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
Introduction
A vector is a mathematical entity that has both magnitude and direction. Vectors are fundamental in mathematics and physics for solving problems related to position, displacement, forces, and other directional quantities.
Vectors can be represented geometrically as directed line segments and algebraically using components in a coordinate system.
Definition of a Vector
A vector is represented as a directed line segment, where:
- The length of the segment represents the magnitude of the vector.
- The arrowhead represents the direction of the vector.
Position Vectors and Displacement Vectors
Position Vector: The position vector describes the position of a point relative to an origin.
For example, the position vectors of points \( A \) and \( B \) are:
\( \vec{OA} = \vec{a} \) and \( \vec{OB} = \vec{b} \)
Displacement Vector: The displacement vector \( \vec{AB} \) from point \( A \) to point \( B \) is calculated as:
\(\vec{AB} = \vec{b} – \vec{a}\)
Representation of Vectors Using Directed Line Segments
Vectors are often represented graphically as arrows, where:
- The tail of the arrow is the starting point.
- The head of the arrow is the terminal point.
The length of the arrow corresponds to the magnitude, and the direction is indicated by the arrowhead.
Components of a Vector
A vector in three-dimensional space can be written in component form:
\[ \vec{v} = \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} = v_1 \vec{i} + v_2 \vec{j} + v_3 \vec{k} \]
- \( v_1 \), \( v_2 \), and \( v_3 \) are the components of the vector along the x, y, and z axes.
- \(\vec{i}\), \(\vec{j}\), and \(\vec{k}\) are the unit base vectors along the x, y, and z axes, respectively.
Algebraic and Geometric Operations on Vectors
Sum and Difference of Two Vectors
The sum of two vectors \( \vec{u} \) and \( \vec{v} \) is obtained by adding their components:
\[ \vec{u} + \vec{v} = \begin{bmatrix} u_1 + v_1 \\ u_2 + v_2 \\ u_3 + v_3 \end{bmatrix} \]
Graphically, the sum is represented using the triangle law or the parallelogram law of vector addition.
The difference of two vectors \( \vec{u} \) and \( \vec{v} \) is given by:
\[ \vec{u} – \vec{v} = \begin{bmatrix} u_1 – v_1 \\ u_2 – v_2 \\ u_3 – v_3 \end{bmatrix} \]
Zero Vector and Negative Vector
- Zero Vector (\(\vec{0}\)): A vector with zero magnitude and no specific direction. \(\vec{0} = [0, 0, 0]\).
- Negative Vector (\(-\vec{v}\)): A vector of the same magnitude as \( \vec{v} \) but in the opposite direction.
Multiplication of a Vector by a Scalar
When a vector \( \vec{v} \) is multiplied by a scalar \( k \):
\[ k \vec{v} = \begin{bmatrix} k v_1 \\ k v_2 \\ k v_3 \end{bmatrix} \]
- If \( k > 0 \), the direction remains the same.
- If \( k < 0 \), the direction is reversed.
Vectors that are scalar multiples of each other are called parallel vectors.
Magnitude of a Vector
The magnitude (or length) of a vector \( \vec{v} = [v_1, v_2, v_3] \) is given by:
\[ |\vec{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2} \]
Unit Vectors
A unit vector has a magnitude of 1 and indicates the direction of a vector. It is obtained by dividing the vector by its magnitude:
\[ \vec{u} = \frac{\vec{v}}{|\vec{v}|} \]
Distance Between Two Points
The distance between two points \( A(x_1, y_1, z_1) \) and \( B(x_2, y_2, z_2) \) is the magnitude of the displacement vector \( \vec{AB} \):
\[ |\vec{AB}| = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 + (z_2 – z_1)^2} \]
IB Mathematics AA SL Concept of a vector Exam Style Worked Out Questions
Question
The diagram below shows a circle with centre O. The points A, B, C lie on the circumference of the circle and [AC] is a diameter.
Let \(\overrightarrow {{\text{OA}}} = {\boldsymbol{a}}\) and \(\overrightarrow {{\text{OB}}} = {\boldsymbol{b}}\) .
a. Write down expressions for \(\overrightarrow {{\text{AB}}} \) and \(\overrightarrow {{\text{CB}}} \) in terms of the vectors \({\boldsymbol{a}}\) and \({\boldsymbol{b}}\) .[2]
b. Hence prove that angle \({\text{A}}\hat {\rm{B}}{\text{C}}\) is a right angle. [3]
▶️Answer/Explanation
Markscheme
a.
\(\overrightarrow {{\text{AB}}} = {\boldsymbol{b}} – {\boldsymbol{a}}\) A1
\(\overrightarrow {{\text{CB}}} = {\boldsymbol{a}} + {\boldsymbol{b}}\) A1 [2 marks]
\(\overrightarrow {{\text{AB}}} \cdot \overrightarrow {{\text{CB}}} = \left( {{\boldsymbol{b}} – {\boldsymbol{a}}} \right) \cdot \left( {{\boldsymbol{b}} + {\boldsymbol{a}}} \right)\) M1
\( = {\left| {\mathbf{b}} \right|^2} – {\left| {\mathbf{a}} \right|^2}\) A1
\( = 0\) since \(\left| {\boldsymbol{b}} \right| = \left| {\boldsymbol{a}} \right|\) R1
Note: Only award the A1 and R1 if working indicates that they understand that they are working with vectors.
so \(\overrightarrow {{\text{AB}}} \) is perpendicular to \(\overrightarrow {{\text{CB}}} \) i.e. \({\text{A}}\hat {\rm{B}}{\text{C}}\) is a right angle AG
[3 marks]
Question
The points A(1, 2, 1) , B(−3, 1, 4) , C(5, −1, 2) and D(5, 3, 7) are the vertices of a tetrahedron.
a. Find the vectors \(\overrightarrow {{\text{AB}}} \) and \(\overrightarrow {{\text{AC}}} \).[2]
b. Find the Cartesian equation of the plane \(\prod \) that contains the face ABC.[4]
▶️Answer/Explanation
Markscheme
a.
\(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}}
{ – 4} \\
{ – 1} \\
3
\end{array}} \right)\), \(\overrightarrow {{\text{AC}}} = \left( {\begin{array}{*{20}{c}}
4 \\
{ – 3} \\
1
\end{array}} \right)\) A1A1
Note: Accept row vectors.
[2 marks]
\(\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AC}}} = \left| {\begin{array}{*{20}{c}}
{\boldsymbol{i}}&{\boldsymbol{j}}&{\boldsymbol{k}} \\
{ – 4}&{ – 1}&3 \\
4&{ – 3}&1
\end{array}} \right| = \left( {\begin{array}{*{20}{c}}
8 \\
{16} \\
{16}
\end{array}} \right)\) M1A1
normal \({\boldsymbol{n}} = \left( {\begin{array}{*{20}{c}}
1 \\
2 \\
2
\end{array}} \right)\) so \({\boldsymbol{r}} \cdot \left( {\begin{array}{*{20}{c}}
1 \\
2 \\
2
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
1 \\
2 \\
1
\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}
1 \\
2 \\
2
\end{array}} \right)\) (M1)
\(x + 2y + 2z = 7\) A1
Note: If attempt to solve by a system of equations:
Award A1 for 3 correct equations, A1 for eliminating a variable and A2 for the correct answer. [4 marks]