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Vectors and Transformations - Magnitude of a Vector

Grade 12A Level

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

🔑Concepts

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The magnitude of a vector v=(xy)\mathbf{v} = \begin{pmatrix} x \\ y \end{pmatrix} represents the scalar length of the vector arrow. Geometrically, it is the hypotenuse of a right-angled triangle where the horizontal component (xx) and vertical component (yy) are the legs.

Vector diagram showing x and y components forming a right-angled triangle with the vector as hypotenuse.
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The notation for magnitude uses vertical bars, written as ∣v∣|\mathbf{v}| or ∣AB⃗∣|\vec{AB}|. It is always a non-negative real number, as distances cannot be negative.

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To calculate the magnitude, we apply the Pythagorean theorem: ∣v∣=x2+y2|\mathbf{v}| = \sqrt{x^2 + y^2}. Squaring the components ensures that negative directions (left or down) still contribute positively to the total length.

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A unit vector is a vector with a magnitude of exactly 1. If a vector a\mathbf{a} has magnitude kk, the vector 1ka\frac{1}{k}\mathbf{a} is a unit vector in the same direction.

📐Formulae

∣v∣=x2+y2|\mathbf{v}| = \sqrt{x^2 + y^2} for a 2D vector 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)

∣v∣=x2+y2+z2|\mathbf{v}| = \sqrt{x^2 + y^2 + z^2} for a 3D vector v=xi+yj+zk\mathbf{v} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}

💡Examples

Problem 1:

Calculate the magnitude of the vector u=(5−12)\mathbf{u} = \begin{pmatrix} 5 \\ -12 \end{pmatrix}.

Solution:

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

Explanation:

Identify the xx and yy components (5 and -12). Square both components, sum them, and take the square root. Note that squaring a negative number results in a positive value.

Problem 2:

Find the magnitude of the vector PQ⃗\vec{PQ} where PP is (1,2)(1, 2) and QQ is (4,6)(4, 6).

Solution:

PQ⃗=(4−16−2)=(34)\vec{PQ} = \begin{pmatrix} 4-1 \\ 6-2 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}. ∣PQ⃗∣=32+42=9+16=25=5|\vec{PQ}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5.

Explanation:

First, find the vector PQ⃗\vec{PQ} by subtracting the coordinates of the initial point PP from the terminal point QQ. Then, apply the magnitude formula to the resulting components.

Problem 3:

If vector v=(k8)\mathbf{v} = \begin{pmatrix} k \\ 8 \end{pmatrix} and ∣v∣=10|\mathbf{v}| = 10, find the possible values of kk.

Solution:

10=k2+82  ⟹  100=k2+64  ⟹  k2=36  ⟹  k=±610 = \sqrt{k^2 + 8^2} \implies 100 = k^2 + 64 \implies k^2 = 36 \implies k = \pm 6.

Explanation:

Set up an equation using the magnitude formula. Square both sides to remove the radical, solve for k2k^2, and remember to include both the positive and negative roots.

Problem 4:

Calculate the magnitude of the vector w=(−68)\mathbf{w} = \begin{pmatrix} -6 \\ 8 \end{pmatrix}.

Vector w starting at origin and ending at (-6, 8).

Solution:

∣w∣=(−6)2+82|\mathbf{w}| = \sqrt{(-6)^2 + 8^2} ∣w∣=36+64|\mathbf{w}| = \sqrt{36 + 64} ∣w∣=100|\mathbf{w}| = \sqrt{100} ∣w∣=10|\mathbf{w}| = 10

Explanation:

Substitute the xx-component (-6) and yy-component (8) into the magnitude formula. Square both terms, add them, and find the square root of the sum.

Problem 5:

Given points M(−2,−1)M(-2, -1) and N(2,2)N(2, 2), find the magnitude of the displacement vector MN⃗\vec{MN}.

Vector from M(-2,-1) to N(2,2) showing a horizontal change of 4 and a vertical change of 3.

Solution:

MN⃗=(2−(−2)2−(−1))=(43)\vec{MN} = \begin{pmatrix} 2 - (-2) \\ 2 - (-1) \end{pmatrix} = \begin{pmatrix} 4 \\ 3 \end{pmatrix} ∣MN⃗∣=42+32|\vec{MN}| = \sqrt{4^2 + 3^2} ∣MN⃗∣=16+9|\vec{MN}| = \sqrt{16 + 9} ∣MN⃗∣=25|\vec{MN}| = \sqrt{25} ∣MN⃗∣=5|\vec{MN}| = 5

Explanation:

First, find the column vector MN⃗\vec{MN} by subtracting the coordinates of the starting point MM from the ending point NN. Then apply the magnitude formula to the resulting components.