Vectors

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Warm up:

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Describing Vectors:

Vectors differ from scalars in that unlike scalars, who have magnitude (size) only, vectors have both magnitude and direction.

Both addition and subtraction of vectors is possible, e.g., $p=a+b$.

$$\left(\begin{array}{c}a\\ b\end{array}\right)+\left(\begin{array}{c}p\\ q\end{array}\right)=\left(\begin{array}{c}a+p\\ b+q\end{array}\right)$$

Multiplication is a little more interesting. We can have scalar multiplication, $p=kb$ where $k$ is a scalar.

$$k\left(\begin{array}{c}a\\ b\end{array}\right)=\left(\begin{array}{c}ka\\ kb\end{array}\right)$$

There are two types of multiplication of vectors by vectors.

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Videos and Questions:

Vectors Video

Practice Questions [MS]

Column Vectors Video

Practice Questions

Textbook Exercise

See also:

Translation Vector Video

Practice Questions

Textbook Exercise

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The Length of a Vector

The length of a vector in two dimensions can be found by using the pythagorean theorem

$${\left|a\right|}^2=a_x^2+a_y^2$$

• where
• $\left|a\right|$ is the length/size/magnitude of the vector;
• and
• $a_x$ and $a_y$ are the components of the vector in the respective axis directions.

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Similarly in three dimensions

$${\left|a\right|}^2=a_x^2+a_y^2+a_z^2$$

Here the $a_x$, $a_y$ and $a_z$ are the components of the vector $a$

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Unit Vectors:

The vector $\hat{a}$ has a magnitude of one and is in the same direction as $a$

$$\hat{a}=\frac{a}{\left|a\right|}$$

There are a set of orthogonal unit vectors called $i$, $j$ and $k$.

$i=\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)$ and points in the direction of the x-axis.

$j=\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)$ and points in the direction of the y-axis.

$k=\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)$ and points in the direction of the z-axis.

These are useful for describing vectors in cartesian space.

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Position Vectors

The vectors $r_A$ and $r_B$ are position vectors. They, like all position vectors, start at the origin, $O$ and finish at the point of interest. $r_{\mathrm{AB}}$ is not a position vector as does not start at the origin.

If the vector $a=xi+yj+zk$ makes an angle ${\theta }_x$ with the positive $x$-axis then $cos{\theta }_x=\frac{x}{\left|a\right|}$ and similarly for the angles ${\theta }_y$ and ${\theta }_z$.

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Comparing Vectors

Two vectors can be equal in which case both their magnitudes (sizes or lengths) and directions are the same, i.e., parallel, $\stackrel{\rightharpoonup }{\mathrm{AB}}=\stackrel{\rightharpoonup }{\mathrm{CD}}$.

They may be in the same direction, but have different magnitudes, these two are said to be parallel. In this case one vector is a scalar multiple of the other.

Vectors with different directions may be of the same magnitude.

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Resolving Vectors (from alevelphysics.co.uk)

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The vector equation of a straight line:

$$r=a+\lambda b$$

• where
• $r$ is the position vector of somewhere along the line;
• $a$ is the position vector of a point, $a$, the line passes through;
• $\lambda$ is a scalar parameter;
• and
• $b$ is a direction vector in which the line points.

The parameter is a measure of the multiples of $b$ that the point $r$ is along the line from $A$.

An alternative equation of a straight line passing through the points $C$ and $D$, with position vectors $c$ and $d$ respectively, is

$$r=c+\lambda (d-c)$$

If $p=\left(\begin{array}{c}{p}_1\\ p_2\\ p_3\end{array}\right)$ and $q=\left(\begin{array}{c}{q}_1\\ q_2\\ q_3\end{array}\right)$, the equation of the line with vector equation

$$r=p+\lambda q$$

can be given in Cartesian form as:

$$\frac{x-p_1}{q_1}=\frac{y-p_2}{q_2}=\frac{z-p_3}{q_3}=\lambda$$

More Notes with Questions

Equation of a Line in Vector Form
(Care of www.savemyexams.co.uk)

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A general point in a plane is given by the vector equation

$$r=a+\lambda b+\mu c$$

• where:
• $r$ is the position vector of a point in the plane;
• $a$ is the position vector of another (reference) point in the plane;
• $b$ and $c$ are non-parallel vector in the plane;
• and,
• $\lambda$ and $\mu$ are parameters.

• The distance from the origin to the point $(x,y,z)$ is $\sqrt{x^2+y^2+z^2}$
• The distance between the points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is $\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2+{(z_1-z_2)}^2}$
• Any 3D vector can be written in column form as $pi+qj+rk=\left(\begin{array}{c}p\\ q\\ r\end{array}\right)$

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Matrices

Matrices Questions
(Care of www.physicsandmathstutor.com)

Matrices Questions
(Care of www.revisely.co.uk)

Multiplying Matrices by a scalar

Practice Questions [MS]

Multiplying Matrices a 2×2 by 2×1

Practice Questions [MS]

Multiplying Matrices a 2×2 by 2×2

Practice Questions [MS]

Identity Matrix

Practice Questions [MS]

Transforming the Unit Square

Practice Questions [MS]

Matrix Transformations

Practice Questions [MS]

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