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Vectors — The scalar product

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Vectors — The Scalar Product

The scalar product, also called the dot product, is a way of combining two vectors to produce a scalar, not a vector.

In Year 13 vector work, the scalar product is used to find angles between vectors, test whether vectors are perpendicular, and work with vectors written in both column-vector form and \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) form.

Key definitions and formulae

1. Definition of the scalar product

If

\[ \mathbf{a}= \begin{pmatrix} a_1\\ a_2\\ a_3 \end{pmatrix}, \qquad \mathbf{b}= \begin{pmatrix} b_1\\ b_2\\ b_3 \end{pmatrix}, \]

then their scalar product is

\[ \mathbf{a}\cdot\mathbf{b}=a_1b_1+a_2b_2+a_3b_3. \]

The answer is a single number.

2. Scalar product in \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) form

If

\[ \mathbf{a}=a_1\mathbf{i}+a_2\mathbf{j}+a_3\mathbf{k}, \qquad \mathbf{b}=b_1\mathbf{i}+b_2\mathbf{j}+b_3\mathbf{k}, \]

then

\[ \mathbf{a}\cdot\mathbf{b}=a_1b_1+a_2b_2+a_3b_3. \]

This works because

\[ \mathbf{i}\cdot\mathbf{i}=1,\qquad \mathbf{j}\cdot\mathbf{j}=1,\qquad \mathbf{k}\cdot\mathbf{k}=1 \] \[ \mathbf{i}\cdot\mathbf{j}=0,\qquad \mathbf{i}\cdot\mathbf{k}=0,\qquad \mathbf{j}\cdot\mathbf{k}=0. \]

So different unit vectors are perpendicular.

3. Scalar product and angle

Another important formula is

\[ \mathbf{a}\cdot\mathbf{b}=|\mathbf{a}||\mathbf{b}|\cos\theta, \]

where \( \theta \) is the angle between the two vectors.

4. Magnitude of a vector

If

\[ \mathbf{a}= \begin{pmatrix} a_1\\ a_2\\ a_3 \end{pmatrix}, \]

then its magnitude is

\[ |\mathbf{a}|=\sqrt{a_1^2+a_2^2+a_3^2}. \]

5. Perpendicular vectors

Two non-zero vectors are perpendicular if and only if

\[ \mathbf{a}\cdot\mathbf{b}=0. \]

This is a very common exam result.

Worked examples

Example 1: Find a scalar product from column vectors

Find \( \mathbf{a}\cdot\mathbf{b} \) if

\[ \mathbf{a}= \begin{pmatrix} 2\\ -1\\ 3 \end{pmatrix}, \qquad \mathbf{b}= \begin{pmatrix} 4\\ 5\\ -2 \end{pmatrix}. \]

Solution

\[ \mathbf{a}\cdot\mathbf{b} =(2)(4)+(-1)(5)+(3)(-2) \] \[ =8-5-6=-3. \]
Therefore, \[ \mathbf{a}\cdot\mathbf{b}=-3. \]

Example 2: Find a scalar product in \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) form

Find \( \mathbf{p}\cdot\mathbf{q} \) if

\[ \mathbf{p}=3\mathbf{i}-2\mathbf{j}+\mathbf{k}, \qquad \mathbf{q}=5\mathbf{i}+\mathbf{j}-4\mathbf{k}. \]

Solution

Multiply matching components and add:

\[ \mathbf{p}\cdot\mathbf{q} =(3)(5)+(-2)(1)+(1)(-4) \] \[ =15-2-4=9. \]
So, \[ \mathbf{p}\cdot\mathbf{q}=9. \]

Example 3: Find the angle between two vectors

Find the angle between

\[ \mathbf{a}= \begin{pmatrix} 1\\ 2\\ 2 \end{pmatrix}, \qquad \mathbf{b}= \begin{pmatrix} 2\\ 1\\ -2 \end{pmatrix}. \]

Solution

First find the scalar product:

\[ \mathbf{a}\cdot\mathbf{b}=(1)(2)+(2)(1)+(2)(-2)=2+2-4=0. \]

Now use

\[ \mathbf{a}\cdot\mathbf{b}=|\mathbf{a}||\mathbf{b}|\cos\theta. \]

Since \( \mathbf{a}\cdot\mathbf{b}=0 \), we have

\[ |\mathbf{a}||\mathbf{b}|\cos\theta=0. \]

As both vectors are non-zero,

\[ \cos\theta=0 \qquad \Rightarrow \qquad \theta=90^\circ. \]
The vectors are perpendicular.

Example 4: Find the angle using \( \cos\theta \)

Find the angle between

\[ \mathbf{u}=2\mathbf{i}+\mathbf{j}+2\mathbf{k}, \qquad \mathbf{v}=\mathbf{i}-2\mathbf{j}+2\mathbf{k}. \]

Solution

First find the scalar product:

\[ \mathbf{u}\cdot\mathbf{v} =(2)(1)+(1)(-2)+(2)(2)=2-2+4=4. \]

Now find the magnitudes:

\[ |\mathbf{u}|=\sqrt{2^2+1^2+2^2}=\sqrt{9}=3, \qquad |\mathbf{v}|=\sqrt{1^2+(-2)^2+2^2}=\sqrt{9}=3. \]

Use the formula:

\[ \cos\theta=\frac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{u}||\mathbf{v}|} =\frac{4}{3\times 3} =\frac{4}{9}. \]
\[ \theta=\cos^{-1}\!\left(\frac{4}{9}\right). \]
Hence, \[ \theta=\cos^{-1}\!\left(\frac{4}{9}\right). \]

Example 5: Show that two vectors are perpendicular

Show that the vectors

\[ \mathbf{a}=4\mathbf{i}-\mathbf{j}+3\mathbf{k}, \qquad \mathbf{b}=\mathbf{i}+5\mathbf{j}+\mathbf{k} \]

are not perpendicular.

Solution

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

Since the scalar product is not zero,

\[ \mathbf{a}\cdot\mathbf{b}\neq 0, \]
the vectors are not perpendicular.

Standard methods

Finding a scalar product

Multiply matching components and add the results.

\[ \mathbf{a}\cdot\mathbf{b}=a_1b_1+a_2b_2+a_3b_3. \]

Finding an angle between two vectors

Use the formula

\[ \cos\theta=\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}. \]

Then use inverse cosine to find \( \theta \).

Testing for perpendicular vectors

If

\[ \mathbf{a}\cdot\mathbf{b}=0, \]

then the vectors are perpendicular, provided both vectors are non-zero.

Common mistakes and exam tips

Common mistakes

  • Thinking that the scalar product gives another vector. It gives a scalar.
  • Forgetting to multiply matching components only.
  • Making sign errors with negative components.
  • Using the angle formula without first finding the magnitudes correctly.
  • Assuming vectors are perpendicular without checking that the scalar product is zero.

Exam tips

  • When vectors are written in \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) form, first identify the three components clearly.
  • For angle questions, write the formula before substituting numbers.
  • If asked to show vectors are perpendicular, your final line should state that the scalar product is zero.
  • Check whether your calculator is in degree mode if the answer should be in degrees.

Summary

\[ \mathbf{a}\cdot\mathbf{b}=a_1b_1+a_2b_2+a_3b_3 \] \[ \mathbf{a}\cdot\mathbf{b}=|\mathbf{a}||\mathbf{b}|\cos\theta \] \[ \mathbf{a}\cdot\mathbf{b}=0 \quad \Longrightarrow \quad \mathbf{a}\perp\mathbf{b} \]

The scalar product is one of the most important tools in Year 13 vectors because it connects algebraic vector methods with geometric ideas such as angle and perpendicularity.

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