IB Mathematics AHL 3.13 Scalar product of two vectors AA HL Paper 3 - Exam Style Questions
The coordinates of points \( A \), \( B \), and \( C \) are given as \( (5, -2, 5) \), \( (5, 4, -1) \), and \( (-1, -2, -1) \) respectively.
(a) Show that \( \text{AB} = \text{AC} \) and that \( \angle \text{BAC} = 60^\circ \). [4]
(b) Find the Cartesian equation of \( \Pi \), the plane passing through \( A \), \( B \), and \( C \). [4]
(c) (i) Find the Cartesian equation of \( \Pi_1 \), the plane perpendicular to \( \overrightarrow{\text{AB}} \) passing through the midpoint of \( [AB] \). [2]
(ii) Find the Cartesian equation of \( \Pi_2 \), the plane perpendicular to \( \overrightarrow{\text{AC}} \) passing through the midpoint of \( [AC] \). [2]
(d) Find the vector equation of \( L \), the line of intersection of \( \Pi_1 \) and \( \Pi_2 \), and show that it is perpendicular to \( \Pi \). [3]
(e) A methane molecule consists of a carbon atom with four hydrogen atoms symmetrically placed around it in three dimensions.
The positions of the centres of three of the hydrogen atoms are \( A \), \( B \), and \( C \) as given. The position of the centre of the fourth hydrogen atom is \( D \). Using the fact that \( \text{AB} = \text{AD} \), show that the coordinates of one of the possible positions of the fourth hydrogen atom is \( (-1, 4, 5) \). [3]
(f) Letting \( D \) be \( (-1, 4, 5) \), show that the coordinates of \( G \), the position of the centre of the carbon atom, are \( (2, 1, 2) \). Hence calculate \( \angle \text{DGA} \), the bonding angle of carbon. [6]
▶️ Answer/Explanation
(a) Show \( \text{AB} = \text{AC} \) and \( \angle \text{BAC} = 60^\circ \):
\[ \overrightarrow{\text{AB}} = (5, 4, -1) – (5, -2, 5) = \begin{pmatrix} 0 \\ 6 \\ -6 \end{pmatrix} \Rightarrow \text{AB} = \sqrt{0^2 + 6^2 + (-6)^2} = \sqrt{72} \quad (A1) \]
\[ \overrightarrow{\text{AC}} = (-1, -2, -1) – (5, -2, 5) = \begin{pmatrix} -6 \\ 0 \\ -6 \end{pmatrix} \Rightarrow \text{AC} = \sqrt{(-6)^2 + 0^2 + (-6)^2} = \sqrt{72} \quad (A1) \]
So \( \text{AB} = \text{AC} \quad (AG) \)
\[ \overrightarrow{\text{AB}} \cdot \overrightarrow{\text{AC}} = (0)(-6) + (6)(0) + (-6)(-6) = 36 \quad (M1) \]
\[ \cos \theta = \frac{\overrightarrow{\text{AB}} \cdot \overrightarrow{\text{AC}}}{|\overrightarrow{\text{AB}}| |\overrightarrow{\text{AC}}|} = \frac{36}{\sqrt{72} \cdot \sqrt{72}} = \frac{36}{72} = \frac{1}{2} \Rightarrow \theta = 60^\circ \quad (A1)(AG) \]
Note: Award (M1)(A1) if candidates find \( \text{BC} = \sqrt{72} \) and claim triangle \( ABC \) is equilateral.
[4 marks]
(b) Cartesian equation of plane \( \Pi \):
Method 1:
\[ \overrightarrow{\text{AB}} \times \overrightarrow{\text{AC}} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 6 & -6 \\ -6 & 0 & -6 \end{vmatrix} = -36 \mathbf{i} + 36 \mathbf{j} + 36 \mathbf{k} \quad (M1)(A1) \]
Normal vector: \( (-36, 36, 36) \), so plane equation: \[ -36x + 36y + 36z = k \quad (M1) \]
Simplify: \[ x – y – z = k \]. Using point \( A(5, -2, 5) \): \( 5 – (-2) – 5 = 2 \), so \[ x – y – z = 2 \quad (A1) \]
Method 2:
General form: \[ x + by + cz = d \quad (M1) \]
Using points \( A \), \( B \), \( C \):
\[ 5 – 2b + 5c = d \]
\[ 5 + 4b – c = d \quad (A1) \]
\[ -1 – 2b – c = d \]
Solving simultaneously: \( b = -1, c = -1, d = 2 \quad (M1) \)
So \[ x – y – z = 2 \quad (A1) \]
[4 marks]
(c) (i) Cartesian equation of \( \Pi_1 \):
Midpoint of \( [AB] \): \[ \left( \frac{5+5}{2}, \frac{-2+4}{2}, \frac{5+(-1)}{2} \right) = (5, 1, 2) \quad (A1) \]
Normal is \( \overrightarrow{\text{AB}} = (0, 6, -6) \), so equation: \[ 0x + 6y – 6z = k \]. Using midpoint: \( 6(1) – 6(2) = -6 \), so \[ y – z = -1 \quad (A1) \]
(c) (ii) Cartesian equation of \( \Pi_2 \):
Midpoint of \( [AC] \): \[ \left( \frac{5+(-1)}{2}, \frac{-2+(-2)}{2}, \frac{5+(-1)}{2} \right) = (2, -2, 2) \quad (A1) \]
Normal is \( \overrightarrow{\text{AC}} = (-6, 0, -6) \), so equation: \[ -6x + 0y – 6z = k \]. Using midpoint: \( -6(2) – 6(2) = -24 \), so \[ x + z = 4 \quad (A1) \]
[4 marks]
(d) Vector equation of line \( L \) and perpendicularity to \( \Pi \):
Either: Solve \( y – z = -1 \) and \( x + z = 4 \): (M1)
\[ \mathbf{r} = \begin{pmatrix} 4 \\ -1 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} \quad (A1) \]
Or: Direction of \( L \) is the cross product of normals of \( \Pi_1 \) \( (0, 1, -1) \) and \( \Pi_2 \) \( (1, 0, 1) \): (M1)
\[ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 1 & -1 \\ 1 & 0 & 1 \end{vmatrix} = \mathbf{i} – \mathbf{j} – \mathbf{k} \quad (A1) \]
Then: Direction \( (-1, 1, 1) \). Dot product with normal of \( \Pi \) \( (1, -1, -1) \): \( (-1)(1) + (1)(-1) + (1)(-1) = -1 – 1 – 1 = -3 \neq 0 \), but perpendicular to both \( \Pi_1 \) and \( \Pi_2 \), which lie in \( \Pi \), confirming perpendicularity: (R1)
[3 marks]
(e) Coordinates of \( D \):
Let \( D = (4 – \lambda, -1 + \lambda, \lambda) \): (M1)
\[ \text{AD}^2 = (4 – \lambda – 5)^2 + (-1 + \lambda – (-2))^2 + (\lambda – 5)^2 = \text{AB}^2 = 72 \quad (M1) \]
\[ (1 + \lambda)^2 + (\lambda – 1)^2 + (\lambda – 5)^2 = 72 \]
\[ 3\lambda^2 – 6\lambda – 45 = 0 \implies \lambda = 5 \text{ or } \lambda = -3 \quad (A1) \]
For \( \lambda = 5 \): \( D = (-1, 4, 5) \quad (AG) \)
Note: Award (M0)(M0)(A0) if candidates only verify \( D(-1, 4, 5) \); award (M1)(M1)(A0) if they also show parametric form.
[3 marks]
(f) Coordinates of \( G \) and bonding angle \( \angle \text{DGA} \):
Either: \( G = (4 – \lambda, -1 + \lambda, \lambda) \), and \( \text{DG} = \text{AG} \): (M1)
\[ (1 + \lambda)^2 + (\lambda – 1)^2 + (\lambda – 5)^2 = (5 – \lambda)^2 + (5 – \lambda)^2 + (5 – \lambda)^2 \quad (M1) \]
\[ (1 + \lambda)^2 = (5 – \lambda)^2 \implies \lambda = 2 \quad (A1) \]
So \( G = (2, 1, 2) \quad (AG) \)
Or: \( G \) is the barycentre of tetrahedron \( ABCD \): (M1)
\[ G = \left( \frac{5 + 5 + (-1) + (-1)}{4}, \frac{-2 + 4 + (-2) + 4}{4}, \frac{5 + (-1) + (-1) + 5}{4} \right) = (2, 1, 2) \quad (M1)(A1) \]
Then:
\[ \overrightarrow{\text{GD}} = (-1 – 2, 4 – 1, 5 – 2) = \begin{pmatrix} -3 \\ 3 \\ 3 \end{pmatrix} \quad (A1) \]
\[ \overrightarrow{\text{GA}} = (5 – 2, -2 – 1, 5 – 2) = \begin{pmatrix} 3 \\ -3 \\ 3 \end{pmatrix} \quad (A1) \]
\[ \cos \theta = \frac{\overrightarrow{\text{GD}} \cdot \overrightarrow{\text{GA}}}{|\overrightarrow{\text{GD}}| |\overrightarrow{\text{GA}}|} = \frac{(-3)(3) + (3)(-3) + (3)(3)}{\sqrt{(-3)^2 + 3^2 + 3^2} \cdot \sqrt{3^2 + (-3)^2 + 3^2}} = \frac{-9}{3 \sqrt{3} \cdot 3 \sqrt{3}} = -\frac{1}{3} \quad (M1) \]
\[ \theta \approx 109^\circ \quad (A1) \]
[6 marks]