Scalar (dot) product of two vectors

What is the scalar product?

In plain words: the scalar product (or dot product) of two vectors multiplies the size of one vector by how much of the other points along it. The answer is an ordinary number, a scalar, with no direction.

On this page, as in NCERT, a vector is written in bold, A, and its magnitude in italics, A. (In your notebook, an arrow over the letter does the same job.)

For two vectors A and B with an angle θ between them:

Here A and B are the magnitudes of the vectors, and θ is the angle between them when they are drawn from the same point; it lies between 0° and 180°. This product is the mathematical tool behind work, the first idea of this chapter: the work done by a force is the scalar product of the force and the displacement.

Projection and the sign of the dot product

The quantity B cos θ is the projection of B on A: the length of the shadow B casts on the line of A, counted as negative when the shadow points opposite to A (when θ is more than 90°). In the same way, A cos θ is the projection of A on B. So

The sign of the dot product tells you how the two vectors point relative to each other:

A dot product of zero, for two vectors that are not zero, means they are perpendicular. This is the quickest test for perpendicular vectors.

Properties of the scalar product

The scalar product is commutative:

It is distributive over addition:

A number λ can be taken outside:

And the dot product of a vector with itself is the square of its magnitude, because the angle is zero:

Together these give a useful identity for the magnitude of a sum or a difference:

In the same way, the square of the magnitude of A − B is the same expression with a minus sign before the last term.

For the unit vectors $\hat i$, $\hat j$ and $\hat k$ along the x, y and z axes, which have magnitude 1 and are at right angles to each other:

Dot product from components

Write the two vectors in components:

Multiply out using the distributive law. The x-term of A gives three products:

which is just A_xB_x, because each Cartesian basis vector dotted with itself gives 1 and with a different perpendicular basis vector gives 0. The y- and z-terms give A_yB_y and A_zB_z in the same way, so only three terms survive:

With B = A, this gives the magnitude of a vector:

Putting the two forms together gives the angle between any two vectors:

This holds for any two non-zero vectors.

For non-zero B, the scalar component (projection) of A along B is A · B divided by its magnitude B. A zero vector does not define a projection direction.

Where the dot product is used

• Work. The work done by a constant force F over a displacement d is W = F · d, which equals Fd cos θ. A force at right angles to the motion does no work.
• Power. The power of a force, the rate at which it does work on a body moving with velocity v, is P = F · v.
• Components of any vector. The component of a vector along any direction is its dot product with the unit vector in that direction: the x-component of A is $\mathbf{A} \cdot \hat i$.
• Checking right angles. Two non-zero vectors are perpendicular when their dot product is zero, which is often quicker than finding the angle.

Solved examples

Common mistakes to avoid

Quick revision summary

• The scalar product of two vectors is a scalar: the magnitude of one times the component of the other along it.
• It is positive for an acute angle, zero for perpendicular vectors and negative for an obtuse angle.
• It is commutative and distributive, and the dot product of a vector with itself is the square of its magnitude.
• Each Cartesian basis unit vector dotted with itself gives 1; different mutually perpendicular basis vectors give 0, so only three component products survive. Two arbitrary different unit vectors need not have zero dot product.
• It gives the angle between two vectors, the projection of one on the other, and the work done by a force.

Questions students ask

What is the difference between the dot product and the cross product?

The dot product A · B = AB cos θ is a scalar: it is largest when the vectors point the same way, zero when they are perpendicular, and A · B = B · A. The cross product A × B is a vector of magnitude AB sin θ, perpendicular to both vectors: it is zero for parallel vectors, largest for perpendicular ones, and A × B = −B × A. Work, F · d, is a dot product; torque, r × F, is a cross product.

Why is the dot product of two vectors a scalar?

The dot product of two vectors is a scalar because AB cos θ is built only from the two magnitudes and the angle between the vectors, and none of these depends on how the axes are chosen or points in any direction. Rotate the axes and the components change, but A_xB_x + A_yB_y + A_zB_z comes out the same. That is why work, the dot product of force and displacement, is a scalar even though force and displacement are vectors.

How do you find the angle between two vectors using the dot product?

The angle θ between two non-zero vectors follows from cos θ = A · B/(AB), with the dot product from components, A_xB_x + A_yB_y + A_zB_z, and each magnitude from A = √(A_x² + A_y² + A_z²). For example, for A = 2î + 2ĵ and B = 3ĵ, A · B = 6, A = 2√2 and B = 3, so cos θ = 1/√2 and θ = 45°.

Is the dot product the same as the scalar product?

Yes, the dot product and the scalar product are two names for the same operation, A · B = AB cos θ: “dot” after the symbol, “scalar” after the kind of answer it gives. In mathematics it is the standard example of an inner product.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. Two vectors have magnitudes 3 and 4, and the angle between them is 60°. Find their scalar product.

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   $$\begin{aligned} \mathbf{A} \cdot \mathbf{B} &= 3 \times 4 \times \cos 60^\circ \\ &= 12 \times 0.5 = 6 \end{aligned}$$

2. Q2. Find the angle between $\hat i + \hat j$ and $\hat j + \hat k$.

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   The dot product is 0 + 1 + 0 = 1, and each vector has magnitude √2, so cos θ = 1/(√2 × √2) = 1/2, and the angle is 60°.

3. Q3. Show that the dot product of A + B with A − B is A^² − B^². What does this tell you about the diagonals of a rhombus?

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   Multiply out with the distributive law:

   $$\begin{aligned} &(\mathbf{A} + \mathbf{B}) \cdot (\mathbf{A} - \mathbf{B}) \\ &= A^{2} - \mathbf{A} \cdot \mathbf{B} + \mathbf{B} \cdot \mathbf{A} - B^{2} \end{aligned}$$

   The two middle terms cancel, because the dot product is commutative, leaving A^² − B^². In a rhombus with sides A and B, the diagonals are A + B and A − B, and A = B, so their dot product is zero: the diagonals of a rhombus are perpendicular.