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Mathematics · Ch 10 — Vector Algebra

Product of Two Vectors

10.6

Product of Two Vectors

10.6 Product of Two Vectors

So far we have studied addition and subtraction of vectors. Multiplication of two vectors is defined in two ways — one where the result is a scalar (the scalar or dot product), and another where the result is a vector (the vector or cross product). Both have wide applications in geometry, mechanics, and engineering.


Scalar (Dot) Product

The scalar product of two vectors a⃗\vec{a} and b⃗\vec{b} is defined as:

a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta

where θ\theta is the angle between a⃗\vec{a} and b⃗\vec{b}, with 0≤θ≤π0 \le \theta \le \pi.

Important

The dot product is a scalar (a real number), not a vector. It is also called the inner product.

If either a⃗=0⃗\vec{a} = \vec{0} or b⃗=0⃗\vec{b} = \vec{0}, then θ\theta is not defined, and we define a⃗⋅b⃗=0\vec{a} \cdot \vec{b} = 0.

Observations
  1. a⃗⋅b⃗\vec{a} \cdot \vec{b} is positive if θ\theta is acute (0≤θ<π20 \le \theta < \frac{\pi}{2}).
  2. a⃗⋅b⃗\vec{a} \cdot \vec{b} is zero if θ=π2\theta = \frac{\pi}{2}.
  3. a⃗⋅b⃗\vec{a} \cdot \vec{b} is negative if θ\theta is obtuse (π2<θ≤π\frac{\pi}{2} < \theta \le \pi).
Note

The dot product of two non-zero vectors is zero if and only if they are perpendicular (orthogonal). This is a key test for orthogonality.

Properties of the Dot Product

Property 1 (Commutativity): a⃗⋅b⃗=b⃗⋅a⃗\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}

›Proof

a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ=∣b⃗∣∣a⃗∣cos⁡θ=b⃗⋅a⃗\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta = |\vec{b}| |\vec{a}| \cos \theta = \vec{b} \cdot \vec{a}, since real-number multiplication is commutative.

Property 2 (Distributivity over addition): a⃗⋅(b⃗+c⃗)=a⃗⋅b⃗+a⃗⋅c⃗\vec{a} \cdot (\vec{b} + \vec{c}) = \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c}

›Proof

Using the dot product as ∣a⃗∣|\vec{a}| times the projection of the other vector onto a⃗\vec{a}: the projection of b⃗+c⃗\vec{b}+\vec{c} onto a⃗\vec{a} equals the sum of the projections of b⃗\vec{b} and c⃗\vec{c} onto a⃗\vec{a}. Multiplying by ∣a⃗∣|\vec{a}| gives the result.

Property 3 (Scalar multiplication): (λa⃗)⋅b⃗=λ(a⃗⋅b⃗)=a⃗⋅(λb⃗)(\lambda \vec{a}) \cdot \vec{b} = \lambda (\vec{a} \cdot \vec{b}) = \vec{a} \cdot (\lambda \vec{b})

›Proof

For λ≥0\lambda \ge 0: (λa⃗)⋅b⃗=∣λ∣∣a⃗∣∣b⃗∣cos⁡θ=λ(a⃗⋅b⃗)(\lambda \vec{a}) \cdot \vec{b} = |\lambda| |\vec{a}| |\vec{b}| \cos \theta = \lambda (\vec{a} \cdot \vec{b}). For λ<0\lambda < 0, the angle between λa⃗\lambda \vec{a} and b⃗\vec{b} is π−θ\pi - \theta, and cos⁡(π−θ)=−cos⁡θ\cos(\pi - \theta) = -\cos \theta, so (λa⃗)⋅b⃗=∣λ∣∣a⃗∣∣b⃗∣(−cos⁡θ)=λ(a⃗⋅b⃗)(\lambda \vec{a}) \cdot \vec{b} = |\lambda| |\vec{a}| |\vec{b}| (-\cos \theta) = \lambda (\vec{a} \cdot \vec{b}). The same reasoning applies to a⃗⋅(λb⃗)\vec{a} \cdot (\lambda \vec{b}).

Property 4 (Dot product with itself): a⃗⋅a⃗=∣a⃗∣2\vec{a} \cdot \vec{a} = |\vec{a}|^2

›Proof

a⃗⋅a⃗=∣a⃗∣∣a⃗∣cos⁡0=∣a⃗∣2\vec{a} \cdot \vec{a} = |\vec{a}| |\vec{a}| \cos 0 = |\vec{a}|^2.

Tip

This gives a convenient way to compute the magnitude: ∣a⃗∣=a⃗⋅a⃗|\vec{a}| = \sqrt{\vec{a} \cdot \vec{a}}.

Dot Product in Component Form

For a⃗=a1i^+a2j^+a3k^\vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k} and b⃗=b1i^+b2j^+b3k^\vec{b} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k}, using i^⋅i^=j^⋅j^=k^⋅k^=1\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1 and i^⋅j^=j^⋅k^=k^⋅i^=0\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0:

a⃗⋅b⃗=a1b1+a2b2+a3b3\vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3

Angle Between Two Vectors

cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣=a1b1+a2b2+a3b3a12+a22+a32b12+b22+b32\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \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}}

Projection of a Vector

The scalar projection of a⃗\vec{a} onto b⃗\vec{b} (the component of a⃗\vec{a} along b⃗\vec{b}) is:

a⃗⋅b⃗∣b⃗∣\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}

The vector projection of a⃗\vec{a} onto b⃗\vec{b} is:

(a⃗⋅b⃗∣b⃗∣2)b⃗\left( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \right) \vec{b}

Note

The scalar projection is a signed quantity — positive if the angle is acute, negative if obtuse. The vector projection points along b⃗\vec{b} (or opposite if the scalar projection is negative).


Vector (Cross) Product

The vector product of two vectors a⃗\vec{a} and b⃗\vec{b} is defined as:

a⃗×b⃗=∣a⃗∣∣b⃗∣sin⁡θ n^\vec{a} \times \vec{b} = |\vec{a}| |\vec{b}| \sin \theta \, \hat{n}

where θ\theta is the angle between a⃗\vec{a} and b⃗\vec{b} (0≤θ≤π0 \le \theta \le \pi), and n^\hat{n} is a unit vector perpendicular to both a⃗\vec{a} and b⃗\vec{b}, such that a⃗,b⃗,n^\vec{a}, \vec{b}, \hat{n} form a right-handed system.

Important

The cross product is a vector. Its magnitude ∣a⃗×b⃗∣=∣a⃗∣∣b⃗∣sin⁡θ|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta equals the area of the parallelogram formed by a⃗\vec{a} and b⃗\vec{b}.

If either a⃗=0⃗\vec{a} = \vec{0} or b⃗=0⃗\vec{b} = \vec{0}, then θ\theta is not defined, and we define a⃗×b⃗=0⃗\vec{a} \times \vec{b} = \vec{0}.

The direction of a⃗×b⃗\vec{a} \times \vec{b} is given by the right-hand rule: if you curl the fingers of your right hand from a⃗\vec{a} to b⃗\vec{b}, your thumb points in the direction of a⃗×b⃗\vec{a} \times \vec{b}.

Watch out

The cross product is not commutative: a⃗×b⃗=−(b⃗×a⃗)\vec{a} \times \vec{b} = - (\vec{b} \times \vec{a}), because reversing the order reverses the direction of n^\hat{n}.

Properties of the Cross Product

Property 1 (Anticommutativity): a⃗×b⃗=−(b⃗×a⃗)\vec{a} \times \vec{b} = - (\vec{b} \times \vec{a})

›Proof

For b⃗×a⃗\vec{b} \times \vec{a}, the right-hand rule gives −n^-\hat{n} (the rotation is opposite), so b⃗×a⃗=∣a⃗∣∣b⃗∣sin⁡θ (−n^)=−(a⃗×b⃗)\vec{b} \times \vec{a} = |\vec{a}| |\vec{b}| \sin \theta \, (-\hat{n}) = - (\vec{a} \times \vec{b}).

Property 2 (Distributivity): a⃗×(b⃗+c⃗)=a⃗×b⃗+a⃗×c⃗\vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}

›Proof

This follows from the geometric interpretation of the cross product as an area vector, or algebraically using components (analogous to the dot-product distributive law, but more involved).

Property 3 (Scalar multiplication): (λa⃗)×b⃗=λ(a⃗×b⃗)=a⃗×(λb⃗)(\lambda \vec{a}) \times \vec{b} = \lambda (\vec{a} \times \vec{b}) = \vec{a} \times (\lambda \vec{b})

›Proof

For λ≥0\lambda \ge 0: (λa⃗)×b⃗=∣λ∣∣a⃗∣∣b⃗∣sin⁡θ n^=λ(a⃗×b⃗)(\lambda \vec{a}) \times \vec{b} = |\lambda| |\vec{a}| |\vec{b}| \sin \theta \, \hat{n} = \lambda (\vec{a} \times \vec{b}). For λ<0\lambda < 0, the angle becomes π−θ\pi - \theta (so sin⁡(π−θ)=sin⁡θ\sin(\pi-\theta) = \sin\theta) but the right-hand rule now gives −n^-\hat{n}, so (λa⃗)×b⃗=∣λ∣∣a⃗∣∣b⃗∣sin⁡θ (−n^)=λ(a⃗×b⃗)(\lambda \vec{a}) \times \vec{b} = |\lambda| |\vec{a}| |\vec{b}| \sin \theta \, (-\hat{n}) = \lambda (\vec{a} \times \vec{b}). The same holds for a⃗×(λb⃗)\vec{a} \times (\lambda \vec{b}).

Property 4 (Cross product with itself): a⃗×a⃗=0⃗\vec{a} \times \vec{a} = \vec{0}

›Proof

a⃗×a⃗=∣a⃗∣∣a⃗∣sin⁡0 n^=0⃗\vec{a} \times \vec{a} = |\vec{a}| |\vec{a}| \sin 0 \, \hat{n} = \vec{0}. …