Definition of Scalar (Dot) Product

The product of two vectors can be defined in two important ways: one gives a scalar quantity and the other gives a vector quantity. The scalar product (or dot product) of two non-zero vectors a⃗\vec{a} and b⃗\vec{b}, written as a⃗⋅b⃗\vec{a} \cdot \vec{b}, is defined by

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

where ∣a⃗∣|\vec{a}| and ∣b⃗∣|\vec{b}| are the magnitudes of the vectors and θ\theta is the angle between them, with 0≤θ≤π0 \le \theta \le \pi.

Important observations

  • The result of a dot product is always a scalar.
  • If either vector is the zero vector, the dot product is defined as 0.
  • If θ=90∘\theta = 90^\circ, then cos⁡90∘=0\cos 90^\circ = 0, so a⃗⋅b⃗=0.\vec{a} \cdot \vec{b} = 0. Thus perpendicular vectors have zero dot product.
  • If θ=0∘\theta = 0^\circ, then a⃗⋅b⃗=∣a⃗∣∣b⃗∣.\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}|.
  • If θ=180∘\theta = 180^\circ, then a⃗⋅b⃗=−∣a⃗∣∣b⃗∣.\vec{a} \cdot \vec{b} = -|\vec{a}| |\vec{b}|.
  • The dot product of a vector with itself is a⃗⋅a⃗=∣a⃗∣2.\vec{a} \cdot \vec{a} = |\vec{a}|^2.

So the dot product gives a measure of how much one vector acts in the direction of another.

Properties of the Scalar Product

Let a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c} be vectors and λ\lambda be a scalar.

  1. Commutativity: a⃗⋅b⃗=b⃗⋅a⃗\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}

  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} and similarly, (a⃗+b⃗)⋅c⃗=a⃗⋅c⃗+b⃗⋅c⃗.(\vec{a} + \vec{b}) \cdot \vec{c} = \vec{a} \cdot \vec{c} + \vec{b} \cdot \vec{c}.

  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})

  4. Dot Product with Zero Vector: a⃗⋅0⃗=0\vec{a} \cdot \vec{0} = 0

These properties make the dot product algebraically very convenient.

Dot Product of Unit Vectors

Consider the mutually perpendicular unit vectors i^,j^,k^\hat{i}, \hat{j}, \hat{k} along the positive coordinate axes.

  • Since the angle between a unit vector and itself is 0∘0^\circ, i^⋅i^=j^⋅j^=k^⋅k^=1.\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1.

  • Since the angle between any two distinct coordinate unit vectors is 90∘90^\circ, i^⋅j^=j^⋅k^=k^⋅i^=0.\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0.

Also, by commutativity, j^⋅i^=0,k^⋅j^=0,i^⋅k^=0.\hat{j} \cdot \hat{i} = 0, \quad \hat{k} \cdot \hat{j} = 0, \quad \hat{i} \cdot \hat{k} = 0.

These identities are the basis of the component formula for the dot product.

Dot Product in Component Form

If a⃗=a1i^+a2j^+a3k^,b⃗=b1i^+b2j^+b3k^,\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}, \qquad \vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}, then using the properties of unit vectors, a⃗⋅b⃗=a1b1+a2b2+a3b3.\vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3.

This is the most useful form in calculations.

Using this, the angle between two non-zero vectors can be found from 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}}.

This formula is especially useful for testing whether two vectors are perpendicular, parallel, or making an acute or obtuse angle.

Projection of a Vector on a Line

The projection of one vector on another tells us how much of one vector lies in the direction of the other.

If a⃗\vec{a} is projected on a non-zero vector b⃗\vec{b}, then:

  • Scalar projection of a⃗\vec{a} on b⃗\vec{b} is a⃗⋅b^=a⃗⋅b⃗∣b⃗∣.\vec{a} \cdot \hat{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}.

This can be positive, negative, or zero depending on the relative direction of the vectors.

  • Vector projection of a⃗\vec{a} on b⃗\vec{b} is (a⃗⋅b⃗∣b⃗∣)b^=(a⃗⋅b⃗∣b⃗∣2)b⃗.\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}\right)\hat{b} = \left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\right)\vec{b}.

The vector projection points along the direction of b⃗\vec{b}.

Example 1: Basic Dot Product

Find the scalar product of two vectors a⃗\vec{a} and b⃗\vec{b} having magnitudes 3 and 4 respectively, and the angle between them is 60°.

Solution: Step 1: Write the formula: a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta

Step 2: Substitute the values: a⃗⋅b⃗=3×4×cos⁡60∘\vec{a} \cdot \vec{b} = 3 \times 4 \times \cos 60^\circ

Step 3: Since cos⁡60∘=12,\cos 60^\circ = \frac{1}{2}, we get a⃗⋅b⃗=12×12=6.\vec{a} \cdot \vec{b} = 12 \times \frac{1}{2} = 6.

Answer: 6

Example 2: Dot Product in Component Form

Find a⃗⋅b⃗\vec{a} \cdot \vec{b} if a⃗=2i^+3j^−k^\vec{a} = 2\hat{i} + 3\hat{j} - \hat{k} and b⃗=−i^+2j^+4k^\vec{b} = -\hat{i} + 2\hat{j} + 4\hat{k}.

Solution: Using the component formula, a⃗⋅b⃗=(2)(−1)+(3)(2)+(−1)(4)\vec{a} \cdot \vec{b} = (2)(-1) + (3)(2) + (-1)(4) =−2+6−4=0.= -2 + 6 - 4 = 0.

Since the dot product is zero and both vectors are non-zero, the vectors are perpendicular.

Answer: 0

Example 3: Finding the Angle Between Vectors

Find the angle between the vectors a⃗=i^+j^−k^\vec{a} = \hat{i} + \hat{j} - \hat{k} and b⃗=i^−j^+k^\vec{b} = \hat{i} - \hat{j} + \hat{k}.

Solution: Step 1: Find the dot product: a⃗⋅b⃗=(1)(1)+(1)(−1)+(−1)(1)=1−1−1=−1.\vec{a} \cdot \vec{b} = (1)(1) + (1)(-1) + (-1)(1) = 1 - 1 - 1 = -1.

Step 2: Find the magnitudes: ∣a⃗∣=12+12+(−1)2=3,|\vec{a}| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{3}, ∣b⃗∣=12+(−1)2+12=3.|\vec{b}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}.

Step 3: Use the angle formula: cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣=−13⋅3=−13.\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{-1}{\sqrt{3}\cdot\sqrt{3}} = -\frac{1}{3}.

So, θ=cos⁡−1(−13).\theta = \cos^{-1}\left(-\frac{1}{3}\right).

Answer: cos⁡−1(−1/3)\cos^{-1}(-1/3)

Example 4: Condition for Orthogonal Vectors

Find the value of λ\lambda for which the vectors a⃗=2i^−3j^+λk^\vec{a} = 2\hat{i} - 3\hat{j} + \lambda\hat{k} and b⃗=3i^+4j^+2k^\vec{b} = 3\hat{i} + 4\hat{j} + 2\hat{k} are perpendicular to each other.

Solution: Two non-zero vectors are perpendicular if and only if their dot product is zero.

So, a⃗⋅b⃗=0.\vec{a} \cdot \vec{b} = 0.

Compute the dot product: (2)(3)+(−3)(4)+(λ)(2)=0(2)(3) + (-3)(4) + (\lambda)(2) = 0 6−12+2λ=06 - 12 + 2\lambda = 0 −6+2λ=0-6 + 2\lambda = 0 2λ=62\lambda = 6 λ=3.\lambda = 3.

Answer: λ=3\lambda = 3

Example 5: Scalar Projection

Find the projection of the vector a⃗=2i^+3j^+2k^\vec{a} = 2\hat{i} + 3\hat{j} + 2\hat{k} on the vector b⃗=i^+2j^+k^\vec{b} = \hat{i} + 2\hat{j} + \hat{k}.

Solution: The scalar projection of a⃗\vec{a} on b⃗\vec{b} is a⃗⋅b⃗∣b⃗∣.\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}.

Step 1: Find the dot product: a⃗⋅b⃗=(2)(1)+(3)(2)+(2)(1)=2+6+2=10.\vec{a} \cdot \vec{b} = (2)(1) + (3)(2) + (2)(1) = 2 + 6 + 2 = 10.

Step 2: Find the magnitude of b⃗\vec{b}: ∣b⃗∣=12+22+12=6.|\vec{b}| = \sqrt{1^2 + 2^2 + 1^2} = \sqrt{6}.

Step 3: Therefore, Projection=106=1066=563.\text{Projection} = \frac{10}{\sqrt{6}} = \frac{10\sqrt{6}}{6} = \frac{5\sqrt{6}}{3}.

Answer: 56/35\sqrt{6}/3

Example 6: Vector Projection

Find the vector projection of u⃗=i^−2j^+k^\vec{u} = \hat{i} - 2\hat{j} + \hat{k} onto v⃗=4i^−4j^+7k^\vec{v} = 4\hat{i} - 4\hat{j} + 7\hat{k}.

Solution: The vector projection of u⃗\vec{u} on v⃗\vec{v} is (u⃗⋅v⃗∣v⃗∣2)v⃗.\left(\frac{\vec{u} \cdot \vec{v}}{|\vec{v}|^2}\right)\vec{v}.

Step 1: Find the dot product: u⃗⋅v⃗=(1)(4)+(−2)(−4)+(1)(7)=4+8+7=19.\vec{u} \cdot \vec{v} = (1)(4) + (-2)(-4) + (1)(7) = 4 + 8 + 7 = 19.

Step 2: Find ∣v⃗∣2|\vec{v}|^2: ∣v⃗∣2=42+(−4)2+72=16+16+49=81.|\vec{v}|^2 = 4^2 + (-4)^2 + 7^2 = 16 + 16 + 49 = 81.

Step 3: Therefore, Vector projection=1981(4i^−4j^+7k^)\text{Vector projection} = \frac{19}{81}(4\hat{i} - 4\hat{j} + 7\hat{k}) =7681i^−7681j^+13381k^.= \frac{76}{81}\hat{i} - \frac{76}{81}\hat{j} + \frac{133}{81}\hat{k}.

Answer: 7681i^−7681j^+13381k^\frac{76}{81}\hat{i} - \frac{76}{81}\hat{j} + \frac{133}{81}\hat{k}

Example 7: Using Dot Product Algebraic Properties

Find ∣a⃗−b⃗∣|\vec{a} - \vec{b}|, if two vectors a⃗\vec{a} and b⃗\vec{b} are such that ∣a⃗∣=2|\vec{a}| = 2, ∣b⃗∣=3|\vec{b}| = 3, and a⃗⋅b⃗=4\vec{a} \cdot \vec{b} = 4.

Solution: Use ∣a⃗−b⃗∣2=(a⃗−b⃗)⋅(a⃗−b⃗).|\vec{a} - \vec{b}|^2 = (\vec{a} - \vec{b}) \cdot (\vec{a} - \vec{b}).

Expand: =a⃗⋅a⃗−a⃗⋅b⃗−b⃗⋅a⃗+b⃗⋅b⃗.= \vec{a}\cdot\vec{a} - \vec{a}\cdot\vec{b} - \vec{b}\cdot\vec{a} + \vec{b}\cdot\vec{b}.

Since dot product is commutative, =∣a⃗∣2−2(a⃗⋅b⃗)+∣b⃗∣2.= |\vec{a}|^2 - 2(\vec{a}\cdot\vec{b}) + |\vec{b}|^2.

Substitute the given values: =22−2(4)+32=4−8+9=5.= 2^2 - 2(4) + 3^2 = 4 - 8 + 9 = 5.

So, ∣a⃗−b⃗∣=5.|\vec{a} - \vec{b}| = \sqrt{5}.

Answer: 5\sqrt{5}

Example 8: Equal Inclination of a Vector

If a⃗\vec{a}, b⃗\vec{b}, and c⃗\vec{c} are three mutually perpendicular vectors of equal magnitude, prove that the vector a⃗+b⃗+c⃗\vec{a} + \vec{b} + \vec{c} is equally inclined to a⃗\vec{a}, b⃗\vec{b}, and c⃗\vec{c}.

Solution: Let the common magnitude be λ\lambda. Then ∣a⃗∣=∣b⃗∣=∣c⃗∣=λ,|\vec{a}| = |\vec{b}| = |\vec{c}| = \lambda, and since they are mutually perpendicular, a⃗⋅b⃗=b⃗⋅c⃗=c⃗⋅a⃗=0.\vec{a}\cdot\vec{b} = \vec{b}\cdot\vec{c} = \vec{c}\cdot\vec{a} = 0.

Let r⃗=a⃗+b⃗+c⃗.\vec{r} = \vec{a} + \vec{b} + \vec{c}.

Step 1: Find ∣r⃗∣2|\vec{r}|^2: ∣r⃗∣2=r⃗⋅r⃗|\vec{r}|^2 = \vec{r}\cdot\vec{r} =(a⃗+b⃗+c⃗)⋅(a⃗+b⃗+c⃗)= (\vec{a}+\vec{b}+\vec{c})\cdot(\vec{a}+\vec{b}+\vec{c}) =∣a⃗∣2+∣b⃗∣2+∣c⃗∣2+2(a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗)= |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) =λ2+λ2+λ2=3λ2.= \lambda^2 + \lambda^2 + \lambda^2 = 3\lambda^2. Thus, ∣r⃗∣=λ3.|\vec{r}| = \lambda\sqrt{3}.

Step 2: Let α\alpha be the angle between r⃗\vec{r} and a⃗\vec{a}. Then cos⁡α=r⃗⋅a⃗∣r⃗∣ ∣a⃗∣.\cos\alpha = \frac{\vec{r}\cdot\vec{a}}{|\vec{r}|\,|\vec{a}|}. Now, r⃗⋅a⃗=(a⃗+b⃗+c⃗)⋅a⃗=∣a⃗∣2+0+0=λ2.\vec{r}\cdot\vec{a} = (\vec{a}+\vec{b}+\vec{c})\cdot\vec{a} = |\vec{a}|^2 + 0 + 0 = \lambda^2. So, cos⁡α=λ2(λ3)(λ)=13.\cos\alpha = \frac{\lambda^2}{(\lambda\sqrt{3})(\lambda)} = \frac{1}{\sqrt{3}}.

Similarly, cos⁡β=13,cos⁡γ=13.\cos\beta = \frac{1}{\sqrt{3}}, \qquad \cos\gamma = \frac{1}{\sqrt{3}}.

Hence the angles made by r⃗\vec{r} with a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c} are equal.

Answer: Proved.

Example 9: Finding the Angle from a Vector Relation

If a⃗+b⃗+c⃗=0⃗\vec{a} + \vec{b} + \vec{c} = \vec{0} and ∣a⃗∣=3|\vec{a}| = 3, ∣b⃗∣=5|\vec{b}| = 5, ∣c⃗∣=7|\vec{c}| = 7, find the angle between a⃗\vec{a} and b⃗\vec{b}.

Solution: From a⃗+b⃗+c⃗=0⃗,\vec{a} + \vec{b} + \vec{c} = \vec{0}, we get a⃗+b⃗=−c⃗.\vec{a} + \vec{b} = -\vec{c}.

Now square both sides using dot product: (a⃗+b⃗)⋅(a⃗+b⃗)=(−c⃗)⋅(−c⃗).(\vec{a}+\vec{b})\cdot(\vec{a}+\vec{b}) = (-\vec{c})\cdot(-\vec{c}).

So, ∣a⃗∣2+∣b⃗∣2+2a⃗⋅b⃗=∣c⃗∣2.|\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a}\cdot\vec{b} = |\vec{c}|^2.

Substitute the magnitudes: 32+52+2(3)(5)cos⁡θ=723^2 + 5^2 + 2(3)(5)\cos\theta = 7^2 9+25+30cos⁡θ=499 + 25 + 30\cos\theta = 49 34+30cos⁡θ=4934 + 30\cos\theta = 49 30cos⁡θ=1530\cos\theta = 15 cos⁡θ=12.\cos\theta = \frac{1}{2}.

Therefore, θ=60∘.\theta = 60^\circ.

Answer: 60∘60^\circ or π/3\pi/3

Example 10: Work Done by a Force

A force F⃗=3i^+2j^−4k^\vec{F} = 3\hat{i} + 2\hat{j} - 4\hat{k} is applied to a particle, displacing it from point A(1,−1,2)A(1, -1, 2) to point B(2,−1,3)B(2, -1, 3). Find the work done by the force.

Solution: Work done is the dot product of force and displacement: W=F⃗⋅d⃗.W = \vec{F} \cdot \vec{d}.

Step 1: Find the displacement vector: d⃗=AB⃗=(2−1)i^+(−1−(−1))j^+(3−2)k^\vec{d} = \vec{AB} = (2-1)\hat{i} + (-1-(-1))\hat{j} + (3-2)\hat{k} d⃗=i^+k^.\vec{d} = \hat{i} + \hat{k}.

Step 2: Compute the dot product: W=(3i^+2j^−4k^)⋅(i^+k^)W = (3\hat{i} + 2\hat{j} - 4\hat{k})\cdot(\hat{i} + \hat{k}) =(3)(1)+(2)(0)+(−4)(1)= (3)(1) + (2)(0) + (-4)(1) =3−4=−1.= 3 - 4 = -1.

Answer: -1 unit of work.

Example 11: Orthogonality of Sum and Difference

Show that for any two non-zero vectors a⃗\vec{a} and b⃗\vec{b}, the vectors ∣a⃗∣b⃗+∣b⃗∣a⃗|\vec{a}|\vec{b} + |\vec{b}|\vec{a} and ∣a⃗∣b⃗−∣b⃗∣a⃗|\vec{a}|\vec{b} - |\vec{b}|\vec{a} are orthogonal.

Solution: Let u⃗=∣a⃗∣b⃗+∣b⃗∣a⃗,v⃗=∣a⃗∣b⃗−∣b⃗∣a⃗.\vec{u} = |\vec{a}|\vec{b} + |\vec{b}|\vec{a}, \qquad \vec{v} = |\vec{a}|\vec{b} - |\vec{b}|\vec{a}. We need to prove that u⃗⋅v⃗=0.\vec{u}\cdot\vec{v} = 0.

Now, u⃗⋅v⃗=(∣a⃗∣b⃗+∣b⃗∣a⃗)⋅(∣a⃗∣b⃗−∣b⃗∣a⃗).\vec{u}\cdot\vec{v} = (|\vec{a}|\vec{b} + |\vec{b}|\vec{a})\cdot(|\vec{a}|\vec{b} - |\vec{b}|\vec{a}).

Expand: =∣a⃗∣2(b⃗⋅b⃗)−∣b⃗∣2(a⃗⋅a⃗)= |\vec{a}|^2(\vec{b}\cdot\vec{b}) - |\vec{b}|^2(\vec{a}\cdot\vec{a}) because the mixed terms cancel.

Now use b⃗⋅b⃗=∣b⃗∣2,a⃗⋅a⃗=∣a⃗∣2.\vec{b}\cdot\vec{b} = |\vec{b}|^2, \qquad \vec{a}\cdot\vec{a} = |\vec{a}|^2. So, u⃗⋅v⃗=∣a⃗∣2∣b⃗∣2−∣b⃗∣2∣a⃗∣2=0.\vec{u}\cdot\vec{v} = |\vec{a}|^2|\vec{b}|^2 - |\vec{b}|^2|\vec{a}|^2 = 0.

Hence the two vectors are orthogonal.

Answer: Proved.