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Geometry and Trigonometry - Introduction to Vectors

Grade 11IB

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

🔑Concepts

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A vector is a mathematical quantity that possesses both magnitude (size) and direction. Visually, it is represented as a directed line segment (an arrow). The length of the arrow corresponds to the magnitude, and the arrowhead indicates the direction of movement from the 'tail' (initial point) to the 'head' (terminal point).

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Component Form and Unit Vectors: A vector can be written as a column vector (xy)\begin{pmatrix} x \\ y \end{pmatrix} or in unit vector form xi+yjx\mathbf{i} + y\mathbf{j}, where i\mathbf{i} is a unit vector of length 1 in the x-direction and j\mathbf{j} is a unit vector of length 1 in the y-direction. Visually, xx represents the horizontal displacement and yy represents the vertical displacement on a coordinate plane.

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Magnitude of a Vector: The magnitude represents the length or distance of the vector. Geometrically, if you draw a vector on a grid, the magnitude is the length of the hypotenuse formed by its horizontal and vertical components. It is denoted by ∣v∣|\mathbf{v}| and is always a non-negative scalar value.

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Vector Addition (Triangle and Parallelogram Laws): To add vectors a\mathbf{a} and b\mathbf{b} visually, use the 'tip-to-tail' method: place the tail of b\mathbf{b} at the tip of a\mathbf{a}. The resultant vector a+b\mathbf{a} + \mathbf{b} is the vector drawn from the start of a\mathbf{a} to the end of b\mathbf{b}. Alternatively, the parallelogram law shows the resultant as the diagonal of a parallelogram formed by a\mathbf{a} and b\mathbf{b} sharing the same tail.

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Scalar Multiplication: Multiplying a vector by a scalar kk changes its magnitude but keeps it on the same line. Visually, if k>1k > 1, the vector is stretched; if 0<k<10 < k < 1, it is shrunk; if kk is negative, the vector's direction is reversed. Algebraically, kv=(kxky)k\mathbf{v} = \begin{pmatrix} kx \\ ky \end{pmatrix}.

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Position Vectors: A position vector OP⃗\vec{OP} describes the location of a point PP relative to the origin O(0,0)O(0,0). If a point has coordinates (x,y)(x, y), its position vector is OP⃗=(xy)\vec{OP} = \begin{pmatrix} x \\ y \end{pmatrix}. This allows us to use vector algebra to solve geometric coordinate problems.

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Displacement Vector between Two Points: To find the vector AB⃗\vec{AB} connecting point AA to point BB, we use the subtraction of their position vectors: AB⃗=OB⃗−OA⃗\vec{AB} = \vec{OB} - \vec{OA}. Visually, this is the path required to move from location AA to location BB.

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Parallel Vectors: Two vectors are parallel if one is a scalar multiple of the other, such that a=kb\mathbf{a} = k\mathbf{b}. Visually, parallel vectors point in the exact same or exact opposite direction and will never intersect if placed on the same plane.

📐Formulae

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

Vector between points A(x1,y1)A(x_1, y_1) and 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}

Vector Addition: (a1a2)+(b1b2)=(a1+b1a2+b2)\begin{pmatrix} a_1 \\ a_2 \end{pmatrix} + \begin{pmatrix} b_1 \\ b_2 \end{pmatrix} = \begin{pmatrix} a_1 + b_1 \\ a_2 + b_2 \end{pmatrix}

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

Unit Vector in the direction of v\mathbf{v}: v^=1∣v∣v\hat{\mathbf{v}} = \frac{1}{|\mathbf{v}|} \mathbf{v}

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

💡Examples

Problem 1:

Given points A(2,−1)A(2, -1) and B(−3,4)B(-3, 4), find the vector AB⃗\vec{AB} in column form and calculate its magnitude ∣AB⃗∣|\vec{AB}|.

Solution:

  1. Find the components of AB⃗\vec{AB} using OB⃗−OA⃗\vec{OB} - \vec{OA}: AB⃗=(−3−24−(−1))=(−55)\vec{AB} = \begin{pmatrix} -3 - 2 \\ 4 - (-1) \end{pmatrix} = \begin{pmatrix} -5 \\ 5 \end{pmatrix}
  2. Calculate the magnitude: ∣AB⃗∣=(−5)2+52=25+25=50|\vec{AB}| = \sqrt{(-5)^2 + 5^2} = \sqrt{25 + 25} = \sqrt{50}
  3. Simplify the radical: ∣AB⃗∣=52|\vec{AB}| = 5\sqrt{2}

Explanation:

To find the vector between two points, subtract the coordinates of the starting point from the coordinates of the end point. The magnitude is then found by treating the components as sides of a right triangle.

Problem 2:

If u=(3−2)\mathbf{u} = \begin{pmatrix} 3 \\ -2 \end{pmatrix} and v=(14)\mathbf{v} = \begin{pmatrix} 1 \\ 4 \end{pmatrix}, find the resultant vector w=2u−3v\mathbf{w} = 2\mathbf{u} - 3\mathbf{v} and find its unit vector w^\hat{\mathbf{w}}.

Solution:

  1. Perform scalar multiplication: 2u=2(3−2)=(6−4)2\mathbf{u} = 2\begin{pmatrix} 3 \\ -2 \end{pmatrix} = \begin{pmatrix} 6 \\ -4 \end{pmatrix} 3v=3(14)=(312)3\mathbf{v} = 3\begin{pmatrix} 1 \\ 4 \end{pmatrix} = \begin{pmatrix} 3 \\ 12 \end{pmatrix}
  2. Subtract the vectors: w=(6−3−4−12)=(3−16)\mathbf{w} = \begin{pmatrix} 6 - 3 \\ -4 - 12 \end{pmatrix} = \begin{pmatrix} 3 \\ -16 \end{pmatrix}
  3. Find the magnitude of w\mathbf{w}: ∣w∣=32+(−16)2=9+256=265|\mathbf{w}| = \sqrt{3^2 + (-16)^2} = \sqrt{9 + 256} = \sqrt{265}
  4. Find the unit vector: w^=1265(3−16)\hat{\mathbf{w}} = \frac{1}{\sqrt{265}} \begin{pmatrix} 3 \\ -16 \end{pmatrix}

Explanation:

This example demonstrates vector arithmetic (scaling and subtraction) followed by the normalization process to find a unit vector, which is a vector of length 1 pointing in the same direction.