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Vectors and Transformations - Vector notation and magnitude

Grade 11A Level

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

🔑Concepts

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A vector represents a displacement with both magnitude (length) and direction. In 2D, it is written as a column vector (xy)\begin{pmatrix} x \\ y \end{pmatrix}, where xx is the horizontal movement and yy is the vertical movement.

A vector AB shown on a coordinate grid representing displacement.
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The magnitude (or modulus) of a vector a=(xy)\mathbf{a} = \begin{pmatrix} x \\ y \end{pmatrix}, denoted by ∣a∣|\mathbf{a}|, is the length of the vector. It is calculated using Pythagoras' Theorem: ∣a∣=x2+y2|\mathbf{a}| = \sqrt{x^2 + y^2}.

Right-angled triangle showing the x and y components of a vector and its magnitude as the hypotenuse.
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Vectors are equal if they have the same magnitude and the same direction, regardless of their starting positions. Parallel vectors are scalar multiples of each other, such as a\mathbf{a} and 2a2\mathbf{a}.

Two parallel lines of equal length representing equal vectors.
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A negative vector, denoted as −a-\mathbf{a}, has the same magnitude as a\mathbf{a} but acts in the exact opposite direction. If AB⃗=a\vec{AB} = \mathbf{a}, then BA⃗=−a\vec{BA} = -\mathbf{a}.

📐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.

Problem 4:

Given vector u=(−86)\mathbf{u} = \begin{pmatrix} -8 \\ 6 \end{pmatrix}, calculate the magnitude ∣u∣|\mathbf{u}|.

Vector u plotted from origin to (-8, 6).

Solution:

  1. Identify the components: x=−8x = -8 and y=6y = 6.
  2. Use the magnitude formula: ∣u∣=x2+y2|\mathbf{u}| = \sqrt{x^2 + y^2}.
  3. Substitute the values: ∣u∣=(−8)2+62|\mathbf{u}| = \sqrt{(-8)^2 + 6^2}.
  4. Calculate the squares: ∣u∣=64+36=100|\mathbf{u}| = \sqrt{64 + 36} = \sqrt{100}.
  5. ∣u∣=10|\mathbf{u}| = 10.

Explanation:

The magnitude represents the straight-line distance from the start to the end of the vector. Since squaring a negative number results in a positive value, the direction (left/right) does not affect the length.

Problem 5:

Point CC is at (−2,−3)(-2, -3) and point DD is at (1,1)(1, 1). Find the column vector CD⃗\vec{CD} and its magnitude ∣CD⃗∣|\vec{CD}|.

Vector CD drawn between points (-2, -3) and (1, 1).

Solution:

  1. Find CD⃗\vec{CD} by subtracting coordinates of CC from DD: CD⃗=(1−(−2)1−(−3))=(34)\vec{CD} = \begin{pmatrix} 1 - (-2) \\ 1 - (-3) \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}.
  2. Calculate the magnitude: ∣CD⃗∣=32+42|\vec{CD}| = \sqrt{3^2 + 4^2}.
  3. ∣CD⃗∣=9+16=25|\vec{CD}| = \sqrt{9 + 16} = \sqrt{25}.
  4. ∣CD⃗∣=5|\vec{CD}| = 5.

Explanation:

To find the vector between two points, calculate the change in xx and change in yy. The magnitude is then found using the distance formula between these coordinates.

Vector notation and magnitude Grade 11 Notes & Examples