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Vectors and Transformations - Vector notation and magnitude

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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Vectors can be represented as column vectors (xy)\binom{x}{y}, where xx is the horizontal displacement and yy is the vertical displacement.

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Notation: Vectors are written as bold lowercase letters (a\mathbf{a}), underlined letters (a‾\underline{a}), or as directed line segments between two points (AB⃗\vec{AB}).

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The magnitude of a vector represents its length and is denoted by ∣a∣|\mathbf{a}| or ∣AB⃗∣|\vec{AB}|.

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A zero vector (00)\binom{0}{0} has a magnitude of 0 and no specific direction.

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Scalar multiplication: Multiplying a vector by a constant kk changes its magnitude but keeps the same direction (if k>0k>0) or opposite direction (if k<0k<0).

📐Formulae

Column Vector: v⃗=(xy)\vec{v} = \begin{pmatrix} x \\ y \end{pmatrix}

Magnitude (Modulus): ∣v⃗∣=x2+y2|\vec{v}| = \sqrt{x^2 + y^2}

Vector from point A(x1,y1)A(x_1, y_1) to B(x2,y2)B(x_2, y_2): AB⃗=(x2−x1y2−y1)\vec{AB} = \begin{pmatrix} x_2 - x_1 \\ y_2 - y_1 \end{pmatrix}

Distance between two points (Magnitude of AB⃗\vec{AB}): ∣AB⃗∣=(x2−x1)2+(y2−y1)2|\vec{AB}| = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}

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

💡Examples

Problem 1:

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

Solution:

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

Explanation:

To find the magnitude, we use the Pythagorean formula on the xx and yy components. Note that squaring a negative number results in a positive value.

Problem 2:

Point AA is (1,4)(1, 4) and point BB is (7,12)(7, 12). Find the vector AB⃗\vec{AB} and its magnitude ∣AB⃗∣|\vec{AB}|.

Solution:

AB⃗=(7−112−4)=(68)\vec{AB} = \begin{pmatrix} 7-1 \\ 12-4 \end{pmatrix} = \begin{pmatrix} 6 \\ 8 \end{pmatrix}. Magnitude: ∣AB⃗∣=62+82=36+64=100=10|\vec{AB}| = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10

Explanation:

First, subtract the coordinates of the starting point (AA) from the coordinates of the end point (BB) to find the column vector. Then apply the magnitude formula.

Problem 3:

If v=(3−2)\mathbf{v} = \begin{pmatrix} 3 \\ -2 \end{pmatrix}, find the magnitude of 3v3\mathbf{v}.

Solution:

3v=3(3−2)=(9−6)3\mathbf{v} = 3\begin{pmatrix} 3 \\ -2 \end{pmatrix} = \begin{pmatrix} 9 \\ -6 \end{pmatrix}. Magnitude: ∣3v∣=92+(−6)2=81+36=117≈10.82|3\mathbf{v}| = \sqrt{9^2 + (-6)^2} = \sqrt{81 + 36} = \sqrt{117} \approx 10.82

Explanation:

Multiply each component of the vector by the scalar 3 first, then calculate the magnitude of the resulting vector.