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Vectors and Transformations - Addition and subtraction of vectors

Grade 11IGCSE

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

🔑Concepts

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A vector is a quantity that has both magnitude (size) and direction.

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Column vectors are written in the form (xy)\begin{pmatrix} x \\ y \end{pmatrix}, where xx is the horizontal displacement and yy is the vertical displacement.

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Vector addition follows the 'nose-to-tail' rule: to add vector b\mathbf{b} to a\mathbf{a}, place the start of b\mathbf{b} at the end of a\mathbf{a}.

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The resultant vector is the single vector that points from the start of the first vector to the end of the last vector.

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Subtracting a vector is the same as adding its negative (a vector of the same magnitude but opposite direction).

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Position vectors represent the position of a point relative to the origin OO, usually denoted as OA⃗=a\vec{OA} = \mathbf{a}.

📐Formulae

Addition of column vectors: (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 of column vectors: (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}

Triangle Law of Addition: AB⃗+BC⃗=AC⃗\vec{AB} + \vec{BC} = \vec{AC}

Vector between two points: PQ⃗=OQ⃗−OP⃗=q−p\vec{PQ} = \vec{OQ} - \vec{OP} = \mathbf{q} - \mathbf{p}

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

💡Examples

Problem 1:

Given a=(4−2)\mathbf{a} = \begin{pmatrix} 4 \\ -2 \end{pmatrix} and b=(−15)\mathbf{b} = \begin{pmatrix} -1 \\ 5 \end{pmatrix}, calculate the resultant vector 3a−2b3\mathbf{a} - 2\mathbf{b}.

Solution:

3(4−2)−2(−15)=(12−6)−(−210)=(12−(−2)−6−10)=(14−16)3\begin{pmatrix} 4 \\ -2 \end{pmatrix} - 2\begin{pmatrix} -1 \\ 5 \end{pmatrix} = \begin{pmatrix} 12 \\ -6 \end{pmatrix} - \begin{pmatrix} -2 \\ 10 \end{pmatrix} = \begin{pmatrix} 12 - (-2) \\ -6 - 10 \end{pmatrix} = \begin{pmatrix} 14 \\ -16 \end{pmatrix}

Explanation:

First, multiply each vector by its respective scalar. Then, subtract the corresponding xx and yy components. Remember that subtracting a negative number results in addition.

Problem 2:

In triangle OABOAB, OA⃗=a\vec{OA} = \mathbf{a} and OB⃗=b\vec{OB} = \mathbf{b}. Point MM is the midpoint of ABAB. Find the vector OM⃗\vec{OM} in terms of a\mathbf{a} and b\mathbf{b}.

Solution:

  1. AB⃗=AO⃗+OB⃗=−a+b\vec{AB} = \vec{AO} + \vec{OB} = -\mathbf{a} + \mathbf{b}.
  2. AM⃗=12AB⃗=12(−a+b)\vec{AM} = \frac{1}{2}\vec{AB} = \frac{1}{2}(-\mathbf{a} + \mathbf{b}).
  3. OM⃗=OA⃗+AM⃗=a+12(−a+b)=a−12a+12b=12a+12b\vec{OM} = \vec{OA} + \vec{AM} = \mathbf{a} + \frac{1}{2}(-\mathbf{a} + \mathbf{b}) = \mathbf{a} - \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b} = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}.

Explanation:

To find OM⃗\vec{OM}, we find a path from OO to MM. We first find AB⃗\vec{AB} using the subtraction of position vectors. Since MM is the midpoint, AM⃗\vec{AM} is half of AB⃗\vec{AB}. Finally, we add OA⃗\vec{OA} and AM⃗\vec{AM} and simplify.