Summary Notes: Vectors

Summary Notes: Vectors

SCALAR PRODUCT (DOT PRODUCT)

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

where \({\theta}\) is the angle between \({\mathbf{a}}\) and \({\mathbf{b}}\) when placed at the same initial points (or the same terminal points), \({0^{\circ} \leq \theta \leq 180^{\circ}}\).

Scalar Product in column vectors:

\({\mathbf{a} \cdot \mathbf{b} = \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix} \cdot \begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix} = a_1 b_1 + a_2 b_2 + a_3 b_3}\)

For example,

\({\begin{pmatrix} -2 \\ 1 \\ 4 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 5 \\ 0 \end{pmatrix} = -4 + 5 + 0 = 1}\)

PROPERTIES OF SCALAR PRODUCT

• \({\mathbf{a} \cdot \mathbf{b}}\) is a real number (scalar), not a vector.

• \({\mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a}}\) (Commutative law).

• If either \({\mathbf{a} = \mathbf{0}}\) or \({\mathbf{b} = \mathbf{0}}\), then \({\mathbf{a} \cdot \mathbf{b} = 0}\) (real number zero).

• \({\mathbf{a} \cdot \mathbf{a} = \left| \mathbf{a} \right|^2}\), since \({\mathbf{a}}\) is parallel to itself \({\Rightarrow \theta = 0^{\circ} \Rightarrow \cos 0^{\circ} = 1}\).

• \({\mathbf{a} \cdot \mathbf{b} > 0}\) when \({0^{\circ} \leq \theta < 90^{\circ}}\); \({\mathbf{a} \cdot \mathbf{b} < 0}\) when \({90^{\circ} < \theta \leq 180^{\circ}}\).

• For \({\lambda, \mu \in \mathbb{R}}\),

1. \({\lambda(\mathbf{a} \cdot \mathbf{b}) = (\lambda \mathbf{a}) \cdot \mathbf{b} = \mathbf{a} \cdot (\lambda \mathbf{b})}\);

2. \({\lambda \mathbf{a} \cdot \mu \mathbf{b} = (\lambda \mu) \mathbf{a} \cdot \mathbf{b}}\).

• Distributive law

1. \({\mathbf{a} \cdot (\mathbf{b} + \mathbf{c}) = \mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{c}}\).

2. \({(\mathbf{a} + \mathbf{b}) \cdot \mathbf{c} = \mathbf{a} \cdot \mathbf{c} + \mathbf{b} \cdot \mathbf{c}}\).

3. \({(\mathbf{a} + \mathbf{b}) \cdot (\mathbf{c} + \mathbf{d}) = \mathbf{a} \cdot \mathbf{c} + \mathbf{a} \cdot \mathbf{d} + \mathbf{b} \cdot \mathbf{c} + \mathbf{b} \cdot \mathbf{d}}\),

specially, \({(\mathbf{a} + \mathbf{b}) \cdot (\mathbf{a} + \mathbf{b}) = \mathbf{a} \cdot \mathbf{a} + 2\mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{b} = \left| \mathbf{a} \right|^2 + 2\mathbf{a} \cdot \mathbf{b} + \left| \mathbf{b} \right|^2}\).

Common Mistake:

The scalar product of vectors does NOT satisfy Associative law or Cancellation law:

\({(\mathbf{a} \cdot \mathbf{b}) \cdot \mathbf{c} \neq \mathbf{a} \cdot (\mathbf{b} \cdot \mathbf{c})}\)

\({\mathbf{a} \cdot \mathbf{b} = \mathbf{a} \cdot \mathbf{c} \not\Rightarrow \mathbf{b} = \mathbf{c}}\)

APPLICATIONS OF SCALAR PRODUCT

PARALLELISM TEST

\({\mathbf{a} \cdot \mathbf{b} = \pm \left| \mathbf{a} \right| \left| \mathbf{b} \right| \iff \mathbf{a} \parallel \mathbf{b}}\)

when \({\theta = 0^{\circ}}\) or \({180^{\circ}}\). Specifically,
when \({\theta = 0^{\circ} \Rightarrow \cos 0^{\circ} = 1}\), \({\mathbf{a}}\) and \({\mathbf{b}}\) are parallel and in the same direction;
when \({\theta = 180^{\circ} \Rightarrow \cos 180^{\circ} = -1}\), \({\mathbf{a}}\) and \({\mathbf{b}}\) are parallel and in the opposite direction.

PERPENDICULARITY TEST

\({\mathbf{a} \cdot \mathbf{b} = 0 \iff \mathbf{a} \perp \mathbf{b}}\)

when \({\theta = 90^{\circ} \Rightarrow \cos 90^{\circ} = 0}\), \({\mathbf{a}}\) and \({\mathbf{b}}\) are perpendicular to each other.

ANGLE BETWEEN TWO VECTORS

\({\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\left| \mathbf{a} \right| \left| \mathbf{b} \right|} \Rightarrow \theta = \cos^{-1} \left( \frac{\mathbf{a} \cdot \mathbf{b}}{\left| \mathbf{a} \right| \left| \mathbf{b} \right|} \right)}\)

LENGTH OF PROJECTION

The length of projection of \({\mathbf{a}}\) onto \({\mathbf{b}}\), \({l}\), is given by \({l = \left| \mathbf{a} \right| \cos \theta = \left| \mathbf{a} \right| \frac{\mathbf{a} \cdot \mathbf{b}}{\left| \mathbf{a} \right| \left| \mathbf{b} \right|}}\), and so

\({l = \frac{\left| \mathbf{a} \cdot \mathbf{b} \right|}{\left| \mathbf{b} \right|} = \left| \mathbf{a} \cdot \hat{\mathbf{b}} \right|}\)

Video Solution:

Video Solution

Video solution locked

A free account saves your progress; this video needs tuition enrolment.

Solution:

Solution locked

680+ questions here come with free worked solutions. A free account saves your questions and progress; this one needs tuition enrolment.

Create a free account to practise similar questions.

Need help? Join our JC Math tuition classes.

Learn more