IB Mathematics AA Vector equation of a line in planes Study Notes
IB Mathematics AA Vector equation of a line in planes Study Notes
IB Mathematics AA Vector equation of a line in planes Notes Offer a clear explanation of Vector equation of a line in planes, 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 Vector equation of a line in planes.
Scalar Product of Two Vectors
Introduction
The scalar product, also known as the dot product, is a mathematical operation that takes two vectors and returns a scalar quantity. This operation is widely used in geometry, physics, and engineering for calculating angles, projections, and more.
Definition of the Scalar Product
The scalar product of two vectors \( \vec{a} \) and \( \vec{b} \) is defined as:
\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \]
Here:
- \( |\vec{a}| \) and \( |\vec{b}| \) are the magnitudes of the vectors \( \vec{a} \) and \( \vec{b} \).
- \( \theta \) is the angle between the two vectors.
The dot (\( \cdot \)) symbol between \( \vec{a} \) and \( \vec{b} \) represents the scalar product.
Component Form of the Scalar Product
Say you have two vectors:
\[ \vec{v} = (x_1, y_1, z_1) \quad \text{and} \quad \vec{w} = (x_2, y_2, z_2) \]
The scalar product is calculated as:
\[ \vec{v} \cdot \vec{w} = x_1 x_2 + y_1 y_2 + z_1 z_2 \]
For example, if:
- \( \vec{v} = (5, 3, 1) \)
- \( \vec{w} = (6, 3, 1) \)
The scalar product would be:
\[ \vec{v} \cdot \vec{w} = (5 \cdot 6) + (3 \cdot 3) + (1 \cdot 1) = 30 + 9 + 1 = 40 \]
Key Properties of the Scalar Product
The scalar product has the following properties:
- \( \vec{v} \cdot \vec{w} = \vec{w} \cdot \vec{v} \) (commutative property).
- \( \vec{u} \cdot (\vec{v} + \vec{w}) = \vec{u} \cdot \vec{v} + \vec{u} \cdot \vec{w} \) (distributive property).
- \( (k\vec{v}) \cdot \vec{w} = k (\vec{v} \cdot \vec{w}) \), where \( k \) is a scalar.
- \( \vec{v} \cdot \vec{v} = |\vec{v}|^2 \).
Geometrical Interpretation
The scalar product can be used to determine the relationship between two vectors:
- If \( \vec{v} \cdot \vec{w} = 0 \), the vectors are perpendicular.
- If \( \vec{v} \cdot \vec{w} > 0 \), the angle between the vectors is acute.
- If \( \vec{v} \cdot \vec{w} < 0 \), the angle between the vectors is obtuse.
Angle Between Two Vectors
The scalar product can be rearranged to find the angle \( \theta \) between two vectors \( \vec{v} \) and \( \vec{w} \):
\[ \cos \theta = \frac{\vec{v} \cdot \vec{w}}{|\vec{v}| |\vec{w}|} \]
Here, \( \theta \) lies between \( 0^\circ \) and \( 180^\circ \).
Applications of the Scalar Product
The scalar product has various practical applications:
- Finding the angle: The angle between two vectors can be calculated using the formula for \( \cos \theta \).
- Projections: The projection of a vector \( \vec{v} \) onto another vector \( \vec{w} \) is given by: \[ \text{Projection of } \vec{v} \text{ onto } \vec{w} = \frac{\vec{v} \cdot \vec{w}}{|\vec{w}|}. \]
- Geometry: Proving properties such as perpendicularity and parallelism in geometrical shapes.
- Physics: Calculating work done by a force, where work \( W = \vec{F} \cdot \vec{d} \).
Examples
Example 1: Magnitude of a Vector
Using the scalar product, the magnitude of a vector \( \vec{v} \) is:
\[ |\vec{v}| = \sqrt{\vec{v} \cdot \vec{v}} \]
Example 2: Finding the Angle Between Vectors
Given \( \vec{v} = [3, 4] \) and \( \vec{w} = [4, -3] \):
\[ \vec{v} \cdot \vec{w} = 3(4) + 4(-3) = 12 – 12 = 0 \]
Since \( \vec{v} \cdot \vec{w} = 0 \), the vectors are perpendicular.
Example 3: Work Done
If a force \( \vec{F} = [10, 0] \) moves an object along a displacement \( \vec{d} = [5, 0] \), the work done is:
\[ W = \vec{F} \cdot \vec{d} = 10(5) + 0(0) = 50 \]
Connections to Other Topics
- Unit Circle: The scalar product relates to the trigonometric functions of the angle between two vectors.
- Odd and Even Functions: The cosine function used in the scalar product is an even function.
- Physics: Concepts of force, work, and energy involve the scalar product.
IB Mathematics AA SL Vector equation of a line in planes Exam Style Worked Out Questions
Question
Consider the vectors a = 6i + 3j + 2k, b = −3j + 4k.
a.(i) Find the cosine of the angle between vectors a and b.
(ii) Find a \( \times \) b.
(iii) Hence find the Cartesian equation of the plane \(\prod \) containing the vectors a and b and passing through the point (1, 1, −1).
(iv) The plane \(\prod \) intersects the x-y plane in the line l. Find the area of the finite triangular region enclosed by l, the x-axis and the y-axis. [11]
(i) show that p\( \cdot \)p = \(|\)p\({|^2}\);
(ii) hence, or otherwise, show that \(|\)p + q\({|^2}\) = \(|\)p\({|^2}\) + 2p\( \cdot \)q + \(|\)q\({|^2}\);
(iii) deduce that \(|\)p + q\(|\) ≤ \(|\)p\(|\) + \(|\)q\(|\). [8]
▶️Answer/Explanation
Markscheme
a.(i) use of a\( \cdot \)b = \(|\)a\(|\)\(|\)b\(|\cos \theta \) (M1)
a\( \cdot \)b = –1 (A1)
\(|\)a\(|\) = 7, \(|\)b\(|\) = 5 (A1)
\(\cos \theta = – \frac{1}{{35}}\) A1
(ii) the required cross product is
\(\left| {\begin{array}{*{20}{c}}
i&j&k \\
6&3&2 \\
0&{ – 3}&4
\end{array}} \right| = \) 18i – 24j -18k M1A1
(iii) using r\( \cdot \)n = p\( \cdot \)n the equation of the plane is (M1)
\(18x – 24y – 18z = 12\,\,\,\,\,(3x – 4y – 3z = 2)\) A1
(iv) recognizing that z = 0 (M1)
x-intercept \( = \frac{2}{3}\), y-intercept \( = – \frac{1}{2}\) (A1)
area \( = \left( {\frac{2}{3}} \right)\left( {\frac{1}{2}} \right)\left( {\frac{1}{2}} \right) = \frac{1}{6}\) A1
[11 marks]
(i) p\( \cdot \)p = \(|\)p\(|\)\(|\)p\(|\cos 0\) M1A1
= \(|\)p\({|^2}\) AG
(ii) consider the LHS, and use of result from part (i)
\(|\)p + q\({|^2}\) = (p + q)\( \cdot \)(p + q) M1
= p\( \cdot \)p + p\( \cdot \)q + q\( \cdot \)p + q\( \cdot \)q (A1)
= p\( \cdot \)p + 2p\( \cdot \)q + q\( \cdot \)q A1
= \(|\)p\({|^2}\) + 2p\( \cdot \)q + \(|\)q\({|^2}\) AG
(iii) EITHER
use of p\( \cdot \)q \( \leqslant \) \(|\)p\(|\)\(|\)q\(|\) M1
so 0 \( \leqslant \) \(|\)p + q\({|^2}\) = \(|\)p\({|^2}\) + 2p\( \cdot \)q + \(|\)q\({|^2}\) \( \leqslant \) \(|\)p\({|^2}\) + 2 \(|\)p\(|\)\(|\)q\(|\) + \(|\)q\({|^2}\) A1
take square root (of these positive quantities) to establish A1
\(|\)p + q\(|\) \( \leqslant \) \(|\)p\(|\) + \(|\)q\(|\) AG
OR
M1M1
Note: Award M1 for correct diagram and M1 for correct labelling of vectors including arrows.
since the sum of any two sides of a triangle is greater than the third side,
\(|\)p\(|\) + \(|\)q\(|\) > \(|\)p + q\(|\) A1
when p and q are collinear \(|\)p\(|\) + \(|\)q\(|\) = \(|\)p + q\(|\)
\( \Rightarrow |\)p + q\(|\) \( \leqslant \) \(|\)p\(|\) + \(|\)q\(|\) AG
[8 marks]
Question
a.For non-zero vectors \({\boldsymbol{a}}\) and \({\boldsymbol{b}}\), show that
(i) if \(\left| {{\boldsymbol{a}} – {\boldsymbol{b}}} \right| = \left| {{\boldsymbol{a}} + {\boldsymbol{b}}} \right|\), then \({\boldsymbol{a}}\) and \({\boldsymbol{b}}\) are perpendicular;
(ii) \({\left| {{\boldsymbol{a}} \times {\boldsymbol{b}}} \right|^2} = {\left| {\boldsymbol{a}} \right|^2}{\left| {\boldsymbol{b}} \right|^2} – {({\boldsymbol{a}} \cdot {\boldsymbol{b}})^2}\). [8]
The points A, B and C have position vectors \({\boldsymbol{a}}\), \({\boldsymbol{b}}\) and \({\boldsymbol{c}}\).
(i) Show that the area of triangle ABC is \(\frac{1}{2}\left| {{\boldsymbol{a}} \times {\boldsymbol{b}} + {\boldsymbol{b}} \times {\boldsymbol{c}} + {\boldsymbol{c}} \times {\boldsymbol{a}}} \right|\).
(ii) Hence, show that the shortest distance from B to AC is
\[\frac{{\left| {{\boldsymbol{a}} \times {\boldsymbol{b}} + {\boldsymbol{b}} \times {\boldsymbol{c}} + {\boldsymbol{c}} \times {\boldsymbol{a}}} \right|}}{{\left| {{\boldsymbol{c}} – {\boldsymbol{a}}} \right|}}{\text{.}}\] [7]
▶️Answer/Explanation
Markscheme
a.(i) \(\left| {{\boldsymbol{a}} – {\boldsymbol{b}}} \right| = \left| {{\boldsymbol{a}} + {\boldsymbol{b}}} \right|\)
\( \Rightarrow \left( {{\boldsymbol{a}} – {\boldsymbol{b}}} \right) \cdot \left( {{\boldsymbol{a}} – {\boldsymbol{b}}} \right) = \left( {{\boldsymbol{a}} + {\boldsymbol{b}}} \right) \cdot \left( {{\boldsymbol{a}} + {\boldsymbol{b}}} \right)\) (M1)
\( \Rightarrow {\left| {\boldsymbol{a}} \right|^2} – 2{\boldsymbol{a}} \cdot {\boldsymbol{b}} + {\left| {\boldsymbol{b}} \right|^2} = {\left| {\boldsymbol{a}} \right|^2} + 2{\boldsymbol{a}} \cdot {\boldsymbol{b}} + {\left| {\boldsymbol{b}} \right|^2}\) A1
\( \Rightarrow 4{\boldsymbol{a}} \cdot {\boldsymbol{b}} = 0 \Rightarrow {\boldsymbol{a}} \cdot {\boldsymbol{b}} = 0\) A1
therefore \({\boldsymbol{a}}\) and \({\boldsymbol{b}}\) are perpendicular R1
Note: Allow use of 2-d components.
Note: Do not condone sloppy vector notation, so we must see something to the effect that \({\left| {\boldsymbol{c}} \right|^2} = {\boldsymbol{c}} \cdot {\boldsymbol{c}}\) is clearly being used for the M1.
Note: Allow a correct geometric argument, for example that the diagonals of a parallelogram have the same length only if it is a rectangle.
(ii) \({\left| {{\boldsymbol{a}} \times {\boldsymbol{b}}} \right|^2} = {\left( {\left| {\boldsymbol{a}} \right|\left| {\boldsymbol{b}} \right|\sin \theta } \right)^2} = {\left| {\boldsymbol{a}} \right|^2}{\left| {\boldsymbol{b}} \right|^2}{\sin ^2}\theta \) M1A1
\({\left| {\boldsymbol{a}} \right|^2}{\left| {\boldsymbol{b}} \right|^2} – {\left( {{\boldsymbol{a}} \cdot {\boldsymbol{b}}} \right)^2} = {\left| {\boldsymbol{a}} \right|^2}{\left| {\boldsymbol{b}} \right|^2} – {\left| {\boldsymbol{a}} \right|^2}{\left| {\boldsymbol{b}} \right|^2}{\cos ^2}\theta \) M1
\( = {\left| {\boldsymbol{a}} \right|^2}{\left| {\boldsymbol{b}} \right|^2}\left( {1 – {{\cos }^2}\theta } \right)\) A1
\( = {\left| {\boldsymbol{a}} \right|^2}{\left| {\boldsymbol{b}} \right|^2}\left( {{{\sin }^2}\theta } \right)\)
\( \Rightarrow {\left| {{\boldsymbol{a}} \times {\boldsymbol{b}}} \right|^2} = {\left| {\boldsymbol{a}} \right|^2}{\left| {\boldsymbol{b}} \right|^2} – {\left( {{\boldsymbol{a}} \cdot {\boldsymbol{b}}} \right)^2}\) AG
[8 marks]
(i) area of triangle \( = \frac{1}{2}\left| {\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AC}}} } \right|\) (M1)
\( = \frac{1}{2}\left| {\left( {{\boldsymbol{b}} – {\boldsymbol{a}}} \right) \times \left( {{\boldsymbol{c}} – {\boldsymbol{a}}} \right)} \right|\) A1
\( = \frac{1}{2}\left| {{\boldsymbol{b}} \times {\boldsymbol{c}} + {\boldsymbol{b}} \times {\boldsymbol{ – a}} + {\boldsymbol{ – a}} \times {\boldsymbol{c}} + {\boldsymbol{ – a}} \times {\boldsymbol{ – a}}} \right|\) A1
\({\boldsymbol{b}} \times {\boldsymbol{ – a}} = {\boldsymbol{a}} \times {\boldsymbol{b}}\); \({\boldsymbol{c}} \times {\boldsymbol{a}} = {\boldsymbol{ – a}} \times {\boldsymbol{c}}\); \({\boldsymbol{ – a}} \times {\boldsymbol{ – a}} = 0\) M1
hence, area of triangle is \(\frac{1}{2}\left| {{\boldsymbol{a}} \times {\boldsymbol{b}} + {\boldsymbol{b}} \times {\boldsymbol{c}} + {\boldsymbol{c}} \times {\boldsymbol{a}}} \right|\) AG
(ii) D is the foot of the perpendicular from B to AC
area of triangle \({\text{ABC}} = \frac{1}{2}\left| {\overrightarrow {{\text{AC}}} } \right|\left| {\overrightarrow {{\text{BD}}} } \right|\) A1
therefore
\(\frac{1}{2}\left| {\overrightarrow {{\text{AC}}} } \right|\left| {\overrightarrow {{\text{BD}}} } \right| = \frac{1}{2}\left| {\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AC}}} } \right|\) M1
hence, \(\left| {\overrightarrow {{\text{BD}}} } \right| = \frac{{\left| {\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AC}}} } \right|}}{{\left| {\overrightarrow {{\text{AC}}} } \right|}}\) A1
\( = \frac{{\left| {{\boldsymbol{a}} \times {\boldsymbol{b}} + {\boldsymbol{b}} \times {\boldsymbol{c}} + {\boldsymbol{c}} \times {\boldsymbol{a}}} \right|}}{{\left| {{\boldsymbol{c}} – {\boldsymbol{a}}} \right|}}\) AG
[7 marks]