| Exam Board | WJEC |
|---|---|
| Module | Further Unit 1 (Further Unit 1) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Perpendicularity conditions |
| Difficulty | Standard +0.3 This is a straightforward application of perpendicularity conditions using dot products of direction vectors. Part (a) involves setting two dot products to zero and solving simple linear equations for n and p. Part (b) requires the standard formula for angle between lines. All techniques are routine for Further Maths students with no novel problem-solving required, making it slightly easier than average. |
| Spec | 4.04c Scalar product: calculate and use for angles4.04d Angles: between planes and between line and plane |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| From lines \(L_1, L_2\): \((2\times3) + (1\times n) + (1\times-3) = 0\), giving \(6 + n - 3 = 0\) | M1 | |
| \(n = -3\) | A1 | convincing |
| From lines \(L_1, L_3\): \((2\times p) + (-3\times3) + (1\times4) = 0\) | (M1) | If not awarded for \(L_1, L_2\) |
| \(p = \frac{5}{2}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\left(3\times\frac{5}{2}\right) + (1\times3) + (-3\times4) = -\frac{3}{2}\) | B1 | si FT their \(p\) for B1B1M1 |
| \( | 3\mathbf{i} + \mathbf{j} - 3\mathbf{k} | = \sqrt{19}\) |
| \(\left | \frac{5}{2}\mathbf{i} + 3\mathbf{j} + 4\mathbf{k}\right | = \sqrt{\frac{125}{4}}\) |
| \(\cos\theta = \dfrac{-\frac{3}{2}}{\sqrt{19}\sqrt{\frac{125}{4}}}\) | M1 | oe |
| \(\theta = 93.5°\), therefore acute angle is \(\theta = 86.5°\) | A1 | cao |
## Question 7:
### Part a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| From lines $L_1, L_2$: $(2\times3) + (1\times n) + (1\times-3) = 0$, giving $6 + n - 3 = 0$ | M1 | |
| $n = -3$ | A1 | convincing |
| From lines $L_1, L_3$: $(2\times p) + (-3\times3) + (1\times4) = 0$ | (M1) | If not awarded for $L_1, L_2$ |
| $p = \frac{5}{2}$ | A1 | |
### Part b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\left(3\times\frac{5}{2}\right) + (1\times3) + (-3\times4) = -\frac{3}{2}$ | B1 | si FT their $p$ for B1B1M1 |
| $|3\mathbf{i} + \mathbf{j} - 3\mathbf{k}| = \sqrt{19}$ | B1 | si Both mods |
| $\left|\frac{5}{2}\mathbf{i} + 3\mathbf{j} + 4\mathbf{k}\right| = \sqrt{\frac{125}{4}}$ | | |
| $\cos\theta = \dfrac{-\frac{3}{2}}{\sqrt{19}\sqrt{\frac{125}{4}}}$ | M1 | oe |
| $\theta = 93.5°$, therefore acute angle is $\theta = 86.5°$ | A1 | cao |
---7. The vector equations of the lines $L _ { 1 } , L _ { 2 } , L _ { 3 }$ are given by
$$\begin{aligned}
& \mathbf { r } = 3 \mathbf { i } + 2 \mathbf { j } + \mathbf { k } + \lambda ( 2 \mathbf { i } + n \mathbf { j } + \mathbf { k } ) \\
& \mathbf { r } = 5 \mathbf { i } - 3 \mathbf { j } - 4 \mathbf { k } + \mu ( 3 \mathbf { i } + \mathbf { j } - 3 \mathbf { k } ) \\
& \mathbf { r } = 6 \mathbf { i } - 3 \mathbf { j } + 2 \mathbf { k } + v ( p \mathbf { i } + 3 \mathbf { j } + 4 \mathbf { k } )
\end{aligned}$$
respectively, where $n$ and $p$ are constants.\\
The line $L _ { 1 }$ is perpendicular to the line $L _ { 2 }$. The line $L _ { 1 }$ is also perpendicular to the line $L _ { 3 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that the value of $n$ is - 3 and find the value of $p$.
\item Find the acute angle between the lines $L _ { 2 }$ and $L _ { 3 }$.
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 1 2022 Q7 [7]}}