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Vectors and Transformations - Vector Notation and Magnitude

Grade 9IGCSE

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, unlike a scalar which only has magnitude.

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Column Vector Notation: Vectors are typically written as (xy)\begin{pmatrix} x \\ y \end{pmatrix}, where xx represents horizontal movement (right is positive, left is negative) and yy represents vertical movement (up is positive, down is negative).

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Magnitude: The magnitude of a vector refers to its length. It is denoted by ∣a∣|\mathbf{a}| or the absolute value symbols surrounding the vector name.

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Position Vector: A vector that starts from the origin O(0,0)O(0,0) to a point A(x,y)A(x,y), written as OA⃗\vec{OA}.

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Scalar Multiplication: Multiplying a vector by a constant kk (a scalar). This changes the magnitude and can reverse the direction if kk is negative.

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

📐Formulae

Magnitude of vector v=(xy):∣v∣=x2+y2\mathbf{v} = \begin{pmatrix} x \\ y \end{pmatrix}: |\mathbf{v}| = \sqrt{x^2 + y^2}

Displacement vector between A(x1,y1)A(x_1, y_1) and B(x2,y2):AB⃗=(x2−x1y2−y1)B(x_2, y_2): \vec{AB} = \begin{pmatrix} x_2 - x_1 \\ y_2 - y_1 \end{pmatrix}

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

Negative Vector: −(xy)=(−x−y)-\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -x \\ -y \end{pmatrix}

💡Examples

Problem 1:

Given the vector u=(5−12)\mathbf{u} = \begin{pmatrix} 5 \\ -12 \end{pmatrix}, calculate the magnitude ∣u∣|\mathbf{u}|.

Solution:

∣u∣=52+(−12)2=25+144=169=13|\mathbf{u}| = \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13.

Explanation:

To find the magnitude, use the formula derived from Pythagoras' Theorem: square both components, add them together, and take the square root of the result.

Problem 2:

Point AA has coordinates (1,2)(1, 2) and point BB has coordinates (4,6)(4, 6). Find the column vector AB⃗\vec{AB} and its magnitude.

Solution:

AB⃗=(4−16−2)=(34)\vec{AB} = \begin{pmatrix} 4 - 1 \\ 6 - 2 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}. Magnitude ∣AB⃗∣=32+42=9+16=5|\vec{AB}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5.

Explanation:

First, find the displacement vector by subtracting the coordinates of the starting point AA from the end point BB. Then, apply the magnitude formula to find the length of the line segment ABAB.

Problem 3:

If a=(2−3)\mathbf{a} = \begin{pmatrix} 2 \\ -3 \end{pmatrix}, find the vector 3a3\mathbf{a}.

Solution:

3a=3(2−3)=(3×23×−3)=(6−9)3\mathbf{a} = 3 \begin{pmatrix} 2 \\ -3 \end{pmatrix} = \begin{pmatrix} 3 \times 2 \\ 3 \times -3 \end{pmatrix} = \begin{pmatrix} 6 \\ -9 \end{pmatrix}.

Explanation:

Multiply both the xx and yy components of the vector by the scalar constant (3 in this case).