If the dot product of two vectors is zero, must one of the vectors be zero?

No. If both vectors are non-zero, a vanishing dot product means the cosine of the angle is zero, so the two vectors are perpendicular. The zero product only forces a zero vector when you additionally know the vectors are parallel.

Scalar (or Dot) Product of Two Vectors

Q: Why is the dot product called the scalar product?

Because the result of the operation is a scalar, that is, a plain number with no direction attached. The cross product is called the vector product for the opposite reason, as its result is a vector.

Q: What is the difference between the scalar projection and the vector projection?

The scalar projection of a on b is the number obtained by dividing the dot product by the magnitude of b, and it can be negative. The vector projection is that number multiplied by the unit vector along b, so it is an actual vector lying along b.

Q: Can the dot product be negative, and what does that mean physically?

Yes, whenever the angle exceeds ninety degrees. In the work formula W equals F dot s, a negative value means the force has a component opposing the displacement, so it removes energy from the body, as friction does.

Q: Is the dot product associative?

Associativity does not even make sense here, because the dot product of a scalar with a vector is undefined. What does hold is mixed associativity with scalars: m times a, dotted with b, equals m times the dot product of a and b.

MathsVector AlgebraFor JEE aspirants

Scalar (or Dot) Product of Two Vectors

Maths · Vector Algebra · Concept 2

The dot product turns two vectors into a single number. That number answers two questions at once: how large is the angle between them, and how much of one vector points along the other. Almost every question about perpendicularity, projections, angles and lengths in vector algebra reduces to a dot product.

Definition

Scalar product

For non-zero vectors $a$ and $b$ with angle $θ$ between them, where $0 \leq θ \leq π$ ,

a \cdot b = ∣ a ∣ ∣ b ∣ cos θ

If either vector is $0$ , the product is defined to be $0$ . The result is a scalar, not a vector.

Geometrical meaning: projection

Let $O A = a$ and $O B = b$ and drop $B L$ perpendicular to $O A$ . Then $O L = ∣ b ∣ cos θ$ is the projection of $b$ on $a$ . Therefore

a \cdot b = ∣ a ∣ (∣ b ∣ cos θ) = (magnitude of a) \times (projection of b on a)

By symmetry it is equally the magnitude of $b$ times the projection of $a$ on $b$ . In words: the dot product is the product of the length of one vector and the length of the projection of the other onto it.

OL is the projection of b on a. The dot product is the length of a multiplied by this projection.

Properties

Algebraic properties

$a \cdot b = b \cdot a$ (commutative)
$a \cdot (b + c) = a \cdot b + a \cdot c$ (distributive over addition)
$(m a) \cdot b = a \cdot (m b) = m (a \cdot b)$ for a scalar $m$
$a \cdot a = ∣ a ∣^{2} = a^{2}$ , so $∣ a ∣ = a \cdot a$
$a \cdot 0 = 0$
$a \cdot b = 0$ with $a \neq = 0$ , $b \neq = 0$ $⟺$ $a ⊥ b$

Products of the base vectors

\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1, \hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0

Hence, in component form, for $a = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}$ and $b = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}$ ,

a \cdot b = a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3}

Why there is no cancellation law

$a \cdot b = a \cdot c$ does not give $b = c$ . It only gives $a \cdot (b - c) = 0$ , so either $a = 0$ or $b - c$ is perpendicular to $a$ . Similarly, $a \cdot b = 0$ does not force one of the vectors to be zero.

Angle between two vectors and the sign of the dot product

Angle formula

cos θ = \frac{a \cdot b}{∣ a ∣ ∣ b ∣} = \frac{a _{1} b _{1} + a _{2} b _{2} + a _{3} b _{3}}{a _{1}^{2} + a _{2}^{2} + a _{3}^{2} b _{1}^{2} + b _{2}^{2} + b _{3}^{2}}

Sign of $a \cdot b$	Angle $θ$	Geometry
$a \cdot b > 0$	$0 \leq θ < 9 0^{\circ}$	Acute, the vectors point broadly the same way
$a \cdot b = 0$	$θ = 9 0^{\circ}$	Perpendicular (orthogonal)
$a \cdot b < 0$	$9 0^{\circ} < θ \leq 18 0^{\circ}$	Obtuse, the vectors point broadly opposite ways

Extreme values

Maximum of $a \cdot b$ is $∣ a ∣∣ b ∣$ , attained when $θ = 0$ (like parallel vectors).

Minimum of $a \cdot b$ is $- ∣ a ∣∣ b ∣$ , attained when $θ = π$ (unlike parallel vectors).

Consequently $a \cdot b \leq ∣ a ∣ ∣ b ∣$ , the Cauchy-Schwarz inequality for vectors.

Example 1

Find $p$ so that $a = 3 \hat{i} + 2 \hat{j} + 9 \hat{k}$ and $b = \hat{i} + p \hat{j} + 3 \hat{k}$ are (i) perpendicular, (ii) parallel.

Solution. (i) $a \cdot b = 0$ gives $3 + 2 p + 27 = 0$ , so $p = - 15$ .

(ii) Parallel needs $\frac{3}{1} = \frac{2}{p} = \frac{9}{3}$ . The first and last ratios are both $3$ , and $\frac{2}{p} = 3$ gives $p = \frac{2}{3}$ .

Example 2

Find all $x$ for which the angle between $a = 2 x^{2} \hat{i} + 4 x \hat{j} + \hat{k}$ and $b = 7 \hat{i} - 2 \hat{j} + x \hat{k}$ is obtuse.

Solution. Since $∣ a ∣, ∣ b ∣ > 0$ , the angle is obtuse exactly when $a \cdot b < 0$ .

$a \cdot b = 14 x^{2} - 8 x + x = 14 x^{2} - 7 x = 7 x (2 x - 1)$ .

$7 x (2 x - 1) < 0 \Rightarrow 0 < x < \frac{1}{2}$ . Hence $x \in (0, \frac{1}{2})$ .

Projections and resolution of a vector

Projection formulas

Scalar projection of $a$ on $b$ (also called the component of $a$ along $b$ ):

a \cdot \hat{b} = \frac{a \cdot b}{∣ b ∣}

Vector projection of $a$ on $b$ :

(a \cdot \hat{b}) \hat{b} = (\frac{a \cdot b}{∣ b ∣ ^{2}}) b

Component of $a$ perpendicular to $b$ (lying in the plane of $a$ and $b$ ):

a - (\frac{a \cdot b}{∣ b ∣ ^{2}}) b

These two pieces add back to $a$ , which is the whole point of resolution: any vector splits uniquely into a part along a given direction and a part perpendicular to it.

Resolution of a along b. The green piece is the vector projection, the blue piece is what is left over and is perpendicular to b.

Example 3

For $a = \hat{i} + \hat{j} + \hat{k}$ and $b = 2 \hat{i} - \hat{j} + 3 \hat{k}$ , find the component of $b$ along $a$ and the component of $b$ perpendicular to $a$ .

Solution. $a \cdot b = 2 - 1 + 3 = 4$ and $∣ a ∣^{2} = 3$ .

Component along $a$ : $\frac{a \cdot b}{∣ a ∣ ^{2}} a = \frac{4}{3} (\hat{i} + \hat{j} + \hat{k})$ .

Component perpendicular to $a$ : $b - \frac{4}{3} (\hat{i} + \hat{j} + \hat{k}) = \frac{1}{3} (2 \hat{i} - 7 \hat{j} + 5 \hat{k})$ .

Check: $(2 \hat{i} - 7 \hat{j} + 5 \hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k}) = 2 - 7 + 5 = 0$ , as required.

Magnitude identities

Expanding squares of vector sums

∣ a \pm b ∣^{2} = ∣ a ∣^{2} + ∣ b ∣^{2} \pm 2 a \cdot b = ∣ a ∣^{2} + ∣ b ∣^{2} \pm 2∣ a ∣∣ b ∣ cos θ

∣ a + b ∣^{2} + ∣ a - b ∣^{2} = 2 (∣ a ∣^{2} + ∣ b ∣^{2}) (parallelogram law)

∣ a + b ∣^{2} - ∣ a - b ∣^{2} = 4 a \cdot b

∣ a + b + c ∣^{2} = ∣ a ∣^{2} + ∣ b ∣^{2} + ∣ c ∣^{2} + 2 (a \cdot b + b \cdot c + c \cdot a)

Resolution along the base vectors

a = (a \cdot \hat{i}) \hat{i} + (a \cdot \hat{j}) \hat{j} + (a \cdot \hat{k}) \hat{k}

This is just the statement that the components of a vector are its projections on the axes, and it is the identity behind the direction cosine formulas.

Example 4

If $a + b + c = 0$ with $∣ a ∣ = 3$ , $∣ b ∣ = 5$ , $∣ c ∣ = 7$ , find the angle between $a$ and $b$ .

Solution. From $a + b = - c$ , take magnitudes squared:

$∣ a ∣^{2} + ∣ b ∣^{2} + 2 a \cdot b = ∣ c ∣^{2}$

$9 + 25 + 2 (3) (5) cos θ = 49 \Rightarrow 30 cos θ = 15 \Rightarrow cos θ = \frac{1}{2}$ .

Hence $θ = \frac{π}{3}$ .

Example 5

$D$ is the midpoint of $B C$ in triangle $A B C$ . Prove that $A B^{2} + A C^{2} = 2 (A D^{2} + B D^{2})$ .

Solution. Write $A B = A D + D B$ and $A C = A D + D C$ . Squaring both,

$A B^{2} = A D^{2} + D B^{2} + 2 A D \cdot D B$ and $A C^{2} = A D^{2} + D C^{2} + 2 A D \cdot D C$ .

Adding, and using $D B + D C = 0$ (since $D$ is the midpoint) together with $D B = D C$ ,

$A B^{2} + A C^{2} = 2 A D^{2} + 2 B D^{2}$ . This is Apollonius' theorem.

Proving classical results with the dot product

Example 6: the cosine rule

In triangle $A B C$ , $A B + B C + C A = 0$ , so $B C + C A = - A B$ . Squaring,

$a^{2} + b^{2} + 2 B C \cdot C A = c^{2}$ .

The angle between $B C$ and $C A$ is $π - C$ , so $B C \cdot C A = ab cos (π - C) = - ab cos C$ . Hence

c^{2} = a^{2} + b^{2} - 2 ab cos C

Example 7: projection rule

Dotting $B C + C A = - A B$ with $A B$ gives $- a c cos B - b c cos A = - c^{2}$ , that is

c = a cos B + b cos A

Example 8: the addition formula for cosine

Let $O P$ make angle $A$ below the $x$ axis and $O Q$ make angle $B$ above it, so the angle between them is $A + B$ . Then

$O P = O P (cos A \hat{i} - sin A \hat{j})$ and $O Q = O Q (cos B \hat{i} + sin B \hat{j})$ .

Computing $O P \cdot O Q$ in both ways:

$O P \cdot O Q cos (A + B) = O P \cdot O Q (cos A cos B - sin A sin B)$ , hence $cos (A + B) = cos A cos B - sin A sin B$ .

JEE Advanced

Mutually perpendicular vectors of equal magnitude. If $a, b, c$ are mutually perpendicular with $∣ a ∣ = ∣ b ∣ = ∣ c ∣ = λ$ , then all pairwise dot products vanish, so

$∣ a + b + c ∣^{2} = 3 λ^{2}$ , giving $∣ a + b + c ∣ = λ 3$ .

The angle that $a + b + c$ makes with each of them is

$cos θ_{1} = \frac{a \cdot ( a + b + c )}{∣ a ∣ ∣ a + b + c ∣} = \frac{λ ^{2}}{λ \cdot λ 3} = \frac{1}{3}$ ,

and the same for the other two. So $a + b + c$ is equally inclined to all three, at $cos^{- 1} \frac{1}{3}$ . This is the vector proof that a cube's diagonal makes equal angles with its three edges.

JEE Advanced

A useful unit-vector identity. If $\overset{a}{^}$ and $\hat{b}$ are unit vectors with angle $θ$ between them, then

$∣ \overset{a}{^} - \hat{b} ∣ = 2 sin \frac{θ}{2}$ and $∣ \overset{a}{^} + \hat{b} ∣ = 2 cos \frac{θ}{2}$ ,

so $tan \frac{θ}{2} = \frac{∣ a ^ - b ^ ∣}{∣ a ^ + b ^ ∣}$ . Both follow from $∣ \overset{a}{^} \pm \hat{b} ∣^{2} = 2 \pm 2 cos θ$ and the half-angle formulas.

Where the dot product is used

Work done by a constant force $F$ over a displacement $s$ is $W = F \cdot s$ . It is a scalar, and it is negative when the force opposes the motion.
Testing perpendicularity of lines, of a line and a normal, or of diagonals of a quadrilateral.
Finding angles between lines, between a line and a plane, and between two planes.
Length calculations in geometry proofs, by squaring a vector relation.

Mistakes that cost marks

Writing $a \cdot b$ as a vector, or writing $a \cdot b \cdot c$ . The triple dot product is meaningless because $a \cdot b$ is already a scalar.
Cancelling $a$ from $a \cdot b = a \cdot c$ .
Dividing the dot product by $∣ b ∣^{2}$ when the scalar projection was asked for. Scalar projection divides by $∣ b ∣$ ; vector projection divides by $∣ b ∣^{2}$ and keeps a factor of $b$ .
Forgetting that "obtuse angle" requires $a \cdot b < 0$ strictly, so endpoints where the dot product is zero must be excluded.
Taking $θ$ outside $[0, π]$ . The angle between vectors is never reflex.

Quick recap

$a \cdot b = ∣ a ∣∣ b ∣ cos θ = a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3}$
$a ⊥ b ⟺ a \cdot b = 0$ ; the sign of the dot product tells you acute or obtuse
Scalar projection $\frac{a \cdot b}{∣ b ∣}$ , vector projection $\frac{a \cdot b}{∣ b ∣ ^{2}} b$
$∣ a \pm b ∣^{2} = ∣ a ∣^{2} + ∣ b ∣^{2} \pm 2 a \cdot b$ is the workhorse identity
$a \cdot b \leq ∣ a ∣∣ b ∣$ with equality only for parallel vectors
$W = F \cdot s$

Frequently asked questions

Why is the dot product called the scalar product?

Because the result of the operation is a scalar, that is, a plain number with no direction attached. The cross product is called the vector product for the opposite reason, as its result is a vector.

If $a \cdot b = 0$ , must one of the vectors be zero?

No. If both vectors are non-zero, a vanishing dot product means $cos θ = 0$ , so the two vectors are perpendicular. The zero product only forces a zero vector when you additionally know the vectors are parallel.

What is the difference between the scalar projection and the vector projection?

The scalar projection of $a$ on $b$ is the number $\frac{a \cdot b}{∣ b ∣}$ , which can be negative. The vector projection is that number multiplied by $\hat{b}$ , so it is an actual vector lying along $b$ .

Can the dot product be negative, and what does that mean physically?

Yes, whenever the angle exceeds ninety degrees. In the work formula $W = F \cdot s$ , a negative value means the force has a component opposing the displacement, so it removes energy from the body, as friction does.

Is the dot product associative?

Associativity does not even make sense here. The expression $(a \cdot b) \cdot c$ asks for a dot product between a scalar and a vector, which is undefined. What does hold is the mixed associativity with scalars, $(m a) \cdot b = m (a \cdot b)$ .

Previous year questions on Scalar (or Dot) Product of Two Vectors

17 questions from past papers, each with a step-by-step solution.

Show all 17 questions

Ready to master Vector Algebra?

Take a full mock test, practice concept-by-concept, and get an AI-powered rank prediction — all on Fundamenthol.

Practice questions →Take a mock test