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Vectors and Transformations - Magnitude of a Vector

Grade 10IGCSE

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

🔑Concepts

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Definition: The magnitude of a vector refers to its length or size. It is a scalar quantity, meaning it has size but no direction.

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Notation: The magnitude of a vector v\mathbf{v} or AB⃗\vec{AB} is denoted by ∣v∣|\mathbf{v}| or ∣AB⃗∣|\vec{AB}|.

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Pythagorean Connection: Calculating the magnitude of a vector in 2D is an application of Pythagoras' Theorem (a2+b2=c2a^2 + b^2 = c^2).

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Scalar Property: Magnitude is always a non-negative real number. Even if the components of the vector are negative, the magnitude will be positive because the components are squared.

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Column Vector Form: If a vector is given as (xy)\begin{pmatrix} x \\ y \end{pmatrix}, xx represents the horizontal displacement and yy represents the vertical displacement.

📐Formulae

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

∣AB⃗∣=(x2−x1)2+(y2−y1)2|\vec{AB}| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} for points A(x1,y1)A(x_1, y_1) and B(x2,y2)B(x_2, y_2)

💡Examples

Problem 1:

Find the magnitude of the vector a=(3−4)\mathbf{a} = \begin{pmatrix} 3 \\ -4 \end{pmatrix}.

Solution:

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

Explanation:

Substitute the xx-component (3) and yy-component (-4) into the magnitude formula. Note that (−4)2(-4)^2 becomes positive 16.

Problem 2:

Given points P(1,2)P(1, 2) and Q(7,10)Q(7, 10), find the magnitude of the vector PQ⃗\vec{PQ}.

Solution:

PQ⃗=(7−110−2)=(68)\vec{PQ} = \begin{pmatrix} 7-1 \\ 10-2 \end{pmatrix} = \begin{pmatrix} 6 \\ 8 \end{pmatrix}. ∣PQ⃗∣=62+82=36+64=100=10|\vec{PQ}| = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10

Explanation:

First, find the components of the vector by subtracting the coordinates of the starting point from the end point. Then, apply the magnitude formula to the resulting components.

Problem 3:

The magnitude of the vector u=(k12)\mathbf{u} = \begin{pmatrix} k \\ 12 \end{pmatrix} is 13. Find the possible values of kk.

Solution:

13=k2+122⇒169=k2+144⇒k2=25⇒k=±513 = \sqrt{k^2 + 12^2} \Rightarrow 169 = k^2 + 144 \Rightarrow k^2 = 25 \Rightarrow k = \pm 5

Explanation:

Set up an equation using the magnitude formula. Square both sides to remove the square root, solve for k2k^2, and take the square root of the result. Remember both positive and negative values satisfy k2=25k^2 = 25.