International Baccalaureate Mathematics

Syllabus Content

The definition of the scalar product of two vectors. The angle between two vectors. Perpendicular vectors; parallel vectors.

Here are some exam-style questions on this statement:

"Consider two perpendicular vectors $p$ and $q$." ... more

"The following diagram shows parallelogram OABC with $ \overrightarrow{OA}\ = \pmb{a} $, $ \overrightarrow{OC}\ = \pmb{c} $ and $|\pmb{c}| = 2|\pmb{a}| $" ... more

"George and Hugo like to fly model airplanes. On one day George's plane takes off from level ground and shortly after that Hugo's plane takes off." ... more

"The line $L_1$ passes through the points A(3, 5, 1) and B(3, 6, 0)." ... more

"Two points $A$ and $B$ have coordinates (1 , 3 , 6) and (8 , 7 , 10) respectively." ... more

"The points A and B have coordinates $(3,-2,1)$ and $(4, 0, -1)$ respectively." ... more

"Two lines, $L_1$ and $L_2$, intersect at point $P$." ... more

"Consider the vectors $\mathbf{a}$ and $\mathbf{b}$ such that $\mathbf{a} = \begin{pmatrix} 16 \\ -12 \end{pmatrix} $ and $ |\mathbf{b}| = 11$." ... more

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Vectors Vectors usually first make an appearance when pupils learn about transformations. A translation is best described with a vector written as a two by one matrix. Following that pupils learn how vectors can be used to prove geometric relationships in simple line diagrams. Ultimately vectors are studied as a major topic for A Level and International Baccalaureate courses where vectors in three dimensions are included.You can think of a vector as what is needed to 'carry' the point A to the point B. The Latin word vector means ‘carrier’ and was first used by 18th century astronomers investigating planet rotation around the Sun.

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Official Guidance, clarification and syllabus links:

Applications of the properties of the scalar product
v·w=w·v;
u·(v+w)=u·v+u·w;
(kv)·w=k(v·w);
v·v=|v|².
v·w=|v||w|cos θ, where θ is the angle between v and w.
For non-zero vectors v·w=0 is equivalent to the vectors being perpendicular;
for parallel vectors |v·w|=|v||w|.

Formula Booklet:

Scalar product

$\mathbf{v} \cdot \mathbf{w} = v_1w_1 + v_2w_2 + v_3w_3, \quad \text{where } \mathbf{v} = \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix} , \text{ and }\mathbf{w} = \begin{pmatrix} w_1 \\ w_2 \\ w_3 \end{pmatrix} $

$\mathbf{v} \cdot \mathbf{w} = |\mathbf{v}|| \mathbf{w}| \cos{\theta}, \text{ where } \theta $ is the angle between $\mathbf{v} $ and $ \mathbf{w}$

Angle between two vectors

$ \cos{\theta} = \dfrac{v_1w_1 + v_2w_2 + v_3w_3}{|\mathbf{v}|| \mathbf{w}|} $

The scalar product of two vectors, also known as the dot product, is a mathematical operation that takes two vectors and returns a scalar quantity. The scalar product is defined as the product of the magnitudes of the two vectors and the cosine of the angle between them.

The formula for the scalar product of two vectors, a and b, is:

$$ \mathbf{a \cdot b} = |\mathbf{a}| |\mathbf{b}| \cos \theta $$

where theta $ \theta $ is the angle between the vectors a and b.

A second way to calculate the scalar product is as follow:

The alternate method of finding the scalar product involves multiplying together the corresponding components of the vectors. Let's say we have two vectors a = $ a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k} $ and b = $ b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k} $, where $ \mathbf{i}, \mathbf{j}, \mathbf{k} $ are the unit vectors in the $x, y,$ and $z$ directions, respectively. Then the scalar product of a and b is given by:

$$\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3$$

This can also be written in a more compact form using summation notation:

$$\mathbf{a} \cdot \mathbf{b} = \sum_{i=1}^3 a_ib_i$$

The dot product can be used find the angle between two vectors.

As an example, let's find the angle between the following two vectors:

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

Rearranging the formula above:

$$ \cos \theta = \frac{\mathbf{a \cdot b}}{|\mathbf{a}| |\mathbf{b}|} = \frac{(2 \times 5) + (3 \times -1) + (4 \times 2)}{\sqrt{2^2 + 3^2 + 4^2} \sqrt{5^2 + (-1)^2 + 2^2}} = 0.5085... $$ $$ \theta \approx 59.4^o $$

Using the absolute value of the scalar product when finding the angle between two vectors ensures that the acute angle is found.

Applications of the properties of the scalar product:

Commutativity: v$\cdot$ w = w $\cdot$ v
Distributivity: u $\cdot$ (v + w) = u $\cdot$ v + u $\cdot$ w
Associativity with scalar multiplication: (kv) $\cdot$ w = k(v $\cdot$ w)
Self-product: v $\cdot$ v = $|\textbf{v}|^2$
Geometric interpretation: v $\cdot$ w = $|\textbf{v}| |\textbf{w}| \cos \theta $, where $ \theta $ is the angle between v and w
Perpendicularity: If v $\cdot$ w = 0, then v and w are perpendicular
Parallelism: If v and w are parallel, then $|\textbf{v} \cdot \textbf{w}| = |\textbf{v}| |\textbf{w}| $

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International Baccalaureate Mathematics

Geometry and Trigonometry

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