{|\vec{a}|} = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2} \vec{a} $$

Since $|\vec{a}|^2 = \vec{a} \cdot \vec{a}$, we can also write this as:

$$ \text{Vector projection of } \vec{b} \text{ onto } \vec{a} = \left( \frac{\vec{a} \cdot \vec{b}}{\vec{a} \cdot \vec{a}} \right) \vec{a} $$

This formula gives the vector component of $\vec{b}$ along the direction of $\vec{a}$. The scalar $\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2}$ is sometimes called the "component multiplier".

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Applications of Dot Product (Angle between Vectors, Work Done)

The scalar product is a versatile tool with numerous applications in geometry, physics, and engineering. Here, we discuss two common applications: finding the angle between vectors and calculating the work done by a force.

1. Finding the Angle Between Two Vectors

One of the most direct applications of the dot product is to determine the angle between two non-zero vectors. We can rearrange the definition of the scalar product, $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$, to solve for $\cos \theta$. Assuming $|\vec{a}| \neq 0$ and $|\vec{b}| \neq 0$, we get:

$$ \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} $$

This formula allows us to find the cosine of the angle using only the dot product and the magnitudes of the vectors, which can be easily calculated if the vectors are given in component form.

If the vectors are $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$, then $\vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3$, $|\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}$, and $|\vec{b}| = \sqrt{b_1^2 + b_2^2 + b_3^2}$. Substituting these into the formula for $\cos \theta$:

$$ \cos \theta = \frac{a_1 b_1 + a_2 b_2 + a_3 b_3}{\sqrt{a_1^2 + a_2^2 + a_3^2} \sqrt{b_1^2 + b_2^2 + b_3^2}} $$

Once the value of $\cos \theta$ is computed, the angle $\theta$ itself can be found using the inverse cosine function (arccosine):

$$ \theta = \arccos\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}\right) $$

Since the angle $\theta$ between two vectors is conventionally taken between $0$ and $\pi$ radians ($0^\circ$ and $180^\circ$), the arccosine function gives a unique value for $\theta$ in this range.

Condition for Perpendicularity (Orthogonality)

As noted earlier, two non-zero vectors $\vec{a}$ and $\vec{b}$ are perpendicular (or orthogonal) if and only if the angle $\theta$ between them is $90^\circ$ ($\pi/2$ radians). For this angle, $\cos 90^\circ = 0$.

Using the dot product definition, $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$, if $\theta = 90^\circ$, then $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| (0) = 0$.

Conversely, if $\vec{a} \cdot \vec{b} = 0$ for non-zero vectors $\vec{a}$ and $\vec{b}$, then $|\vec{a}| |\vec{b}| \cos \theta = 0$. Since $|\vec{a}| \neq 0$ and $|\vec{b}| \neq 0$, it must be that $\cos \theta = 0$. For $0 \le \theta \le \pi$, $\cos \theta = 0$ implies $\theta = \pi/2 = 90^\circ$. Thus, the vectors are perpendicular.

This gives us a simple and powerful algebraic test for perpendicularity using components:

Two non-zero vectors $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$ are perpendicular if and only if the sum of the products of their corresponding components is zero:

$$ \mathbf{\vec{a} \perp \vec{b} \iff a_1 b_1 + a_2 b_2 + a_3 b_3 = 0} $$

(This holds true even if one or both vectors are the zero vector, as the dot product would still be zero, and the zero vector is considered perpendicular to all vectors).

Answer:

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2. Work Done by a Constant Force

In physics, the concept of work done by a force is a direct application of the scalar product. When a constant force $\vec{F}$ acts on an object and causes a displacement $\vec{d}$, the work done (denoted by W) by the force is defined as the scalar product of the force vector and the displacement vector.

$$ W = \vec{F} \cdot \vec{d} $$

Using the definition of the dot product, this formula can also be written as:

$$ W = |\vec{F}| |\vec{d}| \cos \theta $$

where $|\vec{F}|$ is the magnitude of the force, $|\vec{d}|$ is the magnitude of the displacement, and $\theta$ is the angle between the direction of the force and the direction of the displacement.

This definition makes physical sense: work is done only by the component of the force that acts in the direction of the displacement. The component of $\vec{F}$ in the direction of $\vec{d}$ is $|\vec{F}| \cos \theta$ (the scalar projection of $\vec{F}$ onto $\vec{d}$). Multiplying this component by the distance moved ($|\vec{d}|$) gives the work done. Force components perpendicular to the displacement ($\theta = 90^\circ$, $\cos 90^\circ = 0$) do no work.

Work is a scalar quantity, measured in units like Joules (J) in the SI system, where 1 Joule is equal to 1 Newton-meter (1 Nm). This is consistent with the dot product resulting in a scalar.

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