3.13 The scalar product

Adding two vectors gives another vector. The scalar product is a different way of combining two vectors, and its result is a single number. That number measures how far the two vectors point in the same direction: it is large and positive when they point roughly the same way, zero when they are at right angles, and negative when they point in roughly opposite directions.

This makes the scalar product the main tool for angles in three dimensions, where drawing an accurate diagram is difficult. It also appears in physics as the work done by a force, and later in the course it gives the equation of a plane.

The scalar product

The component form

The scalar product of two vectors is found by multiplying corresponding components and adding the results.

The dot between the vectors is part of the notation and must always be written: $\boldsymbol{v} \cdot \boldsymbol{w}$ is read "$\boldsymbol{v}$ dot $\boldsymbol{w}$". The name "scalar product" comes from the fact that the answer is a scalar, not a vector.

The geometric form

The scalar product can also be calculated from the lengths of the two vectors and the angle between them. The angle $\theta$ between two vectors is measured when they are drawn tail to tail, starting from the same point, and it always lies in the interval $0 \le \theta \le \pi$ ($0°$ to $180°$).

Both forms always give the same number. The component form is used to calculate a scalar product, and the geometric form is used to interpret it, most often by rearranging it to find the angle.

Why the two forms agree

Draw $\boldsymbol{v}$ and $\boldsymbol{w}$ from the origin $O$. The third side of the triangle they form runs from the tip of $\boldsymbol{w}$ to the tip of $\boldsymbol{v}$, so it represents $\boldsymbol{v} - \boldsymbol{w}$

The cosine rule (3.2) in this triangle gives

The left-hand side can also be expanded from components:

Comparing the two expressions, the terms $|\boldsymbol{v}|^2 + |\boldsymbol{w}|^2$ match, so

This is a clear case of equivalence: one quantity has an algebraic description (components) and a geometric description (lengths and an angle), and moving between them is what makes the scalar product useful.

Properties of the scalar product

The scalar product follows most of the rules of ordinary multiplication, so expressions with dots can be expanded and simplified much like algebra.

Each property follows from the component form. For example, $\boldsymbol{v} \cdot \boldsymbol{v} = v_1^2 + v_2^2 + v_3^2$

The property $\boldsymbol{v} \cdot \boldsymbol{v} = |\boldsymbol{v}|^2$ is the key to questions that give magnitudes and scalar products but no components. The magnitude of a combination is found by squaring it as a scalar product and expanding.

The angle between two vectors

Finding the angle

Rearranging the geometric form gives the angle between two non-zero vectors directly from their components:

Since $\theta$ is between $0$ and $\pi$, the inverse cosine gives exactly one angle. A negative cosine gives an obtuse angle, so no case analysis is needed.

The method has three steps: find the scalar product, find both magnitudes, then divide and take the inverse cosine.

What the sign of the scalar product shows

The magnitudes $|\boldsymbol{v}|$ and $|\boldsymbol{w}|$ are positive, so the sign of $\boldsymbol{v} \cdot \boldsymbol{w}$ is the sign of $\cos\theta$. This gives information about the angle before any calculation of $\theta$

Exact angles

When the cosine is a value from the unit circle, such as $\frac{1}{2}$ or $-\frac{\sqrt{2}}{2}$

Angles with the coordinate axes

The angle between a vector and the positive $x$-axis is the angle between the vector and $\boldsymbol{i}$. Since $\boldsymbol{v} \cdot \boldsymbol{i} = v_1$ and $|\boldsymbol{i}| = 1$

Angles in triangles and solids

To find an angle in a triangle with known vertices, use the two vectors that start at the vertex where the angle is. For angle $ABC$, the vertex is $B$, so the vectors are $\overrightarrow{BA}$ and $\overrightarrow{BC}$

Perpendicular vectors

The perpendicularity test

Two non-zero vectors are perpendicular when the angle between them is $90°$. Since $\cos 90° = 0$, the geometric form gives $\boldsymbol{v} \cdot \boldsymbol{w} = 0$. Conversely, if the scalar product of two non-zero vectors is zero, then $\cos\theta = 0$

This is one of the most used facts in vector geometry. Perpendicular vectors are also called orthogonal. In two dimensions, $\begin{pmatrix} a \\ b \end{pmatrix}$ is always perpendicular to $\begin{pmatrix} -b \\ a \end{pmatrix}$

A vector perpendicular to two others

In three dimensions, infinitely many directions are perpendicular to a single vector, but only one direction (up to multiples) is perpendicular to two non-parallel vectors. It can be found by writing the unknown vector as $\begin{pmatrix} x \\ y \\ z \end{pmatrix}$

Parallel vectors and the scalar product

Parallel vectors are at an angle of $0$ (same direction) or $\pi$ (opposite directions), where $\cos\theta = \pm 1$. The scalar product then reaches its largest possible size:

For any other angle $|\cos\theta| < 1$, so $|\boldsymbol{v} \cdot \boldsymbol{w}| < |\boldsymbol{v}||\boldsymbol{w}|$.

For example, $\begin{pmatrix} 2 \\ -4 \\ 6 \end{pmatrix} \cdot \begin{pmatrix} -1 \\ 2 \\ -3 \end{pmatrix} = -2 - 8 - 18 = -28$

Proofs with the scalar product

Because a zero scalar product means a right angle, the scalar product turns geometric statements about perpendicular lines into algebra. The method is the same as for other vector proofs: choose a few vectors to describe the figure, write the vectors along the relevant lines in terms of them, and show that the scalar product is zero.

The same few lines prove that the diagonals of a rhombus are perpendicular. In rhombus $ABCD$, let $\overrightarrow{AB} = \boldsymbol{a}$ and $\overrightarrow{AD} = \boldsymbol{b}$

In a parallelogram that is not a rhombus, $|\boldsymbol{a}| \neq |\boldsymbol{b}|$ and the diagonals are not perpendicular, so the proof also shows exactly when the result holds.

Projection: the component of one vector along another

Scalar and vector projection

Shining a light perpendicular to $\boldsymbol{w}$ casts a shadow of $\boldsymbol{v}$ onto the line of $\boldsymbol{w}$. The length of that shadow is $|\boldsymbol{v}|\cos\theta$, from the right-angled triangle formed by $\boldsymbol{v}$

Dividing the scalar product by $|\boldsymbol{w}|$ gives exactly this length, because $\frac{\boldsymbol{v} \cdot \boldsymbol{w}}{|\boldsymbol{w}|} = |\boldsymbol{v}|\cos\theta$

The scalar projection is the scalar product with the unit vector in the direction of $\boldsymbol{w}$. The vector projection multiplies this length by the same unit vector. Projections give the foot of a perpendicular from a point to a line, which is the basis of the shortest-distance methods in 3.15.

Work done by a force

When a constant force $\boldsymbol{F}$ moves an object through a displacement $\boldsymbol{s}$, only the part of the force along the direction of motion does any work. That part is the scalar projection of $\boldsymbol{F}$ onto $\boldsymbol{s}$, and multiplying it by the distance $|\boldsymbol{s}|$ gives

With force in newtons and displacement in metres, the work is in joules (J). A force perpendicular to the motion does no work, which matches $\boldsymbol{F} \cdot \boldsymbol{s} = 0$.

Exam-style questions on the scalar product

The second question is a calculator question about a solid. The vertices come from the dimensions of the cuboid, and the angles come from the vectors along its diagonals.

Self-check

• To do:

  I can calculate a scalar product from components and explain why the result is a number.

• To do:

  I can state the geometric form $\boldsymbol{v} \cdot \boldsymbol{w} = |\boldsymbol{v}||\boldsymbol{w}|\cos\theta$ and explain why it agrees with the component form.

• To do:

  I can find the angle between two vectors, exactly or to 1 decimal place in degrees.

• To do:

  I can find an angle in a triangle using the two vectors that start at its vertex.