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Vectors

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AHL 4.13

The scalar product: definition and component form

AHL 4.13

The scalar product: definition and component form

Target: I can find the scalar product of two vectors from their magnitudes and the angle between them. (+3 more)

By the end of this section

Key points

The definition

  • The scalar product (also called the dot product) of \(\mathbf{a}\) and \(\mathbf{b}\) is \(\mathbf{a}\cdot\mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta\).
  • \(\theta\) is the angle between the vectors when both arrows leave the same point. Set the calculator to degrees.
  • The result is a number, not a vector. Acute angle: positive. Angle of \(90^\circ\): zero. Obtuse angle: negative.

Component form

  • If \(\mathbf{a} = (a_1, a_2, a_3)\) and \(\mathbf{b} = (b_1, b_2, b_3)\) then \(\mathbf{a}\cdot\mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3\).
  • The same rule works for \(a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}\). Write a missing term as 0. In 2D use two terms.
  • Watch the signs: multiply each pair first, then add.

Using it

  • \(\mathbf{a}\cdot\mathbf{b} = \mathbf{b}\cdot\mathbf{a}\), and \(\mathbf{a}\cdot\mathbf{a} = |\mathbf{a}|^2\).
  • \(\mathbf{a}\cdot(\mathbf{b}+\mathbf{c}) = \mathbf{a}\cdot\mathbf{b} + \mathbf{a}\cdot\mathbf{c}\).
  • Work done by a force \(\mathbf{F}\) through a displacement \(\mathbf{d}\) is \(\mathbf{F}\cdot\mathbf{d}\).

Explore: the scalar product and the angle

Drag the tip of a and the tip of b. Both vectors leave the origin. Watch the sign of a · b and the angle between them.

a · b
Angle θ
The angle is

Click a tip, then use the arrow keys to move it one unit. The tips stay on whole numbers, so you can check each line by hand.

Drag a tip to change a vector.

  1. Keep a still and swing b all the way round. Where does a · b change from positive to negative? What is the angle there?
  2. Make a · b equal to 0 with neither vector being the zero vector. What is the same about every way of doing it?
  3. Make b twice as long without changing its direction. What happens to a · b, to cos θ and to θ? Why?
  4. Make the vectors point in opposite directions. What are cos θ and θ now, and how does the working show that they are parallel?

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
Scalar product from magnitudes and angle

The vectors \(\mathbf{a}\) and \(\mathbf{b}\) leave the same point. \(|\mathbf{a}| = 6\), \(|\mathbf{b}| = 11\), and the angle between them is \(35^\circ\). Find \(\mathbf{a}\cdot\mathbf{b}\).

|a| = 6|b| = 1135°OAB
Working
  1. The definition

    \(\mathbf{a}\cdot\mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta\)

    Both vectors point away from the same point, so θ is the angle shown.

  2. Substitute

    \(6 \times 11 \times \cos 35^\circ\)

    Use degrees mode.

  3. Evaluate

    \(\mathbf{a}\cdot\mathbf{b} = 54.1\)

    A number, not a vector. It is positive because the angle is acute.

Video walkthrough coming
An obtuse angle gives a negative

\(|\mathbf{a}| = 8\), \(|\mathbf{b}| = 5\) and the angle between \(\mathbf{a}\) and \(\mathbf{b}\) is \(125^\circ\). Find \(\mathbf{a}\cdot\mathbf{b}\).

|a| = 8|b| = 5125°OAB
Working
  1. Substitute

    \(\mathbf{a}\cdot\mathbf{b} = 8 \times 5 \times \cos 125^\circ\)

    θ is the angle between the two arrows when they start at the same point.

  2. Evaluate

    \(= 40 \times (-0.574) = -22.9\)

    cos 125° is negative, so the scalar product is negative. Acute angle: positive. 90°: zero. Obtuse: negative.

Video walkthrough coming
Scalar product of two column vectors

Given \(\mathbf{a} = \begin{pmatrix}4\\-1\\3\end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix}2\\5\\-2\end{pmatrix}\), calculate \(\mathbf{a}\cdot\mathbf{b}\).

Working
  1. Multiply matching components

    \(\mathbf{a}\cdot\mathbf{b} = (4)(2) + (-1)(5) + (3)(-2)\)

    First with first, second with second, third with third.

  2. Add

    \(= 8 - 5 - 6 = -3\)

    The result is a single number.

Video walkthrough coming
Scalar product in i, j, k form

Given \(\mathbf{a} = 2\mathbf{i} + 5\mathbf{j} - 3\mathbf{k}\) and \(\mathbf{b} = \mathbf{i} - 4\mathbf{j} + 6\mathbf{k}\), find \(\mathbf{a}\cdot\mathbf{b}\).

Working
  1. Write the components

    \(\mathbf{a} = \begin{pmatrix}2\\5\\-3\end{pmatrix},\quad \mathbf{b} = \begin{pmatrix}1\\-4\\6\end{pmatrix}\)

    The coefficients of i, j and k are the components. A missing term has component 0.

  2. Multiply and add

    \(\mathbf{a}\cdot\mathbf{b} = (2)(1) + (5)(-4) + (-3)(6) = 2 - 20 - 18 = -36\)

    Negative: the angle between the vectors is obtuse.

Video walkthrough coming
Work done by a force

A force \(\mathbf{F} = \begin{pmatrix}12\\-5\\8\end{pmatrix}\) N moves an object through the displacement \(\mathbf{d} = \begin{pmatrix}3\\4\\-1\end{pmatrix}\) m. The work done, in joules, is \(\mathbf{F}\cdot\mathbf{d}\). Find the work done.

Working
  1. Scalar product

    \(\mathbf{F}\cdot\mathbf{d} = (12)(3) + (-5)(4) + (8)(-1)\)

    Multiply matching components.

  2. Add

    \(= 36 - 20 - 8 = 8\)

    The work done is 8 J.

Video walkthrough coming