IB Mathematics AA Concept of a vector Study Notes
IB Mathematics AA Concept of a vector Study Notes
IB Mathematics AA Concept of a vector Study Notes Offer a clear explanation of Concept of a vector, including various formula, rules, exam style questions as example to explain the topics. Worked Out examples and common problem types provided here will be sufficient to cover for topic Concept of a vector.
Concept of a Vector
Concept of a Vector
A vector is a quantity that has both magnitude (size) and direction. Vectors are represented graphically by arrows where:
- The length of the arrow corresponds to the magnitude.
- The direction of the arrow indicates the direction of the vector.
Examples of vector quantities: displacement, velocity, acceleration, force.
Position Vectors
A position vector specifies the position of a point relative to the origin of a coordinate system.
- In 2D: \( \vec{r} = \begin{pmatrix} x \\ y \end{pmatrix} \)
- In 3D: \( \vec{r} = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \)
It tells how far and in what direction the point is from the origin.
Displacement Vectors
A displacement vector represents the change in position from one point to another:
\( \vec{d} = \vec{r}_2 – \vec{r}_1 \)
- \( \vec{r}_1 \): position vector of initial point
- \( \vec{r}_2 \): position vector of final point
The displacement vector shows both the distance and direction from the starting point to the endpoint.
Example :
Write the position vector of point P(3, 4).
▶️ Answer/Explanation
\( \overrightarrow{OP} = \begin{pmatrix}3 \\ 4\end{pmatrix} \)
Example :
Find the displacement vector from A(1,2) to B(4,6).
▶️Answer/Explanation
\( \overrightarrow{AB} = \overrightarrow{OB} – \overrightarrow{OA} = \begin{pmatrix}4 – 1 \\ 6 – 2\end{pmatrix} = \begin{pmatrix}3 \\ 4\end{pmatrix} \)
Representation of Vectors Using Directed Line Segments
Representation of Vectors Using Directed Line Segments
A vector can be represented geometrically by a directed line segment. A directed line segment is drawn as an arrow from one point to another.
- The tail (or initial point) of the vector is where the arrow starts.
- The head (or terminal point) of the vector is where the arrow points to.
- The length of the segment represents the magnitude of the vector.
- The arrowhead indicates the direction of the vector.
For example, if a vector starts at point \( A \) and ends at point \( B \), it is denoted as:
\( \overrightarrow{AB} \)
The vector \( \overrightarrow{AB} \) has:
- Magnitude: equal to the distance between \( A \) and \( B \).
- Direction: from \( A \) towards \( B \).
Directed line segments can represent vectors anywhere in space. Vectors with the same magnitude and direction are equal, even if their initial points differ (they are free vectors).
Example :
Find the vector representing the directed line segment from A(1,2) to B(4,7).
▶️ Answer/Explanation
\( \overrightarrow{AB} = \overrightarrow{OB} – \overrightarrow{OA} = (4-1)\mathbf{i} + (7-2)\mathbf{j} \)
\( \overrightarrow{AB} = 3\mathbf{i} + 5\mathbf{j} \)
Conclusion: The directed line segment AB is \( 3\mathbf{i} + 5\mathbf{j} \).
Base & Components of a Vectors
Base Vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \)
In 2D or 3D Cartesian coordinate systems, vectors can be expressed using base vectors:
- \( \mathbf{i} \): unit vector in the x-direction
- \( \mathbf{j} \): unit vector in the y-direction
- \( \mathbf{k} \): unit vector in the z-direction (only for 3D)
Each base vector has a magnitude of 1 and points along its respective axis.
Components of a Vector
Any vector can be written as a combination of its components along the coordinate axes:
- In 2D: \( \mathbf{v} = v_x \mathbf{i} + v_y \mathbf{j} \)
- In 3D: \( \mathbf{v} = v_x \mathbf{i} + v_y \mathbf{j} + v_z \mathbf{k} \)
where:
- \( v_x, v_y, v_z \) are the scalar components of \( \mathbf{v} \) in x, y, z directions respectively.
The magnitude of \( \mathbf{v} \) is given by:
- 2D: \( |\mathbf{v}| = \sqrt{v_x^2 + v_y^2} \)
- 3D: \( |\mathbf{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2} \)
Example :
Express the vector from the origin O(0,0) to point P(5, -3) in terms of i and j.
▶️ Answer/Explanation
The position vector of P is:
\( \overrightarrow{OP} = 5 \mathbf{i} – 3 \mathbf{j} \)
Conclusion: The vector is \( 5\mathbf{i} – 3\mathbf{j} \).
Vector Operations: Algebraic and Geometric Approaches
Sum and Difference of Two Vectors
Algebraic: If \( \mathbf{u} = \begin{pmatrix} u_1 \\ u_2 \end{pmatrix} \), \( \mathbf{v} = \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} \), then
\( \mathbf{u} + \mathbf{v} = \begin{pmatrix} u_1 + v_1 \\ u_2 + v_2 \end{pmatrix} \),
\( \mathbf{u} – \mathbf{v} = \begin{pmatrix} u_1 – v_1 \\ u_2 – v_2 \end{pmatrix} \).
Geometric:
Sum: place the tail of \( \mathbf{v} \) at the head of \( \mathbf{u} \). The resultant vector goes from the tail of \( \mathbf{u} \) to the head of \( \mathbf{v} \).
Difference: \( \mathbf{u} – \mathbf{v} \) is the vector from the head of \( \mathbf{v} \) to the head of \( \mathbf{u} \).
The Zero Vector \( \mathbf{0} \), Negative Vector \( -\mathbf{v} \)
- \( \mathbf{0} \) has zero magnitude and no direction.
- \( -\mathbf{v} \) has the same magnitude as \( \mathbf{v} \) but the opposite direction.
Multiplication by a Scalar
- Algebraically: \( k\mathbf{v} = \begin{pmatrix} kv_1 \\ kv_2 \end{pmatrix} \)
- Geometrically: stretches or compresses \( \mathbf{v} \) by a factor of \( |k| \). If \( k < 0 \), reverses direction.
- Vectors that are scalar multiples of one another are parallel.
Magnitude of a Vector
- In 2D: \( |\mathbf{v}| = \sqrt{v_1^2 + v_2^2} \)
- In 3D: \( |\mathbf{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2} \)
Unit Vector
- The unit vector in the direction of \( \mathbf{v} \) is:
\( \frac{\mathbf{v}}{|\mathbf{v}|} \).
Position and Displacement Vectors
- Position vectors:
\( \overrightarrow{OA} = \mathbf{a} \), \( \overrightarrow{OB} = \mathbf{b} \) - Displacement vector from A to B:
\( \overrightarrow{AB} = \mathbf{b} – \mathbf{a} \)
Example :
Given \( \mathbf{u} = \begin{pmatrix}2 \\ -1\end{pmatrix} \) and \( \mathbf{v} = \begin{pmatrix}3 \\ 4\end{pmatrix} \), find \( \mathbf{u} + \mathbf{v} \) and \( \mathbf{u} – \mathbf{v} \).
▶️ Answer/Explanation
\( \mathbf{u} + \mathbf{v} = \begin{pmatrix}2+3 \\ -1+4\end{pmatrix} = \begin{pmatrix}5 \\ 3\end{pmatrix} \)
\( \mathbf{u} – \mathbf{v} = \begin{pmatrix}2-3 \\ -1-4\end{pmatrix} = \begin{pmatrix}-1 \\ -5\end{pmatrix} \)
Example :
Show that \( \mathbf{p} = \begin{pmatrix}2 \\ 3\end{pmatrix} \) and \( \mathbf{q} = \begin{pmatrix}4 \\ 6\end{pmatrix} \) are parallel.
▶️ Answer/Explanation
\( \mathbf{q} = 2 \mathbf{p} \Rightarrow \) vectors are parallel (scalar multiple)
Example :
Find the magnitude and unit vector of \( \mathbf{r} = \begin{pmatrix}3 \\ 4\end{pmatrix} \).
▶️ Answer/Explanation
\( |\mathbf{r}| = \sqrt{3^2 + 4^2} = 5 \)
Unit vector = \( \frac{1}{5}\begin{pmatrix}3 \\ 4\end{pmatrix} = \begin{pmatrix}0.6 \\ 0.8\end{pmatrix} \)
Proofs of Geometrical Properties Using Vectors
Proofs of Geometrical Properties Using Vectors
Vectors provide a powerful method for proving geometrical properties because they allow precise algebraic manipulation of points, lines, and shapes in a coordinate system.
Typical Properties Proven Using Vectors:
- Collinearity: Three points A, B, C are collinear if \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \) are parallel, i.e., one is a scalar multiple of the other.
- Midpoints and diagonals: The diagonals of a parallelogram bisect each other if their midpoints are equal.
- Right angle: Two vectors are perpendicular if their dot product is zero.
- Parallel lines: Two vectors represent parallel lines if they are scalar multiples of each other.
- Triangle properties: Proofs involving medians, centroids, or equal sides can be done using vector expressions.
General Strategy for Vector Proofs:
- Assign position vectors to key points (e.g., \( \mathbf{a}, \mathbf{b}, \mathbf{c} \)).
- Write down vectors representing sides, medians, or diagonals.
- Use properties such as:
- Parallelism: \( \mathbf{u} = k \mathbf{v} \)
- Perpendicularity: \( \mathbf{u} \cdot \mathbf{v} = 0 \)
- Midpoint formula: \( \frac{\mathbf{a} + \mathbf{b}}{2} \)
- Simplify and interpret the result to prove the desired property.
Example Statements That Can Be Proven Using Vectors:
- The diagonals of a parallelogram bisect each other.
- The medians of a triangle intersect at the centroid.
- The opposite sides of a parallelogram are equal and parallel.
- Three points lie on a straight line.
Example :
Show that the points A(1, 2), B(3, 6) and C(5, 10) are collinear.
▶️ Answer/Explanation
\( \overrightarrow{AB} = \begin{pmatrix}3-1 \\ 6-2\end{pmatrix} = \begin{pmatrix}2 \\ 4\end{pmatrix} \)
\( \overrightarrow{AC} = \begin{pmatrix}5-1 \\ 10-2\end{pmatrix} = \begin{pmatrix}4 \\ 8\end{pmatrix} \)
Notice: \( \overrightarrow{AC} = 2 \overrightarrow{AB} \) → AC is a scalar multiple of AB
Conclusion: A, B, and C are collinear.
Example :
Prove that the diagonals of parallelogram ABCD bisect each other where:
A(0,0), B(4,0), C(6,3), D(2,3).
▶️ Answer/Explanation
Midpoint of AC:
\( M_1 = \frac{1}{2}(A + C) = \frac{1}{2}\begin{pmatrix}0+6 \\ 0+3\end{pmatrix} = \begin{pmatrix}3 \\ 1.5\end{pmatrix} \)
Midpoint of BD:
\( M_2 = \frac{1}{2}(B + D) = \frac{1}{2}\begin{pmatrix}4+2 \\ 0+3\end{pmatrix} = \begin{pmatrix}3 \\ 1.5\end{pmatrix} \)
Conclusion: Diagonals bisect each other.
Example :
Show that vectors \( \mathbf{u} = \begin{pmatrix}3 \\ -2\end{pmatrix} \) and \( \mathbf{v} = \begin{pmatrix}4 \\ 6\end{pmatrix} \) are perpendicular.
▶️ Answer/Explanation
Compute dot product:
\( \mathbf{u} \cdot \mathbf{v} = 3 \times 4 + (-2) \times 6 = 12 – 12 = 0 \)
Conclusion: The vectors are perpendicular.
Example :
Show that \( \mathbf{p} = \begin{pmatrix}2 \\ 3\end{pmatrix} \) and \( \mathbf{q} = \begin{pmatrix}4 \\ 6\end{pmatrix} \) are parallel.
▶️Answer/Explanation
\( \mathbf{q} = 2 \mathbf{p} \)
Conclusion: The vectors are parallel (one is a scalar multiple of the other).