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Vectors and Transformations - Vector Addition, Subtraction, and Scalar Multiplication

Grade 10IGCSE

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

🔑Concepts

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Definition of a Vector: A quantity that has both magnitude (size) and direction.

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Column Vectors: Represented as (xy)\begin{pmatrix} x \\ y \end{pmatrix}, where xx is the horizontal displacement and yy is the vertical displacement.

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Vector Addition: Geometrically, this follows the 'nose-to-tail' rule. Algebraically, it involves adding the corresponding xx and yy components.

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Vector Subtraction: Subtracting a vector is the same as adding its opposite. Geometrically, A⃗−B⃗\vec{A} - \vec{B} is the vector from the tip of BB to the tip of AA.

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Scalar Multiplication: Multiplying a vector by a real number kk changes its magnitude. If kk is negative, it reverses the direction.

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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

Addition: (x1y1)+(x2y2)=(x1+x2y1+y2)\text{Addition: } \begin{pmatrix} x_1 \\ y_1 \end{pmatrix} + \begin{pmatrix} x_2 \\ y_2 \end{pmatrix} = \begin{pmatrix} x_1 + x_2 \\ y_1 + y_2 \end{pmatrix}

Subtraction: (x1y1)−(x2y2)=(x1−x2y1−y2)\text{Subtraction: } \begin{pmatrix} x_1 \\ y_1 \end{pmatrix} - \begin{pmatrix} x_2 \\ y_2 \end{pmatrix} = \begin{pmatrix} x_1 - x_2 \\ y_1 - y_2 \end{pmatrix}

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

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

💡Examples

Problem 1:

Given a=(4−2)\mathbf{a} = \begin{pmatrix} 4 \\ -2 \end{pmatrix} and b=(−15)\mathbf{b} = \begin{pmatrix} -1 \\ 5 \end{pmatrix}, calculate 3a+b3\mathbf{a} + \mathbf{b}.

Solution:

3(4−2)+(−15)=(12−6)+(−15)=(11−1)3 \begin{pmatrix} 4 \\ -2 \end{pmatrix} + \begin{pmatrix} -1 \\ 5 \end{pmatrix} = \begin{pmatrix} 12 \\ -6 \end{pmatrix} + \begin{pmatrix} -1 \\ 5 \end{pmatrix} = \begin{pmatrix} 11 \\ -1 \end{pmatrix}

Explanation:

First, perform scalar multiplication by multiplying each component of vector a\mathbf{a} by 3. Then, add the resulting xx-components (12+(−1)12 + (-1)) and yy-components (−6+5-6 + 5).

Problem 2:

Are the vectors u=(23)\mathbf{u} = \begin{pmatrix} 2 \\ 3 \end{pmatrix} and v=(1015)\mathbf{v} = \begin{pmatrix} 10 \\ 15 \end{pmatrix} parallel?

Solution:

Yes, they are parallel because v=5u\mathbf{v} = 5\mathbf{u}.

Explanation:

To check if vectors are parallel, determine if one is a scalar multiple of the other. Since 10=5×210 = 5 \times 2 and 15=5×315 = 5 \times 3, vector v\mathbf{v} is exactly 5 times vector u\mathbf{u}, confirming they share the same direction.

Problem 3:

Find the magnitude of the vector PQ⃗\vec{PQ} where P=(1,2)P = (1, 2) and Q=(4,6)Q = (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 column vector by subtracting the coordinates of the starting point from the end point. Then, use the Pythagorean theorem formula for magnitude.