IBDP Maths AHL 3.14 Vector equation of a line in planes AA SL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
To find \(\overrightarrow{AB}\):
\(\overrightarrow{AB} = \overrightarrow{OB} – \overrightarrow{OA} = \begin{pmatrix} 9 \\ m \\ -6 \end{pmatrix} – \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 8 \\ m-1 \\ -8 \end{pmatrix}\)
[2 marks]
Since point B lies on line L, we can substitute its coordinates into the line equation:
\(\begin{pmatrix} 9 \\ m \\ -6 \end{pmatrix} = \begin{pmatrix} -3 \\ -19 \\ 24 \end{pmatrix} + s\begin{pmatrix} 2 \\ 4 \\ -5 \end{pmatrix}\)
This gives us three equations:
1. \(9 = -3 + 2s \Rightarrow s = 6\)
2. \(m = -19 + 4s\)
3. \(-6 = 24 – 5s \Rightarrow s = 6\) (consistent)
Substituting \(s = 6\) into the second equation:
\(m = -19 + 4(6) = -19 + 24 = 5\)
[5 marks]
First, find the value of \(p\) for the unit vector \(\mathbf{u}\):
For a unit vector: \(|\mathbf{u}| = 1\)
\(\sqrt{p^2 + \left(-\frac{2}{3}\right)^2 + \left(\frac{1}{3}\right)^2} = 1\)
\(\sqrt{p^2 + \frac{4}{9} + \frac{1}{9}} = 1\)
\(p^2 + \frac{5}{9} = 1\)
\(p^2 = \frac{4}{9}\)
Since \(p > 0\), \(p = \frac{2}{3}\)
Now, the unit vector is:
\(\mathbf{u} = \frac{2}{3}\mathbf{i} – \frac{2}{3}\mathbf{j} + \frac{1}{3}\mathbf{k} = \begin{pmatrix} \frac{2}{3} \\ -\frac{2}{3} \\ \frac{1}{3} \end{pmatrix}\)
Find \(\overrightarrow{BC}\):
\(\overrightarrow{BC} = 9\mathbf{u} = 9\begin{pmatrix} \frac{2}{3} \\ -\frac{2}{3} \\ \frac{1}{3} \end{pmatrix} = \begin{pmatrix} 6 \\ -6 \\ 3 \end{pmatrix}\)
Find coordinates of C:
\(\overrightarrow{OC} = \overrightarrow{OB} + \overrightarrow{BC} = \begin{pmatrix} 9 \\ 5 \\ -6 \end{pmatrix} + \begin{pmatrix} 6 \\ -6 \\ 3 \end{pmatrix} = \begin{pmatrix} 15 \\ -1 \\ -3 \end{pmatrix}\)
Therefore, the coordinates of C are (15, -1, -3).
[8 marks]
- Vector subtraction to find displacement vectors
- Substituting points into line equations
- Properties of unit vectors
- Vector addition to find point coordinates
▶️ Answer/Explanation
Substitute \((-1, 0, 3)\) into the Cartesian equation:
\(\frac{-1+1}{2} = 0 = 3 – 3\)
\(0 = 0 = 0\) ✓
The point satisfies all three equations, so it lies on \(L_1\).
From the Cartesian equation \(\frac{x+1}{2} = y = 3 – z = \lambda\):
Let \(\frac{x+1}{2} = y = 3 – z = \lambda\)
Then:
\(x = 2\lambda – 1\)
\(y = \lambda\)
\(z = 3 – \lambda\)
Vector equation:
\(\mathbf{r} = \begin{pmatrix} -1 \\ 0 \\ 3 \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}\)
[4 marks]
Using scalar product formula:
\(\begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix} \cdot \begin{pmatrix} a \\ 1 \\ -1 \end{pmatrix} = \sqrt{6} \cdot \sqrt{a^2+2} \cdot \cos 45^\circ\)
\(2a + 2 = \sqrt{6(a^2+2)} \cdot \frac{\sqrt{2}}{2}\)
Simplify:
\(2a + 2 = \sqrt{3(a^2+2)}\)
Square both sides:
\(4a^2 + 8a + 4 = 3a^2 + 6\)
Simplify:
\(a^2 + 8a – 2 = 0\)
Solve using quadratic formula:
\(a = \frac{-8 \pm \sqrt{72}}{2} = -4 \pm 3\sqrt{2}\)
Verified solution: \(a = -4 + 3\sqrt{2}\) or \(a = -4 – 3\sqrt{2}\)
[8 marks]
Equate parametric forms:
\(\begin{cases} 2\lambda – 1 = ta \\ \lambda = 1 + t \\ 3 – \lambda = 2 – t \end{cases}\)
Solve the system:
From second equation: \(t = \lambda – 1\)
Substitute into first equation:
\(2\lambda – 1 = (\lambda – 1)a\)
\(\Rightarrow \lambda(a – 2) = a – 1\)
\(\Rightarrow \lambda = \frac{a-1}{a-2}\)
Unique solution exists when \(a \neq 2\), so \(k = 2\)
Coordinates of A:
\(\left( \frac{a}{a-2}, \frac{a-1}{a-2}, \frac{2a-5}{a-2} \right)\)
Verified solution: \(k = 2\) and \(A\left(\frac{a}{a-2}, \frac{a-1}{a-2}, \frac{2a-5}{a-2}\right)\)
[7 marks]
- Converting between Cartesian and vector equations of lines
- Calculating angles between lines using direction vectors
- Conditions for intersecting and parallel lines
- Solving systems of equations to find intersection points