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Vectors and Transformations - Vector Addition and Subtraction

Grade 9IGCSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A vector represents a quantity with both magnitude (size) and direction.

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Column Vector Notation: A vector is written as (xy)\begin{pmatrix} x \\ y \end{pmatrix}, where xx is the horizontal displacement and yy is the vertical displacement.

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Resultant Vector: The single vector that has the same effect as two or more vectors combined.

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Triangle Law of Addition: To add vectors a\mathbf{a} and b\mathbf{b} geometrically, place the 'tail' of b\mathbf{b} at the 'head' of a\mathbf{a}. The resultant is the vector from the start of a\mathbf{a} to the end of b\mathbf{b}.

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Negative Vectors: The vector −a-\mathbf{a} has the same magnitude as a\mathbf{a} but acts in the opposite direction.

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Vector Subtraction: Subtracting a vector is the same as adding its negative: a−b=a+(−b)\mathbf{a} - \mathbf{b} = \mathbf{a} + (-\mathbf{b}).

📐Formulae

Addition: (x1y1)+(x2y2)=(x1+x2y1+y2)\begin{pmatrix} x_1 \\ y_1 \end{pmatrix} + \begin{pmatrix} x_2 \\ y_2 \end{pmatrix} = \begin{pmatrix} x_1 + x_2 \\ y_1 + y_2 \end{pmatrix}

Subtraction: (x1y1)−(x2y2)=(x1−x2y1−y2)\begin{pmatrix} x_1 \\ y_1 \end{pmatrix} - \begin{pmatrix} x_2 \\ y_2 \end{pmatrix} = \begin{pmatrix} x_1 - x_2 \\ y_1 - y_2 \end{pmatrix}

Scalar Multiplication: k(xy)=(kxky)k \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} kx \\ ky \end{pmatrix}

Magnitude (Length): ∣a∣=x2+y2|\mathbf{a}| = \sqrt{x^2 + y^2}

Displacement between points: AB⃗=OB⃗−OA⃗\vec{AB} = \vec{OB} - \vec{OA} (where OO is the origin)

💡Examples

Problem 1:

Given vectors a=(3−2)\mathbf{a} = \begin{pmatrix} 3 \\ -2 \end{pmatrix} and b=(15)\mathbf{b} = \begin{pmatrix} 1 \\ 5 \end{pmatrix}, calculate 2a+b2\mathbf{a} + \mathbf{b}.

Solution:

2a+b=2(3−2)+(15)=(6−4)+(15)=(71)2\mathbf{a} + \mathbf{b} = 2\begin{pmatrix} 3 \\ -2 \end{pmatrix} + \begin{pmatrix} 1 \\ 5 \end{pmatrix} = \begin{pmatrix} 6 \\ -4 \end{pmatrix} + \begin{pmatrix} 1 \\ 5 \end{pmatrix} = \begin{pmatrix} 7 \\ 1 \end{pmatrix}

Explanation:

First, multiply vector a\mathbf{a} by the scalar 2 by multiplying both components. Then, add the resulting xx-components and yy-components separately.

Problem 2:

In triangle OABOAB, OA⃗=a\vec{OA} = \mathbf{a} and OB⃗=b\vec{OB} = \mathbf{b}. Find the vector AB⃗\vec{AB} in terms of a\mathbf{a} and b\mathbf{b}.

Solution:

AB⃗=AO⃗+OB⃗=−a+b\vec{AB} = \vec{AO} + \vec{OB} = -\mathbf{a} + \mathbf{b} (or b−a\mathbf{b} - \mathbf{a})

Explanation:

To get from AA to BB, you can travel from AA back to the origin OO (which is −a-\mathbf{a}) and then from the origin to BB (which is b\mathbf{b}).

Problem 3:

If c=(−43)\mathbf{c} = \begin{pmatrix} -4 \\ 3 \end{pmatrix}, find the magnitude ∣c∣|\mathbf{c}|.

Solution:

∣c∣=(−4)2+32=16+9=25=5|\mathbf{c}| = \sqrt{(-4)^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5

Explanation:

The magnitude is found using Pythagoras' theorem on the xx and yy components of the column vector.