AOS5 Topic 1: Vector Basics

Definition:

A vector is a quantity that has both magnitude and direction. It is represented geometrically by an arrow.

Notation:

Vectors are typically denoted by boldface letters or with an arrow on top, e.g., a or →a.

Vector Representation

Components: A vector in 2D or 3D space can be represented by its components along the coordinate axes.

Position Vector: A vector from the origin to a point in space.



Unit Vectors in Vectors

In vector notation, \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \) are unit vectors that represent the directions of the \( x \), \( y \), and \( z \) axes, respectively, in a three-dimensional Cartesian coordinate system. These unit vectors are typically used to denote the standard basis vectors in three-dimensional Euclidean space.

Here's how they are defined:

\( \mathbf{i} \) represents the unit vector along the positive \( x \)-axis, which is \( \langle 1, 0, 0 \rangle \).

\( \mathbf{j} \) represents the unit vector along the positive \( y \)-axis, which is \( \langle 0, 1, 0 \rangle \).

\( \mathbf{k} \) represents the unit vector along the positive \( z \)-axis, which is \( \langle 0, 0, 1 \rangle \).

Together, these unit vectors form the standard basis for three-dimensional space. Any vector in three-dimensional space can be represented as a linear combination of these unit vectors. For example, a vector \( \mathbf{v} \) can be expressed as \( \mathbf{v} = v_x \mathbf{i} + v_y \mathbf{j} + v_z \mathbf{k} \).

Displacement Vector: A vector representing the change in position from one point to another.


In the simulation below. b is the unit vector of a. b always has a length of 1 and is always in the direction of a.


Created with GeoGebra®, by Nicholas Howe, Link


Vector Operations

Addition

Geometrically: The sum of two vectors is the diagonal of the parallelogram formed by the vectors.

Component-wise: Add corresponding components of the vectors.

\[ \mathbf{a} + \mathbf{b} = (a_1 + b_1, a_2 + b_2, a_3 + b_3) \]

Subtraction

Geometrically: The difference of two vectors is the diagonal of the parallelogram formed by the vectors with one reversed.

Component-wise: Subtract corresponding components of the vectors.

\[ \mathbf{a} - \mathbf{b} = (a_1 - b_1, a_2 - b_2, a_3 - b_3) \]

Scalar Multiplication

  • Multiply a vector by a scalar.

    • Geometrically: Scales the magnitude of the vector.

    Component-wise: Multiply each component by the scalar.

    \[ k \mathbf{a} = (k a_1, k a_2, k a_3) \]


    Vector Properties

    Magnitude: The length of the vector, denoted by \( |\mathbf{a}| \).

    Unit Vector: A vector with magnitude 1, obtained by dividing a vector by its magnitude.

    Direction Cosines: Cosines of the angles a vector makes with the coordinate axes.


    The video below explains all theses concepts in depth.


    Dot Product (Scalar Product)

    Definition: The dot product of two vectors \(\mathbf{a}\) and \(\mathbf{b}\) is given by \(\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \theta\), where \(\theta\) is the angle between them.

    Properties: Commutative, distributive, and can be used to find the angle between vectors.


    Watch the video below for an in-depth understanding of the dot product.


    Below is the graphical representation of the dot product.

    Created with GeoGebra®, by Dave Nero, Link


    Point of Intersection between Two Lines

    Given two lines in 3D space, represented by their parametric equations:

    \[ L_1 : \begin{cases} x = x_1 + at_1 \\ y = y_1 + bt_1 \\ z = z_1 + ct_1 \end{cases} \] \[ L_2 : \begin{cases} x = x_2 + at_2 \\ y = y_2 + bt_2 \\ z = z_2 + ct_2 \end{cases} \]

    The point of intersection can be found by setting the equations equal to each other and solving for the parameters $t_1$ and $t_2$.

    Parallel and Skewed Lines

    Parallel Lines: Two lines are parallel if their direction vectors are scalar multiples of each other. If two lines are parallel, they either never intersect or coincide.

    Skewed Lines: Two lines are skewed if they are not parallel and do not intersect. They lie in different planes and are not coplanar.

    Example 1

    Addition of vectors:

    Given two vectors \(\mathbf{a} = \langle 2, -3 \rangle\) and \(\mathbf{b} = \langle -1, 4 \rangle\), find the sum \(\mathbf{a} + \mathbf{b}\).

  • Solution:
    \(\mathbf{a} + \mathbf{b} = \langle 2, -3 \rangle + \langle -1, 4 \rangle = \langle 1, 1 \rangle\)

    • Example 2

    Finding magnitude:

    Find the magnitude of the vector \(\mathbf{v} = \langle 3, -4 \rangle\).

    Solution:

    \(|\mathbf{v}| = \sqrt{3^2 + (-4)^2} = 5\)

    Example 3

    Finding unit vector:

    Given the vector \(\mathbf{u} = \langle 2, -1 \rangle\), find a unit vector in the direction of \(\mathbf{u}\).

    Solution:

    A unit vector in the direction of \(\mathbf{u}\) is \(\mathbf{u_{\text{unit}}} = \frac{\mathbf{u}}{|\mathbf{u}|} = \frac{\langle 2, -1 \rangle}{\sqrt{5}}\)

    Example 4

    Finding the angle between vectors

    Determine the angle between vectors: \(\mathbf{a} = \langle 2, 3 \rangle\) and \(\mathbf{b} = \langle -1, 4 \rangle\).

    Solution:

    \(\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|} = \frac{10}{\sqrt{13} \cdot \sqrt{17}}\), so \(\theta = \cos^{-1}\left(\frac{10}{\sqrt{13} \cdot \sqrt{17}}\right)\)

    Example 5

    Cross Product

    Given vectors \(\mathbf{a} = \langle 1, -2, 3 \rangle\) and \(\mathbf{b} = \langle -2, 1, 0 \rangle\), find the cross product \(\mathbf{a} \times \mathbf{b}\).

    Solution:

    Multiply the corresponding components of the vectors:

    \(\mathbf{a} \times \mathbf{b} = \langle (2 \times 0) - (3 \times 1), (3 \times -2) - (1 \times 0), (1 \times 1) - (-2 \times -2) \rangle\)

    \(\mathbf{a} \times \mathbf{b} = \langle -3, -6, 1 \rangle\)

    Example 6

    Finding Polar coordinates

    Find the polar coordinates of the vector \(\mathbf{v} = \langle 3, -4 \rangle\) in 2D space.

    Solution

    Identify the Components: Given the vector \(\mathbf{v} = \langle 3, -4 \rangle\), the components along the \(x\) and \(y\) axes are \(3\) and \(-4\) respectively.

    Calculate the Magnitude: To find the magnitude \(r\), use the formula: \[ r = |\mathbf{v}| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

    Determine the Angle: The angle $\theta$ can be calculated using the arctan function: \[ \theta = \arctan\left(\frac{-4}{3}\right) \]

    Example 7

    Find the angle between the vectors 𝐯 = ⟨3, -2⟩ and 𝐰 = ⟨-1, 4⟩.

    Solution:

  • Given Vectors: $\mathbf{a} = \langle 2, 1 \rangle$ and $\mathbf{b} = \langle -1, 3 \rangle$.
  • Calculate Dot Product: The dot product of $\mathbf{a}$ and $\mathbf{b}$ is given by: \[ \mathbf{a} \cdot \mathbf{b} = (2)(-1) + (1)(3) = -2 + 3 = 1 \]
  • Find Magnitudes: Calculate the magnitudes of $\mathbf{a}$ and $\mathbf{b}$: \[ |\mathbf{a}| = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5} \] \[ |\mathbf{b}| = \sqrt{(-1)^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10} \]

  • Calculate Angle: Use the formula for the angle between vectors: \[ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|} = \frac{1}{\sqrt{5} \cdot \sqrt{10}} \] \[ \theta = \arccos\left(\frac{1}{\sqrt{5} \cdot \sqrt{10}}\right) \]

    • Example 8

    Point of Intersection between Two Lines

    Given two lines \(L_1\) and \(L_2\) represented parametrically as: \[ L_1 : \begin{cases} x = 1 + t \\ y = 2 - 2t \\ z = 3 + 3t \end{cases} \] \[ L_2 : \begin{cases} x = 2 + 2s \\ y = 3 + s \\ z = 4 - 3s \end{cases} \]

    Find the point of intersection, if any.

    Solution:

    Set the equations of \(L_1\) and \(L_2\) equal to each other: \[ 1 + t = 2 + 2s \] \[ 2 - 2t = 3 + s \] \[ 3 + 3t = 4 - 3s \]

    Now, solve this system of equations:

  • From the first equation, we get \(t = 2s + 1\).
  • Substitute \(t = 2s + 1\) into the second equation:
    • \[ 2 - 2(2s + 1) = 3 + s \] \[ 2 - 4s - 2 = 3 + s \] \[ -4s = 3 + s - 4 \] \[ -4s = s - 1 \] \[ 5s = 1 \] \[ s = \frac{1}{5} \]
  • Substitute \(s = \frac{1}{5}\) into the first equation:
    • \[ t = 2\left(\frac{1}{5}\right) + 1 = \frac{2}{5} + 1 = \frac{7}{5} \]
  • Substitute \(t\) and \(s\) into the third equation to check for consistency:
    • \[ 3 + 3\left(\frac{7}{5}\right) = 4 - 3\left(\frac{1}{5}\right) \] \[ 3 + \frac{21}{5} = 4 - \frac{3}{5} \] \[ \frac{15}{5} + \frac{21}{5} = \frac{20}{5} - \frac{3}{5} \] \[ \frac{36}{5} = \frac{17}{5} \]

    Since the equation \(\frac{36}{5} = \frac{17}{5}\) is not true, the lines do not intersect. They are skewed.

    Exercise &&1&& (&&1&& Question)

    Given two vectors \(\mathbf{a} = \langle 4, -6 \rangle\) and \(\mathbf{b} = \langle -3, -4 \rangle\), find the sum \(\mathbf{a} + \mathbf{b}\).

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    Exercise &&2&& (&&1&& Question)

    Find the magnitude of the vector \(\mathbf{v} = \langle 5, -4 \rangle\).

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    Exercise &&3&& (&&1&& Question)

    Find the dot product of vectors \(\mathbf{a} = \langle 2, -1 \rangle\) and \(\mathbf{b} = \langle -3, 4 \rangle\).

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    Exercise &&4&& (&&1&& Question)

    Find a unit vector in the direction of \(\mathbf{u} = \langle 3, 4 \rangle\).

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    Exercise &&5&& (&&1&& Question)

    A vector from the origin to a point in space is called:

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    Exercise &&6&& (&&1&& Question)

    Multiply the vector \(\mathbf{v} = \langle 2, -3 \rangle\) by the scalar \(k = 4\).

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    Exercise &&7&& (&&1&& Question)

    Given two vectors \( \mathbf{a} \) and \( \mathbf{b} \) with magnitudes \( |\mathbf{a}| = 4 \) and \( |\mathbf{b}| = 2 \) respectively, and an angle \( \theta = 60^\circ \), calculate their dot product.

    7
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