Dot Product of Vectors

Angle between Two Vectors

When we need to measure the angle between two vectors \(\overrightarrow{\mathbf{a}}\) and \(\overrightarrow{\mathbf{b}}\) in three-dimensional space, we first place them co-initial, meaning we position them so that they share the same starting point. Once the vectors are co-initial, the angle between them is the smallest angle through which one vector must be rotated to align with the other, measured in the plane containing both vectors. This angle is always taken between \(0\) and \(\pi\), ensuring that it represents the true directional difference between the vectors, independent of their magnitudes.

If the two vectors \(\overrightarrow{\mathbf{a}}\) and \(\overrightarrow{\mathbf{b}}\) are collinear, meaning they lie along the same line, the angle between them is either \(0\) or \(\pi\). If they point in the same direction, the angle is \(0\), and if they point in opposite directions, the angle is \(\pi\).

On the other hand, if the vectors are perpendicular, meaning they meet at a right angle, the angle between them is exactly \(\frac{\pi}{2}\). If the angle is less than \(\frac{\pi}{2}\), the vectors form an acute angle, meaning they are closer to pointing in the same direction. If the angle is greater than \(\frac{\pi}{2}\) but less than \(\pi\), they form an obtuse angle, meaning they are more spread apart but not completely opposite in direction.

Thus, we classify the angle between two vectors as:

• \(0 \leq \theta < \frac{\pi}{2}\) → Acute angle
• \(\theta = \frac{\pi}{2}\) → Right angle (perpendicular vectors)
• \(\frac{\pi}{2} < \theta \leq \pi\) → Obtuse angle

The nature of the angle gives insight into how much the vectors are inclined towards or away from each other.

Measuring the Angle between Two Vectors

To measure the angle between two vectors algebraically, we first express them in their resolved form. Let

\[ \overrightarrow{\mathbf{u}} = x_1 \mathbf{i} + y_1 \mathbf{j} + z_1 \mathbf{k}, \quad \overrightarrow{\mathbf{v}} = x_2 \mathbf{i} + y_2 \mathbf{j} + z_2 \mathbf{k} \]

We position them co-initial at the origin by moving their initial points to the origin. Define:

\[ \overrightarrow{\mathbf{OP}} = \overrightarrow{\mathbf{u}}, \quad \overrightarrow{\mathbf{OQ}} = \overrightarrow{\mathbf{v}} \]

Let the angle between \(\overrightarrow{\mathbf{OP}}\) and \(\overrightarrow{\mathbf{OQ}}\) be \(\theta\). The magnitudes of these vectors are:

\[ |\overrightarrow{\mathbf{OP}}| = \sqrt{x_1^2 + y_1^2 + z_1^2}, \quad |\overrightarrow{\mathbf{OQ}}| = \sqrt{x_2^2 + y_2^2 + z_2^2} \]

The vector \(\overrightarrow{\mathbf{PQ}}\) is given by:

\[ \overrightarrow{\mathbf{PQ}} = \overrightarrow{\mathbf{OQ}} - \overrightarrow{\mathbf{OP}} = (x_2 - x_1) \mathbf{i} + (y_2 - y_1) \mathbf{j} + (z_2 - z_1) \mathbf{k} \]

Its magnitude is:

\[ |\overrightarrow{\mathbf{PQ}}| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]

Applying the cosine law in \(\triangle OPQ\):

\[ \cos \theta = \frac{|\overrightarrow{\mathbf{OP}}|^2 + |\overrightarrow{\mathbf{OQ}}|^2 - |\overrightarrow{\mathbf{PQ}}|^2}{2 |\overrightarrow{\mathbf{OP}}| |\overrightarrow{\mathbf{OQ}}|} \]

Substituting the values:

\[ \cos \theta = \frac{x_1^2 + y_1^2 + z_1^2 + x_2^2 + y_2^2 + z_2^2 - [(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2]}{2 |\overrightarrow{\mathbf{OP}}| |\overrightarrow{\mathbf{OQ}}|} \]

Expanding \((x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2\):

\[ (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2 = x_2^2 - 2x_1x_2 + x_1^2 + y_2^2 - 2y_1y_2 + y_1^2 + z_2^2 - 2z_1z_2 + z_1^2 \]

Substituting this back:

\[ \cos \theta = \frac{x_1^2 + y_1^2 + z_1^2 + x_2^2 + y_2^2 + z_2^2 - (x_2^2 - 2x_1x_2 + x_1^2 + y_2^2 - 2y_1y_2 + y_1^2 + z_2^2 - 2z_1z_2 + z_1^2)}{2 |\overrightarrow{\mathbf{OP}}| |\overrightarrow{\mathbf{OQ}}|} \]

Canceling terms:

\[ \cos \theta = \frac{2x_1x_2 + 2y_1y_2 + 2z_1z_2}{2 |\overrightarrow{\mathbf{OP}}| |\overrightarrow{\mathbf{OQ}}|} \]

\[ \cos \theta = \frac{x_1 x_2 + y_1 y_2 + z_1 z_2}{|\overrightarrow{\mathbf{OP}}| |\overrightarrow{\mathbf{OQ}}|} \]

Thus, we obtain:

\[ \cos \theta = \frac{x_1 x_2 + y_1 y_2 + z_1 z_2}{\sqrt{x_1^2 + y_1^2 + z_1^2} \sqrt{x_2^2 + y_2^2 + z_2^2}} \]

which is the formula for the angle between two vectors.

Dot Product of Vectors

The quantity in the numerator of our formula—\( x_1x_2 + y_1y_2 + z_1z_2 \)—is not just any sum; it is something fundamental. This expression appears naturally when computing the angle between two vectors, and it captures an essential relationship between them.

This motivates us to define a new operation: the dot product of two vectors. We write:

\[ \overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{v}} = x_1x_2 + y_1y_2 + z_1z_2 \]

This definition is not arbitrary—it arises precisely because it allows us to express the angle between two vectors in a simple and elegant way. Substituting this into our formula for \(\cos \theta\), we obtain:

\[ \cos \theta = \frac{\overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{v}}}{|\overrightarrow{\mathbf{u}}| |\overrightarrow{\mathbf{v}}|} \]

This equation is profound. It tells us that the dot product encodes both the magnitudes of the vectors and the angle between them. Rewriting it, we arrive at a beautiful result:

\[ \overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{v}} = |\overrightarrow{\mathbf{u}}| |\overrightarrow{\mathbf{v}}| \cos \theta \]

About Notation

The dot product of two vectors \(\overrightarrow{\mathbf{u}}\) and \(\overrightarrow{\mathbf{v}}\) is denoted by the symbol "\(\cdot\)" (a centered dot). Using this notation, we write:

\[ \overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{v}} \]

which is read as "\(\overrightarrow{\mathbf{u}}\) dot \(\overrightarrow{\mathbf{v}}\)."

Properties of Dot Product

So, we have introduced a new operator on vectors—the dot product. Whenever we learn a new operation, we investigate its properties to understand how it interacts with other vector operations, such as addition and scalar multiplication.

Dot Product Produces a Scalar

The dot product (or scalar product) of two vectors is not a vector—it results in a scalar. That is, the operation is not closed under vectors:

\[ \overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{v}} = x_1x_2 + y_1y_2 + z_1z_2 \]

The result is a real number, not another vector. However, despite being a scalar, this number carries important geometric information about the vectors.

Sign of the Dot Product and the Angle Between Vectors

The value of \(\overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{v}}\) tells us about the angle \(\theta\) between the two vectors:

• If \(\overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{v}} > 0\), the angle \(\theta\) is acute (\(0 < \theta < \frac{\pi}{2}\)).
• If \(\overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{v}} < 0\), the angle \(\theta\) is obtuse (\(\frac{\pi}{2} < \theta < \pi\)).
• If \(\overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{v}} = 0\), then either one of the vectors is the zero vector or the vectors are perpendicular.

To see why this is true, recall the fundamental formula:

\[ \overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{v}} = |\overrightarrow{\mathbf{u}}| |\overrightarrow{\mathbf{v}}| \cos \theta \]

Since the angle \(\theta\) is always in the range \(0 \leq \theta \leq \pi\), the value of \(\cos \theta\) determines the sign of the dot product:

• For \(0 < \theta < \frac{\pi}{2}\), we have \(\cos \theta > 0\), so the dot product is positive.
• For \(\frac{\pi}{2} < \theta < \pi\), we have \(\cos \theta < 0\), so the dot product is negative.
• For \(\theta = \frac{\pi}{2}\), we have \(\cos \theta = 0\), so the dot product is zero. This means the vectors are perpendicular.

Thus, from just the sign of the dot product, we can determine whether two vectors form an acute, right, or obtuse angle with each other.

Commutativity and Associativity

Another fundamental property of the dot product is that it is commutative. That is, for any two vectors \(\overrightarrow{\mathbf{u}}\) and \(\overrightarrow{\mathbf{v}}\),

\[ \overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{v}} = \overrightarrow{\mathbf{v}} \cdot \overrightarrow{\mathbf{u}} \]

This follows directly from the definition:

\[ \overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{v}} = x_1x_2 + y_1y_2 + z_1z_2 \]

\[ \overrightarrow{\mathbf{v}} \cdot \overrightarrow{\mathbf{u}} = x_2x_1 + y_2y_1 + z_2z_1 \]

Since real numbers satisfy the commutative property of multiplication (\(a b = b a\)), it follows that:

\[ x_1x_2 + y_1y_2 + z_1z_2 = x_2x_1 + y_2y_1 + z_2z_1 \]

Thus,

\[ \overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{v}} = \overrightarrow{\mathbf{v}} \cdot \overrightarrow{\mathbf{u}} \]

This means that the order in which we take the dot product does not matter—just like how multiplying two numbers does not depend on order. Geometrically, this reflects the fact that the angle between two vectors remains the same regardless of which vector is considered first.

Another important observation is that the dot product is not associative. That is, in general, we cannot write:

\[ (\overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{v}}) \cdot \overrightarrow{\mathbf{w}} = \overrightarrow{\mathbf{u}} \cdot (\overrightarrow{\mathbf{v}} \cdot \overrightarrow{\mathbf{w}}) \]

Such an expression is not even defined, because the dot product produces a scalar, not a vector. Once we take the dot product of two vectors, the result is a number, and taking the dot product of a number with another vector does not make sense.

Thus, unlike vector addition or scalar multiplication, associativity does not apply to the dot product. The dot product only works between two vectors at a time, and no further dot product operation can be applied to the result.

Dot product and Scalar Multiplication

Another key property of the dot product is compatibility with scalar multiplication. That is, for any scalar \( k \) and vectors \(\overrightarrow{\mathbf{u}}\) and \(\overrightarrow{\mathbf{v}}\), we have:

\[ k (\overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{v}}) = (k\overrightarrow{\mathbf{u}}) \cdot \overrightarrow{\mathbf{v}} = \overrightarrow{\mathbf{u}} \cdot (k\overrightarrow{\mathbf{v}}) \]

This property tells us that multiplying one or both vectors by a scalar before taking the dot product is equivalent to first computing the dot product and then multiplying the result by the same scalar.

Verification

1. Left-hand side: Scaling the Dot Product

   \[ k (\overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{v}}) = k (x_1x_2 + y_1y_2 + z_1z_2) = kx_1x_2 + ky_1y_2 + kz_1z_2 \]

2. Scaling \(\overrightarrow{\mathbf{u}}\) First:

   \[ (k\overrightarrow{\mathbf{u}}) \cdot \overrightarrow{\mathbf{v}} = (kx_1, ky_1, kz_1) \cdot (x_2, y_2, z_2) \]
   \[ = kx_1x_2 + ky_1y_2 + kz_1z_2 \]

3. Scaling \(\overrightarrow{\mathbf{v}}\) First:

   \[ \overrightarrow{\mathbf{u}} \cdot (k\overrightarrow{\mathbf{v}}) = (x_1, y_1, z_1) \cdot (kx_2, ky_2, kz_2) \]
   \[ = kx_1x_2 + ky_1y_2 + kz_1z_2 \]

Since all three expressions are identical, we conclude:

\[ k (\overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{v}}) = (k\overrightarrow{\mathbf{u}}) \cdot \overrightarrow{\mathbf{v}} = \overrightarrow{\mathbf{u}} \cdot (k\overrightarrow{\mathbf{v}}) \]

This property confirms that the dot product distributes over scalar multiplication, maintaining linearity in each argument.

Distributive Property over Vector Addition

Another fundamental property of the dot product is its distributive property over vector addition:

\[ \overrightarrow{\mathbf{u}} \cdot (\overrightarrow{\mathbf{v}} + \overrightarrow{\mathbf{w}}) = \overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{v}} + \overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{w}} \]

This means that taking the dot product with a sum of vectors distributes over addition, much like multiplication distributes over addition in algebra.

Verification

Let the vectors be expressed in unit vector notation:

\[ \overrightarrow{\mathbf{u}} = x_1 \mathbf{i} + y_1 \mathbf{j} + z_1 \mathbf{k}, \]

\[ \overrightarrow{\mathbf{v}} = x_2 \mathbf{i} + y_2 \mathbf{j} + z_2 \mathbf{k}, \]

\[ \overrightarrow{\mathbf{w}} = x_3 \mathbf{i} + y_3 \mathbf{j} + z_3 \mathbf{k} \]

First, compute the left-hand side:

\[ \overrightarrow{\mathbf{v}} + \overrightarrow{\mathbf{w}} = (x_2 + x_3) \mathbf{i} + (y_2 + y_3) \mathbf{j} + (z_2 + z_3) \mathbf{k} \]

Now, take the dot product with \(\overrightarrow{\mathbf{u}}\):

\[ \overrightarrow{\mathbf{u}} \cdot (\overrightarrow{\mathbf{v}} + \overrightarrow{\mathbf{w}}) = [x_1 \mathbf{i} + y_1 \mathbf{j} + z_1 \mathbf{k}] \cdot [(x_2 + x_3) \mathbf{i} + (y_2 + y_3) \mathbf{j} + (z_2 + z_3) \mathbf{k}] \]

Expanding using the definition of the dot product:

\[ = x_1 (x_2 + x_3) + y_1 (y_2 + y_3) + z_1 (z_2 + z_3) \]

Distributing each term:

\[ = x_1 x_2 + x_1 x_3 + y_1 y_2 + y_1 y_3 + z_1 z_2 + z_1 z_3 \]

Rewriting:

\[ = (x_1 x_2 + y_1 y_2 + z_1 z_2) + (x_1 x_3 + y_1 y_3 + z_1 z_3) \]

Recognizing these as dot products:

\[ = (x_1 \mathbf{i} + y_1 \mathbf{j} + z_1 \mathbf{k}) \cdot (x_2 \mathbf{i} + y_2 \mathbf{j} + z_2 \mathbf{k}) + (x_1 \mathbf{i} + y_1 \mathbf{j} + z_1 \mathbf{k}) \cdot (x_3 \mathbf{i} + y_3 \mathbf{j} + z_3 \mathbf{k}) \]

\[ = \overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{v}} + \overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{w}} \]

Thus, the distributive property of the dot product is verified:

\[ \overrightarrow{\mathbf{u}} \cdot (\overrightarrow{\mathbf{v}} + \overrightarrow{\mathbf{w}}) = \overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{v}} + \overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{w}} \]

This property ensures that the dot product interacts linearly with vector addition.

Cauchy–Schwarz Inequality

A fundamental property of the dot product is the Cauchy–Schwarz inequality, which states that for any two vectors:

\[ |\overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{v}}| \leq |\overrightarrow{\mathbf{u}}| |\overrightarrow{\mathbf{v}}| \]

This inequality establishes an upper bound on the absolute value of the dot product, ensuring that it never exceeds the product of the magnitudes of the two vectors.

Proof

Let

\[ \overrightarrow{\mathbf{u}} = x_1 \mathbf{i} + y_1 \mathbf{j} + z_1 \mathbf{k}, \quad \overrightarrow{\mathbf{v}} = x_2 \mathbf{i} + y_2 \mathbf{j} + z_2 \mathbf{k} \]

From the fundamental formula for the dot product:

\[ \overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{v}} = x_1x_2 + y_1y_2 + z_1z_2 \]

we also have the expression:

\[ \overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{v}} = |\overrightarrow{\mathbf{u}}| |\overrightarrow{\mathbf{v}}| \cos \theta \]

Since \(|\cos \theta| \leq 1\) for any angle \(\theta\), it follows that:

\[ |\overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{v}}| = |\overrightarrow{\mathbf{u}}| |\overrightarrow{\mathbf{v}}| |\cos \theta| \leq |\overrightarrow{\mathbf{u}}| |\overrightarrow{\mathbf{v}}| \]

which is exactly the Cauchy–Schwarz inequality.

When does equality hold?

The equality holds in the Cauchy–Schwarz inequality, i.e.,

\[ |\overrightarrow{\mathbf{u}} \cdot \overrightarrow{\mathbf{v}}| = |\overrightarrow{\mathbf{u}}| |\overrightarrow{\mathbf{v}}| \]

if and only if \(\overrightarrow{\mathbf{u}}\) and \(\overrightarrow{\mathbf{v}}\) are collinear, meaning one is a scalar multiple of the other:

\[ \overrightarrow{\mathbf{v}} = k \overrightarrow{\mathbf{u}} \quad \text{for some scalar } k \]

Dot Products of Standard Unit Vectors along axes

\[ \mathbf{i} \cdot \mathbf{i} = \mathbf{j} \cdot \mathbf{j} = \mathbf{k} \cdot \mathbf{k} = 1 \]

This follows directly from the definition:

\[ \mathbf{i} \cdot \mathbf{i} = |\mathbf{i}| |\mathbf{i}| \cos 0 = 1 \cdot 1 \cdot 1 = 1 \]

since \(\cos 0 = 1\), and similarly for \(\mathbf{j} \cdot \mathbf{j}\) and \(\mathbf{k} \cdot \mathbf{k}\).

For the dot products of different unit vectors, we use the fact that these vectors are perpendicular to each other:

\[ \mathbf{i} \cdot \mathbf{j} = \mathbf{j} \cdot \mathbf{k} = \mathbf{k} \cdot \mathbf{i} = 0 \]

Since the angle between any two different unit vectors is \(90^\circ\), we have:

\[ \cos 90^\circ = 0 \quad \Rightarrow \quad \mathbf{i} \cdot \mathbf{j} = |\mathbf{i}| |\mathbf{j}| \cos 90^\circ = 1 \cdot 1 \cdot 0 = 0 \]

and similarly for \(\mathbf{j} \cdot \mathbf{k}\) and \(\mathbf{k} \cdot \mathbf{i}\).

Thus, the fundamental dot product relations among the unit vectors are:

\[ \mathbf{i} \cdot \mathbf{i} = \mathbf{j} \cdot \mathbf{j} = \mathbf{k} \cdot \mathbf{k} = 1, \quad \mathbf{i} \cdot \mathbf{j} = \mathbf{j} \cdot \mathbf{k} = \mathbf{k} \cdot \mathbf{i} = 0 \]

Square of a Vector

A particularly useful property of the dot product arises when a vector is dotted with itself:

\[ \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{a}} = |\overrightarrow{\mathbf{a}}| |\overrightarrow{\mathbf{a}}| \cos 0 = |\overrightarrow{\mathbf{a}}|^2 \]

Since \(\cos 0 = 1\), this simplifies to:

\[ \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{a}} = |\overrightarrow{\mathbf{a}}|^2 \]

Thus, taking the dot product of a vector with itself gives the square of its magnitude. We often write:

\[ \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{a}} = \overrightarrow{\mathbf{a}}^2 \]

but be careful—this notation does not extend naturally to higher powers. There is no meaningful way to define something like \( \overrightarrow{\mathbf{a}}^3 \) in the same way.

Using this property, we derive several key vector identities that resemble algebraic identities in real numbers:

Expansion of \((\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}})^2\)

\[ |\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}}|^2 = (\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}}) \cdot (\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}}) \]

Expanding using the distributive property:

\[ = \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{b}} \]

Since the dot product is commutative:

\[ = \overrightarrow{\mathbf{a}}^2 + \overrightarrow{\mathbf{b}}^2 + 2 (\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}) \]

Thus, we obtain the parallelogram identity:

\[ |\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}}|^2 = \overrightarrow{\mathbf{a}}^2 + \overrightarrow{\mathbf{b}}^2 + 2 \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} \]

Expansion of \((\overrightarrow{\mathbf{a}} - \overrightarrow{\mathbf{b}})^2\)

Similarly, applying the same steps to \(\overrightarrow{\mathbf{a}} - \overrightarrow{\mathbf{b}}\):

\[ |\overrightarrow{\mathbf{a}} - \overrightarrow{\mathbf{b}}|^2 = (\overrightarrow{\mathbf{a}} - \overrightarrow{\mathbf{b}}) \cdot (\overrightarrow{\mathbf{a}} - \overrightarrow{\mathbf{b}}) \]

Expanding:

\[ = \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{a}} - \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} - \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{b}} \]

\[ = \overrightarrow{\mathbf{a}}^2 + \overrightarrow{\mathbf{b}}^2 - 2 \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} \]

Product of Sum and Difference

\[ (\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}}) \cdot (\overrightarrow{\mathbf{a}} - \overrightarrow{\mathbf{b}}) \]

Expanding:

\[ = \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{a}} - \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{a}} - \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{b}} \]

Since \(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} = \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{a}}\), we get:

\[ = \overrightarrow{\mathbf{a}}^2 - \overrightarrow{\mathbf{b}}^2 \]

Thus:

\[ (\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}}) \cdot (\overrightarrow{\mathbf{a}} - \overrightarrow{\mathbf{b}}) = \overrightarrow{\mathbf{a}}^2 - \overrightarrow{\mathbf{b}}^2 \]

No Higher Powers Like \( |\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}}|^3 \)

While these identities look similar to algebraic identities, there is no direct analog to expressions like:

\[ |\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}}|^3 \]

because dot product expansion works only for quadratic expressions. There is no meaningful way to generalize it to cubic or higher powers in the same way as algebraic multiplication.

Expansion of the square of sum of Three Vectors

\[ |\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{c}}|^2 = |\overrightarrow{\mathbf{a}}|^2 + |\overrightarrow{\mathbf{b}}|^2 + |\overrightarrow{\mathbf{c}}|^2 + 2 (\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}} + \overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}}) \]

Proof

Expanding the squared magnitude using the dot product definition:

\[ |\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{c}}|^2 = (\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{c}}) \cdot (\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{c}}) \]

Applying the distributive property:

\[ = \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{c}} \]

\[ + \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}} \]

\[ + \overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{c}} \]

Since the dot product is commutative (\(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} = \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{a}}\)), we rewrite:

\[ = |\overrightarrow{\mathbf{a}}|^2 + |\overrightarrow{\mathbf{b}}|^2 + |\overrightarrow{\mathbf{c}}|^2 + 2 (\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}} + \overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}}) \]

This identity generalizes the earlier two-vector expansion:

\[ |\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}}|^2 = |\overrightarrow{\mathbf{a}}|^2 + |\overrightarrow{\mathbf{b}}|^2 + 2 \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} \]

to three vectors, showing how the squared magnitude of a sum involves both the squared magnitudes of individual vectors and pairwise dot products.

Parallelogram Identity

A fundamental identity in vector algebra is:

\[ |\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}}|^2 + |\overrightarrow{\mathbf{a}} - \overrightarrow{\mathbf{b}}|^2 = 2 (|\overrightarrow{\mathbf{a}}|^2 + |\overrightarrow{\mathbf{b}}|^2) \]

This identity is known as the parallelogram identity, as it arises naturally from the geometry of a parallelogram formed by two vectors. It establishes a relationship between the sums and differences of vectors and their magnitudes.

Proof

Expanding both squared terms using the previously derived formulas:

\[ |\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}}|^2 = \overrightarrow{\mathbf{a}}^2 + \overrightarrow{\mathbf{b}}^2 + 2 \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} \]

\[ |\overrightarrow{\mathbf{a}} - \overrightarrow{\mathbf{b}}|^2 = \overrightarrow{\mathbf{a}}^2 + \overrightarrow{\mathbf{b}}^2 - 2 \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} \]

Adding both equations:

\[ |\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}}|^2 + |\overrightarrow{\mathbf{a}} - \overrightarrow{\mathbf{b}}|^2 = (\overrightarrow{\mathbf{a}}^2 + \overrightarrow{\mathbf{b}}^2 + 2 \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}) + (\overrightarrow{\mathbf{a}}^2 + \overrightarrow{\mathbf{b}}^2 - 2 \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}) \]

Since \( + 2 \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} \) and \( - 2 \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} \) cancel out:

\[ |\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}}|^2 + |\overrightarrow{\mathbf{a}} - \overrightarrow{\mathbf{b}}|^2 = 2 \overrightarrow{\mathbf{a}}^2 + 2 \overrightarrow{\mathbf{b}}^2 \]

\[ = 2 (|\overrightarrow{\mathbf{a}}|^2 + |\overrightarrow{\mathbf{b}}|^2) \]

This proves the identity. It confirms that the sum of the squares of the diagonals of a parallelogram equals the sum of the squares of all four sides, a result that also holds in classical Euclidean geometry.

Triangle Inequality

For any two vectors \(\overrightarrow{\mathbf{a}}\) and \(\overrightarrow{\mathbf{b}}\), the triangle inequality states:

\[ ||\overrightarrow{\mathbf{a}}| - |\overrightarrow{\mathbf{b}}|| \leq |\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}}| \leq |\overrightarrow{\mathbf{a}}| + |\overrightarrow{\mathbf{b}}| \]

This inequality expresses the fundamental geometric fact that the length of one side of a triangle cannot exceed the sum of the lengths of the other two sides, nor can it be smaller than the absolute difference of the two.

Proof

We start with the identity:

\[ |\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}}|^2 = |\overrightarrow{\mathbf{a}}|^2 + |\overrightarrow{\mathbf{b}}|^2 + 2 \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} \]

Since the dot product satisfies:

\[ \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} = |\overrightarrow{\mathbf{a}}| |\overrightarrow{\mathbf{b}}| \cos \theta \]

where \(-1 \leq \cos \theta \leq 1\), we substitute:

\[ |\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}}|^2 = |\overrightarrow{\mathbf{a}}|^2 + |\overrightarrow{\mathbf{b}}|^2 + 2 |\overrightarrow{\mathbf{a}}| |\overrightarrow{\mathbf{b}}| \cos \theta \]

Since \(-1 \leq \cos \theta \leq 1\), we obtain two inequalities:

\[ |\overrightarrow{\mathbf{a}}|^2 + |\overrightarrow{\mathbf{b}}|^2 - 2 |\overrightarrow{\mathbf{a}}| |\overrightarrow{\mathbf{b}}| \leq |\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}}|^2 \leq |\overrightarrow{\mathbf{a}}|^2 + |\overrightarrow{\mathbf{b}}|^2 + 2 |\overrightarrow{\mathbf{a}}| |\overrightarrow{\mathbf{b}}| \]

Rewriting these:

\[ (|\overrightarrow{\mathbf{a}}| - |\overrightarrow{\mathbf{b}}|)^2 \leq |\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}}|^2 \leq (|\overrightarrow{\mathbf{a}}| + |\overrightarrow{\mathbf{b}}|)^2 \]

Taking square roots on all sides:

\[ ||\overrightarrow{\mathbf{a}}| - |\overrightarrow{\mathbf{b}}|| \leq |\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}}| \leq |\overrightarrow{\mathbf{a}}| + |\overrightarrow{\mathbf{b}}| \]

which is the required result. \(\blacksquare\)

Equality Conditions in the Triangle Inequality

The right-hand equality in the triangle inequality:

\[ |\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}}| = |\overrightarrow{\mathbf{a}}| + |\overrightarrow{\mathbf{b}}| \]

occurs when the vectors \(\overrightarrow{\mathbf{a}}\) and \(\overrightarrow{\mathbf{b}}\) point in the same direction, meaning the angle between them is \(0\). This happens when:

\[ \cos \theta = 1 \quad \Rightarrow \quad \overrightarrow{\mathbf{b}} = k\overrightarrow{\mathbf{a}}, \quad k > 0 \]

which means the vectors are collinear and positively proportional.

The left-hand equality:

\[ |\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}}| = ||\overrightarrow{\mathbf{a}}| - |\overrightarrow{\mathbf{b}}|| \]

occurs when the vectors \(\overrightarrow{\mathbf{a}}\) and \(\overrightarrow{\mathbf{b}}\) point in exactly opposite directions, meaning the angle between them is \(\pi\). This happens when:

\[ \cos \theta = -1 \quad \Rightarrow \quad \overrightarrow{\mathbf{b}} = k\overrightarrow{\mathbf{a}}, \quad k < 0 \]

which means the vectors are collinear and oppositely directed.

Thus, the triangle inequality sharpens to an equality only when the two vectors lie along the same line, either in the same direction (\(\theta = 0\)) or in opposite directions (\(\theta = \pi\)).

This inequality confirms that the length of the resultant vector from adding two vectors always lies between the absolute difference and the sum of their magnitudes, reinforcing the fundamental triangle inequality in Euclidean geometry.

Example

Suppose we have three unit vectors \(\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}, \overrightarrow{\mathbf{c}}\). Without changing their length, only their direction, in what orientations does \(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}} + \overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}}\) attain its minimum and maximum values? What are these values?

Solution:

We start by using the fundamental identity:

\[ |\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{c}}|^2 = |\overrightarrow{\mathbf{a}}|^2 + |\overrightarrow{\mathbf{b}}|^2 + |\overrightarrow{\mathbf{c}}|^2 + 2(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}} + \overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}}) \]

Since \(\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}, \overrightarrow{\mathbf{c}}\) are unit vectors, we have:

\[ |\overrightarrow{\mathbf{a}}|^2 = |\overrightarrow{\mathbf{b}}|^2 = |\overrightarrow{\mathbf{c}}|^2 = 1 \]

Substituting this:

\[ |\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{c}}|^2 = 3 + 2(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}} + \overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}}) \]

Since a magnitude is always non-negative:

\[ |\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{c}}|^2 \geq 0 \]

\[ 3 + 2(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}} + \overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}}) \geq 0 \]

Rearranging:

\[ \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}} + \overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}} \geq -\frac{3}{2} \]

Thus, the minimum possible value is \(-\frac{3}{2}\).

Equality occurs when:

\[ |\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{c}}| = 0 \]

which implies:

\[ \overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{c}} = 0 \]

Thus:

\[ \overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} = -\overrightarrow{\mathbf{c}} \]

Taking squared magnitudes on both sides:

\[ |\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}}|^2 = |\overrightarrow{\mathbf{c}}|^2 \]

\[ |\overrightarrow{\mathbf{a}}|^2 + |\overrightarrow{\mathbf{b}}|^2 + 2\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} = |\overrightarrow{\mathbf{c}}|^2 \]

Since \(|\overrightarrow{\mathbf{c}}| = 1\):

\[ 1 + 1 + 2\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} = 1 \]

\[ 2\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} = -1 \]

\[ \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} = -\frac{1}{2} \]

Since \(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} = |\overrightarrow{\mathbf{a}}| |\overrightarrow{\mathbf{b}}| \cos \theta\), and both vectors are unit vectors, this implies:

\[ \cos \theta = -\frac{1}{2} \]

\[ \theta = 120^\circ \]

By symmetry, the same reasoning applies to \(\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}}\) and \(\overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}}\), so:

\[ \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}} = \overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}} = -\frac{1}{2} \]

Thus, the minimum value \(-\frac{3}{2}\) occurs when the three vectors are coplanar and form angles of \(120^\circ\) with each other as shown in the figure below:

We now find the maximum value of \(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}} + \overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}}\).

We use the inequality:

\[ |\overrightarrow{\mathbf{a}} - \overrightarrow{\mathbf{b}}|^2 + |\overrightarrow{\mathbf{b}} - \overrightarrow{\mathbf{c}}|^2 + |\overrightarrow{\mathbf{c}} - \overrightarrow{\mathbf{a}}|^2 \geq 0 \]

Expanding each term:

\[ (|\overrightarrow{\mathbf{a}}|^2 + |\overrightarrow{\mathbf{b}}|^2 - 2 \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}) + (|\overrightarrow{\mathbf{b}}|^2 + |\overrightarrow{\mathbf{c}}|^2 - 2 \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}}) + (|\overrightarrow{\mathbf{c}}|^2 + |\overrightarrow{\mathbf{a}}|^2 - 2 \overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}}) \geq 0 \]

Since \(|\overrightarrow{\mathbf{a}}| = |\overrightarrow{\mathbf{b}}| = |\overrightarrow{\mathbf{c}}| = 1\), substituting these magnitudes gives:

\[ (1 + 1 - 2 \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}) + (1 + 1 - 2 \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}}) + (1 + 1 - 2 \overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}}) \geq 0 \]

\[ 6 - 2 (\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}} + \overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}}) \geq 0 \]

Rearranging:

\[ \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}} + \overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}} \leq 3 \]

Thus, the maximum possible value is 3.

When does equality occur?

Equality occurs when:

\[ |\overrightarrow{\mathbf{a}} - \overrightarrow{\mathbf{b}}|^2 + |\overrightarrow{\mathbf{b}} - \overrightarrow{\mathbf{c}}|^2 + |\overrightarrow{\mathbf{c}} - \overrightarrow{\mathbf{a}}|^2 = 0 \]

Since each term is a sum of squares, this implies each individual term must be zero:

\[ |\overrightarrow{\mathbf{a}} - \overrightarrow{\mathbf{b}}| = 0, \quad |\overrightarrow{\mathbf{b}} - \overrightarrow{\mathbf{c}}| = 0, \quad |\overrightarrow{\mathbf{c}} - \overrightarrow{\mathbf{a}}| = 0 \]

which means:

\[ \overrightarrow{\mathbf{a}} - \overrightarrow{\mathbf{b}} = 0, \quad \overrightarrow{\mathbf{b}} - \overrightarrow{\mathbf{c}} = 0, \quad \overrightarrow{\mathbf{c}} - \overrightarrow{\mathbf{a}} = 0 \]

Thus,

\[ \overrightarrow{\mathbf{a}} = \overrightarrow{\mathbf{b}} = \overrightarrow{\mathbf{c}} \]

This means the maximum value 3 occurs when all three vectors are identical, i.e., they point in the same direction. In this case, each dot product evaluates to:

\[ \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} = \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}} = \overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}} = 1 \]

and summing these gives the maximum value:

\[ 1 + 1 + 1 = 3 \]

Thus, the maximum value of \( \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}} + \overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}} \) is 3, which occurs when \( \overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}, \overrightarrow{\mathbf{c}} \) are the same vector.

Projections: The Shadow of a Vector

We came up with the idea of the dot product when we tried to measure the angle between two vectors. This remains one of its most fundamental and important applications. However, there is another powerful use of the dot product—one that plays a crucial role in geometry, physics, and engineering. This is the idea of projections.

Let \(\overrightarrow{\mathbf{a}}\) and \(\overrightarrow{\mathbf{b}}\) be two nonzero vectors making some angle between them. Imagine the head of \(\overrightarrow{\mathbf{a}}\). From this point, drop a perpendicular onto the line of support of \(\overrightarrow{\mathbf{b}}\). Let the foot of this perpendicular be N.

Now, from the tail of \(\overrightarrow{\mathbf{a}}\), drop another perpendicular onto the line of support of \(\overrightarrow{\mathbf{a}}\). Let the foot of this perpendicular be M.

The length of MN is what we call the projection of \(\overrightarrow{\mathbf{a}}\) on \(\overrightarrow{\mathbf{b}}\). It represents the part of \(\overrightarrow{\mathbf{a}}\) that "aligns" with \(\overrightarrow{\mathbf{b}}\).

It is like the shadow of \(\overrightarrow{\mathbf{a}}\) cast onto the direction of \(\overrightarrow{\mathbf{b}}\), if the "light" were shining perpendicular to \(\overrightarrow{\mathbf{b}}\). This concept of projection captures how much of \(\overrightarrow{\mathbf{a}}\) points along \(\overrightarrow{\mathbf{b}}\), making it one of the most powerful applications of the dot product.

Scalar Projection

To compute the projection of a vector \(\overrightarrow{\mathbf{a}}\) along another vector \(\overrightarrow{\mathbf{b}}\), we first make them co-initial at a common point, say \(O\). This simplifies calculations and allows us to directly analyze their geometric relationship. Define:

\[ \overrightarrow{\mathbf{a}} = \overrightarrow{\mathbf{OA}}, \quad \overrightarrow{\mathbf{b}} = \overrightarrow{\mathbf{OB}} \]

From the head of \(\overrightarrow{\mathbf{a}}\), i.e., point \(A\), drop a perpendicular onto the line of support of \(\overrightarrow{\mathbf{b}}\). Let the foot of this perpendicular be N. The quantity we seek to compute is ON, the projection of \(\overrightarrow{\mathbf{a}}\) along \(\overrightarrow{\mathbf{b}}\).

Since the vectors form an angle \(\theta\), we apply the definition of cosine in right-angled \(\triangle OAN\):

\[ \cos \theta = \frac{\text{ON}}{\text{OA}} \]

Since \(\text{OA} = |\overrightarrow{\mathbf{a}}|\), this gives:

\[ \text{ON} = |\overrightarrow{\mathbf{a}}| \cos \theta \]

Thus, the projection of \(\overrightarrow{\mathbf{a}}\) along \(\overrightarrow{\mathbf{b}}\) is:

\[ |\overrightarrow{\mathbf{a}}| \cos \theta \]

However, this is only half the truth. The projection is not merely a length—it is a signed quantity that depends on the sign of \(\cos \theta\):

• If \(0^\circ \leq \theta < 90^\circ\), then \(\cos \theta > 0\), making the projection positive.
• If \(90^\circ < \theta \leq 180^\circ\), then \(\cos \theta < 0\), making the projection negative.
• If \(\theta = 90^\circ\), then \(\cos \theta = 0\), meaning the projection vanishes—there is no component of \(\overrightarrow{\mathbf{a}}\) along \(\overrightarrow{\mathbf{b}}\).

Thus, the projection is not just a length, but a signed scalar quantity that can be positive, negative, or zero, depending on the angle between the vectors.

Projection in Terms of the Dot Product

Since the dot product satisfies:

\[ \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} = |\overrightarrow{\mathbf{a}}| |\overrightarrow{\mathbf{b}}| \cos \theta \]

dividing by \( |\overrightarrow{\mathbf{b}}| \) gives:

\[ |\overrightarrow{\mathbf{a}}| \cos \theta = \frac{\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}}{|\overrightarrow{\mathbf{b}}|} \]

Now, recalling that the unit vector along \(\overrightarrow{\mathbf{b}}\) is:

\[ \hat{\mathbf{b}} = \frac{\overrightarrow{\mathbf{b}}}{|\overrightarrow{\mathbf{b}}|} \]

we rewrite the projection as:

\[ \overrightarrow{\mathbf{a}} \cdot \hat{\mathbf{b}} \]

This formula expresses the signed projection of \(\overrightarrow{\mathbf{a}}\) along \(\overrightarrow{\mathbf{b}}\) purely in terms of the dot product. It tells us exactly how much of \(\overrightarrow{\mathbf{a}}\) lies along \(\overrightarrow{\mathbf{b}}\), considering both magnitude and direction. This is one of the most powerful applications of the dot product.

Example

Find the projection of \( \overrightarrow{\mathbf{u}} = \mathbf{i} + \mathbf{j} - 2\mathbf{k} \) along \( \overrightarrow{\mathbf{v}} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k} \).

Solution:

The projection of \( \overrightarrow{\mathbf{u}} \) along \( \overrightarrow{\mathbf{v}} \) is given by:

\[ \overrightarrow{\mathbf{u}} \cdot \hat{\mathbf{v}} \]

where \( \hat{\mathbf{v}} \) is the unit vector along \( \overrightarrow{\mathbf{v}} \), given by:

\[ \hat{\mathbf{v}} = \frac{\overrightarrow{\mathbf{v}}}{|\overrightarrow{\mathbf{v}}|} \]

First, compute the magnitude of \( \overrightarrow{\mathbf{v}} \):

\[ |\overrightarrow{\mathbf{v}}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]

Thus, the unit vector along \( \overrightarrow{\mathbf{v}} \) is:

\[ \hat{\mathbf{v}} = \frac{1}{3} (\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}) \]

Now, compute \( \overrightarrow{\mathbf{u}} \cdot \hat{\mathbf{v}} \):

\[ \overrightarrow{\mathbf{u}} \cdot \hat{\mathbf{v}} = \left( 1 \mathbf{i} + 1 \mathbf{j} - 2 \mathbf{k} \right) \cdot \left( \frac{1}{3} \mathbf{i} + \frac{2}{3} \mathbf{j} + \frac{2}{3} \mathbf{k} \right) \]

\[ = \frac{1}{3} (1 \cdot 1) + \frac{2}{3} (1 \cdot 1) + \frac{2}{3} (-2 \cdot 1) \]

\[ = \frac{1}{3} + \frac{2}{3} - \frac{4}{3} = -\frac{1}{3} \]

Thus, the projection of \( \overrightarrow{\mathbf{u}} \) along \( \overrightarrow{\mathbf{v}} \) is:

\[ -\frac{1}{3} \]

Vector Projection

The projection of \( \overrightarrow{\mathbf{a}} \) along \( \overrightarrow{\mathbf{b}} \) is a signed scalar quantity, which is why it is often referred to as the scalar projection of \( \overrightarrow{\mathbf{a}} \) along \( \overrightarrow{\mathbf{b}} \). This scalar value tells us how much of \( \overrightarrow{\mathbf{a}} \) lies in the direction of \( \overrightarrow{\mathbf{b}} \), including its sign, which depends on whether the vectors form an acute or obtuse angle.

In contrast, the vector projection of \( \overrightarrow{\mathbf{a}} \) along \( \overrightarrow{\mathbf{b}} \) gives us the actual vector in the direction of \( \overrightarrow{\mathbf{b}} \) that represents the component of \( \overrightarrow{\mathbf{a}} \) along \( \overrightarrow{\mathbf{b}} \). Geometrically, this is the vector ON, where \( N \) is the foot of the perpendicular dropped from the head of \( \overrightarrow{\mathbf{a}} \) onto the line of support of \( \overrightarrow{\mathbf{b}} \).

Since the scalar projection is given by:

\[ \overrightarrow{\mathbf{a}} \cdot \hat{\mathbf{b}} \]

the vector projection of \( \overrightarrow{\mathbf{a}} \) along \( \overrightarrow{\mathbf{b}} \) is obtained by multiplying this scalar by \( \hat{\mathbf{b}} \), ensuring the result is a vector in the direction of \( \overrightarrow{\mathbf{b}} \):

\[ \text{Proj}_{\overrightarrow{\mathbf{b}}} (\overrightarrow{\mathbf{a}}) = (\overrightarrow{\mathbf{a}} \cdot \hat{\mathbf{b}}) \hat{\mathbf{b}} \]

This formula expresses the vector ON, which is the component of \( \overrightarrow{\mathbf{a}} \) in the direction of \( \overrightarrow{\mathbf{b}} \). Unlike the scalar projection, which only tells us the magnitude and sign, the vector projection retains both magnitude and direction. This makes it useful in applications where we need to decompose vectors into components along different directions, such as resolving forces in mechanics or analyzing motion in physics.

Example

Find the vector projection of \( \overrightarrow{\mathbf{u}} = \mathbf{i} + \mathbf{j} \) along \( \overrightarrow{\mathbf{v}} = 3\mathbf{j} + 4\mathbf{k} \).

Solution:

The formula for the vector projection of \( \overrightarrow{\mathbf{u}} \) along \( \overrightarrow{\mathbf{v}} \) is:

\[ \text{Proj}_{\overrightarrow{\mathbf{v}}} (\overrightarrow{\mathbf{u}}) = (\overrightarrow{\mathbf{u}} \cdot \hat{\mathbf{v}}) \hat{\mathbf{v}} \]

where \( \hat{\mathbf{v}} \) is the unit vector along \( \overrightarrow{\mathbf{v}} \):

\[ \hat{\mathbf{v}} = \frac{\overrightarrow{\mathbf{v}}}{|\overrightarrow{\mathbf{v}}|} \]

The magnitude of \( \overrightarrow{\mathbf{v}} \) is:

\[ |\overrightarrow{\mathbf{v}}| = \sqrt{0^2 + 3^2 + 4^2} = \sqrt{9 + 16} = 5 \]

Thus, the unit vector along \( \overrightarrow{\mathbf{v}} \) is:

\[ \hat{\mathbf{v}} = \frac{1}{5} (0\mathbf{i} + 3\mathbf{j} + 4\mathbf{k}) = \frac{3}{5} \mathbf{j} + \frac{4}{5} \mathbf{k} \]

Now, compute \( \overrightarrow{\mathbf{u}} \cdot \hat{\mathbf{v}} \) by taking the dot product:

\[ \overrightarrow{\mathbf{u}} \cdot \hat{\mathbf{v}} = (1 \cdot 0) + (1 \cdot \frac{3}{5}) + (0 \cdot \frac{4}{5}) \]

\[ = 0 + \frac{3}{5} + 0 = \frac{3}{5} \]

Thus, the projection vector is:

\[ \text{Proj}_{\overrightarrow{\mathbf{v}}} (\overrightarrow{\mathbf{u}}) = \frac{3}{5} \hat{\mathbf{v}} \]

\[ = \frac{3}{5} \times \left(\frac{3}{5} \mathbf{j} + \frac{4}{5} \mathbf{k} \right) \]

\[ = \frac{9}{25} \mathbf{j} + \frac{12}{25} \mathbf{k} \]

Thus, the vector projection of \( \overrightarrow{\mathbf{u}} \) along \( \overrightarrow{\mathbf{v}} \) is:

\[ \frac{9}{25} \mathbf{j} + \frac{12}{25} \mathbf{k} \]

Decomposition of a Vector into Parallel and Orthogonal Components

Given two nonzero vectors \( \overrightarrow{\mathbf{a}} \) and \( \overrightarrow{\mathbf{b}} \), the vector \( \overrightarrow{\mathbf{a}} \) can be expressed as the sum of two components:

1. A parallel component along \( \overrightarrow{\mathbf{b}} \), which represents the part of \( \overrightarrow{\mathbf{a}} \) that lies in the same direction as \( \overrightarrow{\mathbf{b}} \).
2. An orthogonal component with respect to \( \overrightarrow{\mathbf{b}} \), which represents the part of \( \overrightarrow{\mathbf{a}} \) that is perpendicular to \( \overrightarrow{\mathbf{b}} \).

To obtain this decomposition, place the vectors so that they start from the same point \( O \). From the head of \( \overrightarrow{\mathbf{a}} \) (point \( A \)), drop a perpendicular onto the line of support of \( \overrightarrow{\mathbf{b}} \), meeting it at \( N \). The vector \( \overrightarrow{\mathbf{ON}} \) represents the parallel component of \( \overrightarrow{\mathbf{a}} \) along \( \overrightarrow{\mathbf{b}} \).

Now, from \( A \), drop another perpendicular to a line perpendicular to the line of support of \( \overrightarrow{\mathbf{b}} \), within the plane containing \( \overrightarrow{\mathbf{a}} \) and \( \overrightarrow{\mathbf{b}} \). Let the foot of this perpendicular be \( M \). Since both \( \overrightarrow{\mathbf{NA}} \) and \( \overrightarrow{\mathbf{OM}} \) are perpendicular to the line of support of \( \overrightarrow{\mathbf{b}} \), they are equal in magnitude and direction, meaning:

\[ \overrightarrow{\mathbf{NA}} = \overrightarrow{\mathbf{OM}} \]

Thus, using parallelogram law of vectors, the vector \( \overrightarrow{\mathbf{a}} \) can be written as:

\[ \overrightarrow{\mathbf{a}} = \overrightarrow{\mathbf{ON}} + \overrightarrow{\mathbf{OM}} \]

where:

• \( \overrightarrow{\mathbf{ON}} \), the parallel component, is given by:

  \[ \overrightarrow{\mathbf{ON}} = \text{Proj}_{\overrightarrow{\mathbf{b}}} (\overrightarrow{\mathbf{a}}) = (\overrightarrow{\mathbf{a}} \cdot \hat{\mathbf{b}}) \hat{\mathbf{b}} \]

  This vector lies in the direction of \( \overrightarrow{\mathbf{b}} \).

• \( \overrightarrow{\mathbf{OM}} \), the orthogonal component, is given by:

  \[ \overrightarrow{\mathbf{OM}} = \overrightarrow{\mathbf{a}} - \overrightarrow{\mathbf{ON}} = \overrightarrow{\mathbf{a}}-(\overrightarrow{\mathbf{a}} \cdot \hat{\mathbf{b}}) \hat{\mathbf{b}} \]

  This vector is perpendicular to \( \overrightarrow{\mathbf{b}} \), satisfying:

  \[ \overrightarrow{\mathbf{OM}} \cdot \overrightarrow{\mathbf{b}} = 0 \]

Since \( \overrightarrow{\mathbf{OM}} \) represents the component of \( \overrightarrow{\mathbf{a}} \) that is independent of \( \overrightarrow{\mathbf{b}} \), this decomposition separates \( \overrightarrow{\mathbf{a}} \) into two distinct parts:

• The parallel part, fully aligned with \( \overrightarrow{\mathbf{b}} \).
• The orthogonal part, completely independent of \( \overrightarrow{\mathbf{b}} \).

Later on, we will see that the magnitude of \( \overrightarrow{\mathbf{OM}} \) is given by

\[ |\overrightarrow{\mathbf{OM}}| = \left| \overrightarrow{\mathbf{a}} - (\overrightarrow{\mathbf{a}} \cdot \hat{\mathbf{b}}) \hat{\mathbf{b}} \right| = |\overrightarrow{\mathbf{a}} \times \hat{\mathbf{b}}| \]

when we study the cross product.