krit.club logo

Vectors and Transformations - Vector Addition, Subtraction, and Scaling

Grade 12IGCSE

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

🔑Concepts

•

Definition of a Vector: A quantity that has both magnitude and direction, often represented by a directed line segment or a column vector.

•

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

•

Vector Addition: Geometrically performed using the 'head-to-tail' rule or the parallelogram law. Algebraically, it involves adding corresponding components.

•

Vector Subtraction: Calculated as a−b=a+(−b)\mathbf{a} - \mathbf{b} = \mathbf{a} + (-\mathbf{b}), representing the vector from the tip of b\mathbf{b} to the tip of a\mathbf{a}.

•

Scalar Multiplication (Scaling): Multiplying a vector by a real number kk changes its magnitude by factor ∣k∣|k|; if k<0k < 0, the direction is reversed.

•

Parallel Vectors: Two vectors are parallel if one is a scalar multiple of the other (i.e., a=kb\mathbf{a} = k\mathbf{b}).

•

Position Vectors: A vector that starts from the origin O(0,0)O(0,0) to a point P(x,y)P(x,y), denoted as OP⃗\vec{OP}.

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

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

Displacement between two points: AB⃗=OB⃗−OA⃗=(xB−xAyB−yA)\vec{AB} = \vec{OB} - \vec{OA} = \begin{pmatrix} x_B - x_A \\ y_B - y_A \end{pmatrix}

💡Examples

Problem 1:

Given vectors a=(3−2)\mathbf{a} = \begin{pmatrix} 3 \\ -2 \end{pmatrix} and b=(−14)\mathbf{b} = \begin{pmatrix} -1 \\ 4 \end{pmatrix}, calculate the resultant vector c=2a+3b\mathbf{c} = 2\mathbf{a} + 3\mathbf{b}.

Solution:

c=2(3−2)+3(−14)=(6−4)+(−312)=(6+(−3)−4+12)=(38)\mathbf{c} = 2\begin{pmatrix} 3 \\ -2 \end{pmatrix} + 3\begin{pmatrix} -1 \\ 4 \end{pmatrix} = \begin{pmatrix} 6 \\ -4 \end{pmatrix} + \begin{pmatrix} -3 \\ 12 \end{pmatrix} = \begin{pmatrix} 6 + (-3) \\ -4 + 12 \end{pmatrix} = \begin{pmatrix} 3 \\ 8 \end{pmatrix}

Explanation:

First, scale each vector by its respective scalar (2 and 3) by multiplying each component. Then, add the resulting xx components and yy components together.

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 OM⃗\vec{OM} in terms of a\mathbf{a} and b\mathbf{b}.

Solution:

AB⃗=OB⃗−OA⃗=b−a\vec{AB} = \vec{OB} - \vec{OA} = \mathbf{b} - \mathbf{a}. Since MM is the midpoint, AM⃗=12AB⃗=12(b−a)\vec{AM} = \frac{1}{2}\vec{AB} = \frac{1}{2}(\mathbf{b} - \mathbf{a}). Therefore, OM⃗=OA⃗+AM⃗=a+12(b−a)=12a+12b\vec{OM} = \vec{OA} + \vec{AM} = \mathbf{a} + \frac{1}{2}(\mathbf{b} - \mathbf{a}) = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}.

Explanation:

To find OM⃗\vec{OM}, we find the displacement vector AB⃗\vec{AB} first. Then we move from the origin to AA, and then halfway along the vector AB⃗\vec{AB} to reach MM.

Problem 3:

Determine if the vectors u=(4−6)\mathbf{u} = \begin{pmatrix} 4 \\ -6 \end{pmatrix} and v=(−69)\mathbf{v} = \begin{pmatrix} -6 \\ 9 \end{pmatrix} are parallel.

Solution:

Check if v=ku\mathbf{v} = k\mathbf{u}. Comparing xx-components: −6=4k  ⟹  k=−1.5-6 = 4k \implies k = -1.5. Comparing yy-components: 9=−6k  ⟹  k=−1.59 = -6k \implies k = -1.5. Since kk is consistent, v=−1.5u\mathbf{v} = -1.5\mathbf{u}.

Explanation:

Vectors are parallel if one can be expressed as a scalar multiple of the other. Since both components share the same ratio (−1.5-1.5), the vectors are parallel and point in opposite directions.