IBDP Maths SL 3.1 The distance between two points AA HL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
Detailed Solution:
(a)(i) Midpoint of AC:
Given points A(6,0) and C(3,3√3), the midpoint M is: \[ M = \left( \frac{6+3}{2}, \frac{0+3\sqrt{3}}{2} \right) = \left( \frac{9}{2}, \frac{3\sqrt{3}}{2} \right) \]
(a)(ii) Equation of line OB:
Since OABC is symmetrical about OB, point M lies on OB. Using points O(0,0) and M: \[ \text{Slope } m = \frac{\frac{3\sqrt{3}}{2}-0}{\frac{9}{2}-0} = \frac{\sqrt{3}}{3} \] Thus, the equation is: \[ y = \frac{\sqrt{3}}{3}x \]
(b) Area of quadrilateral OABC:
1. Find point B by substituting x=6 into the OB equation: \[ y = \frac{\sqrt{3}}{3} \times 6 = 2\sqrt{3} \] So B(6,2√3)
2. Since OA is perpendicular to AB: \[ \text{Slope of OA} = 0 \] \[ \text{Slope of AB} = \frac{2\sqrt{3}-0}{6-6} \text{ (undefined)} \] This confirms the perpendicularity.
3. Area calculation: \[ \text{Area of } \triangle OAB = \frac{1}{2} \times OA \times AB = \frac{1}{2} \times 6 \times 2\sqrt{3} = 6\sqrt{3} \] Due to symmetry, total area = \( 2 \times 6\sqrt{3} = 12\sqrt{3} \)
Markscheme:
(a)(i)
• Correct midpoint \(\left( \frac{9}{2}, \frac{3\sqrt{3}}{2} \right)\) A1
(a)(ii)
• Correct slope calculation M1
• Correct equation \( y = \frac{\sqrt{3}}{3}x \) A1
(b)
• Correct coordinates for B(6,2√3) M1
• Correct area of triangle OAB (6√3) A1
• Correct total area (12√3) A1
Total: [6 marks]
Geometric Properties:
1. The quadrilateral is kite-shaped due to symmetry about OB
2. The angle between OA and AB is 90°
3. The line OB bisects angle AOC
▶️ Answer/Explanation
Detailed Solution:
(a) Showing planes do not intersect:
METHOD 1:
Attempt to eliminate variables from \( \Pi_1 \) and \( \Pi_2 \):
We get either:
\( -3x + z = -3 \) and \( -3x + z = 44 \)
or \( -5x + y = -7 \) and \( -5x + y = 40 \)
These pairs are inconsistent (e.g., \( -3 \neq 44 \)), so the planes do not intersect.
METHOD 2:
Find line of intersection \( L \) of \( \Pi_1 \) and \( \Pi_2 \):
Direction vector: \( \begin{pmatrix}1\\5\\3\end{pmatrix} \)
Parametric equations: \( (1 + \lambda, -2 + 5\lambda, 3\lambda) \)
Substitute into \( \Pi_3 \):
\( -9(1 + \lambda) + 3(-2 + 5\lambda) – 2(3\lambda) = 32 \Rightarrow -15 = 32 \) (contradiction)
Thus, no common intersection point.
(b)
(i) Verification for point \( P(1, -2, 0) \):
\( \Pi_1 \): \( 2(1) – (-2) + 0 = 4 \) ✔️
\( \Pi_2 \): \( 1 – 2(-2) + 3(0) = 5 \) ✔️
(ii) Line of intersection \( L \):
Direction vector from cross product of normals:
\( \begin{pmatrix}2\\-1\\1\end{pmatrix} \times \begin{pmatrix}1\\-2\\3\end{pmatrix} = \begin{pmatrix}-1\\-5\\-3\end{pmatrix} \)
Vector equation:
\( \mathbf{r} = \begin{pmatrix}1\\-2\\0\end{pmatrix} + \lambda\begin{pmatrix}1\\5\\3\end{pmatrix} \)
(c) Distance between \( L \) and \( \Pi_3 \):
METHOD 1:
1. Find point where normal from \( P \) meets \( \Pi_3 \):
Line equation: \( \mathbf{r} = \begin{pmatrix}1\\-2\\0\end{pmatrix} + t\begin{pmatrix}-9\\3\\-2\end{pmatrix} \)
Substitute into \( \Pi_3 \): \( -9(1-9t) + 3(-2+3t) – 2(-2t) = 32 \)
Solve: \( t = \frac{1}{2} \Rightarrow \) Point \( Q\left(-\frac{7}{2}, -\frac{1}{2}, -1\right) \)
2. Distance \( PQ \):
\( \frac{1}{2}\sqrt{(-9)^2 + 3^2 + (-2)^2} = \frac{\sqrt{94}}{2} \)
METHOD 2:
Using plane parallel to \( \Pi_3 \) through \( P \):
Normal vector equation of \( \Pi_4 \):
\( \begin{pmatrix}-9\\3\\-2\end{pmatrix} \cdot \begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}-9\\3\\-2\end{pmatrix} \cdot \begin{pmatrix}1\\-2\\0\end{pmatrix} = -15 \)
Distance between planes:
\( \frac{|32 – (-15)|}{\sqrt{(-9)^2 + 3^2 + (-2)^2}} = \frac{47}{\sqrt{94}} = \frac{\sqrt{94}}{2} \)
