| Exam Board | WJEC |
|---|---|
| Module | Further Unit 1 (Further Unit 1) |
| Year | 2019 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Cartesian equation of a plane |
| Difficulty | Standard +0.3 This is a standard Further Maths question requiring finding two direction vectors from three points, computing their cross product to get a normal vector, then using the scalar product form and converting to Cartesian. While it involves multiple steps and vector manipulation, it follows a well-established procedure taught explicitly in Further Maths courses with no novel problem-solving required. |
| Spec | 4.04b Plane equations: cartesian and vector forms4.04h Shortest distances: between parallel lines and between skew lines |
| Answer | Marks | Guidance |
|---|---|---|
| Let \(r \cdot n = 1\) where \(n = pi + qj + rk\). \(\therefore A: 3p + 5q + 6r = 1\), \(B: 5p - q + 7r = 1\), \(C: -p + 7q = 1\) | M1, A2 | A1 for any 1; A2 for all 3 |
| Substituting \(p = 7q - 1\) into \(A\) and \(B\) gives: \(26q + 6r = 4\), \(34q + 7r = 6\) | m1 | Accept working with \(q\) or \(r\) instead of \(p\) |
| Solving: \(q = \frac{4}{11}\), \(r = -\frac{11}{11}\), \(p = \frac{17}{11}\) | M1, A2 | A1 for 2 variables provided m1 awarded |
| Therefore, \(r.\left(\frac{17}{11}i + \frac{4}{11}j - \frac{10}{11}k\right) = 1\) oe, \(\frac{17}{11}x + \frac{4}{11}y - \frac{10}{11}z = 1\) oe | B1, B1 | FT \(p,q,r\); FT equation of the plane |
| Answer | Marks | Guidance |
|---|---|---|
| Let \(r \cdot n = 1\) where \(n = pi + qj + rk\). \(AB = 2i - 6j + k\), \(BC = -6i + 8j - 7k\), \(CA = 4i - 2j + 6k\). \(AB \cdot n \rightarrow 2p - 6q + r = 0\) (1), \(BC \cdot n \rightarrow -6p + 8q - 7r = 0\) (2), \(CA \cdot n \rightarrow 4p - 2q + 6r = 0\) (3) | (M1), (A2) | A1 for any 1; A2 for all 3 |
| Row operations: (2) + 3(1): \(0 - 10q - 4r = 0 \rightarrow r = -\frac{5}{2}q\), (2) + 7(1): \(8p - 34q + 0 = 0 \rightarrow p = \frac{17}{4}q\). Let \(q = 4\), \(\therefore p = 17\), \(q = 4\), \(r = -10\) oe | (m1), (A2) | A1 for 2 variables |
| \(a \cdot n = (3 \times 17) + (5 \times 4) + (6 \times -10) = 11\) | (B1) | |
| Therefore, \(p = \frac{17}{11}\), \(q = \frac{4}{11}\), \(r = -\frac{10}{11}\), \(r.\left(\frac{17}{11}i + \frac{4}{11}j - \frac{10}{11}k\right) = 1\) oe, \(\frac{17}{11}x + \frac{4}{11}y - \frac{10}{11}z = 1\) oe | (B1), (B1) | FT \(p,q,r\); FT equation of the plane |
**Method 1:**
Let $r \cdot n = 1$ where $n = pi + qj + rk$. $\therefore A: 3p + 5q + 6r = 1$, $B: 5p - q + 7r = 1$, $C: -p + 7q = 1$ | M1, A2 | A1 for any 1; A2 for all 3
Substituting $p = 7q - 1$ into $A$ and $B$ gives: $26q + 6r = 4$, $34q + 7r = 6$ | m1 | Accept working with $q$ or $r$ instead of $p$
Solving: $q = \frac{4}{11}$, $r = -\frac{11}{11}$, $p = \frac{17}{11}$ | M1, A2 | A1 for 2 variables provided m1 awarded
Therefore, $r.\left(\frac{17}{11}i + \frac{4}{11}j - \frac{10}{11}k\right) = 1$ oe, $\frac{17}{11}x + \frac{4}{11}y - \frac{10}{11}z = 1$ oe | B1, B1 | FT $p,q,r$; FT equation of the plane
**Method 2:**
Let $r \cdot n = 1$ where $n = pi + qj + rk$. $AB = 2i - 6j + k$, $BC = -6i + 8j - 7k$, $CA = 4i - 2j + 6k$. $AB \cdot n \rightarrow 2p - 6q + r = 0$ (1), $BC \cdot n \rightarrow -6p + 8q - 7r = 0$ (2), $CA \cdot n \rightarrow 4p - 2q + 6r = 0$ (3) | (M1), (A2) | A1 for any 1; A2 for all 3
Row operations: (2) + 3(1): $0 - 10q - 4r = 0 \rightarrow r = -\frac{5}{2}q$, (2) + 7(1): $8p - 34q + 0 = 0 \rightarrow p = \frac{17}{4}q$. Let $q = 4$, $\therefore p = 17$, $q = 4$, $r = -10$ oe | (m1), (A2) | A1 for 2 variables
$a \cdot n = (3 \times 17) + (5 \times 4) + (6 \times -10) = 11$ | (B1) |
Therefore, $p = \frac{17}{11}$, $q = \frac{4}{11}$, $r = -\frac{10}{11}$, $r.\left(\frac{17}{11}i + \frac{4}{11}j - \frac{10}{11}k\right) = 1$ oe, $\frac{17}{11}x + \frac{4}{11}y - \frac{10}{11}z = 1$ oe | (B1), (B1) | FT $p,q,r$; FT equation of the plane
---8. The plane $\Pi$ contains the three points $A ( 3,5,6 ) , B ( 5 , - 1,7 )$ and $C ( - 1,7,0 )$.
Find the vector equation of the plane $\Pi$ in the form r.n $= d$.\\
Express this equation in Cartesian form.\\
\hfill \mbox{\textit{WJEC Further Unit 1 2019 Q8 [9]}}