| Exam Board | WJEC |
|---|---|
| Module | Further Unit 1 (Further Unit 1) |
| Year | 2019 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Parallel and perpendicular lines |
| Difficulty | Moderate -0.3 This is a straightforward Further Maths vectors question requiring standard techniques: finding direction vectors by subtraction, writing vector equations in standard form, and checking perpendicularity via dot product. All steps are routine with no problem-solving insight needed, making it slightly easier than average even for Further Maths. |
| Spec | 1.10b Vectors in 3D: i,j,k notation4.04a Line equations: 2D and 3D, cartesian and vector forms4.04c Scalar product: calculate and use for angles |
| Answer | Marks | Guidance |
|---|---|---|
| \(AB: r = 2i + 3j - k + \lambda(4j + 5k - 2i - 3j + k)\), \(r = 2i + 3j - k + \lambda(-2i + j + 6k)\) oe | M1, A1 | Award M1 for either AB or CD; if no marks, award SC1 for both AB and CD; condone 1st omission of \(r =\); penalise \(-1\) for 2nd omission of \(r =\) |
| \(CD: r = 7i - 3k + \mu(-3i - j - 5k - 7i + 3k)\), \(r = 7i - 3k + \mu(-10i - j - 2k)\) oe | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Directions: \(-2i + j + 6k\) and \(-10i - j - 2k\) | B1 | FT (a) |
| Checking for perpendicularity: \((-2i + j + 6k) \cdot (-10i - j - 2k) = 20 - 1 - 12 = 7\). Because \(7 \neq 0\), AB and CD are not perpendicular. | M1, A1, E1 | For checking scalar product; must refer to \(\neq 0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Directions: \(-2i + j + 6k\) and \(-10i - j - 2k\) | (B1) | FT (a) |
| Angle between vectors: \(\cos\theta = \frac{a \cdot b}{ | a | |
| Because \(\cos\theta \neq 0\) (OR \(\theta \neq 90°\)), AB and CD are not perpendicular. | (E1) | Must refer to \(\neq 0\) (OR \(\neq 90°\)) |
**Part (a):**
$AB: r = 2i + 3j - k + \lambda(4j + 5k - 2i - 3j + k)$, $r = 2i + 3j - k + \lambda(-2i + j + 6k)$ oe | M1, A1 | Award M1 for either AB or CD; if no marks, award SC1 for both AB and CD; condone 1st omission of $r =$; penalise $-1$ for 2nd omission of $r =$
$CD: r = 7i - 3k + \mu(-3i - j - 5k - 7i + 3k)$, $r = 7i - 3k + \mu(-10i - j - 2k)$ oe | A1 |
**Part (b):**
Directions: $-2i + j + 6k$ and $-10i - j - 2k$ | B1 | FT (a)
Checking for perpendicularity: $(-2i + j + 6k) \cdot (-10i - j - 2k) = 20 - 1 - 12 = 7$. Because $7 \neq 0$, AB and CD are not perpendicular. | M1, A1, E1 | For checking scalar product; must refer to $\neq 0$
**Method 2:**
Directions: $-2i + j + 6k$ and $-10i - j - 2k$ | (B1) | FT (a)
Angle between vectors: $\cos\theta = \frac{a \cdot b}{|a||b|}$, $\cos\theta = \frac{7}{\sqrt{41}\sqrt{105}}$ OR $\theta = 83.9°$ | (M1), (A1) | Use
Because $\cos\theta \neq 0$ (OR $\theta \neq 90°$), AB and CD are not perpendicular. | (E1) | Must refer to $\neq 0$ (OR $\neq 90°$)
---2. The position vectors of the points $A , B , C , D$ are given by\\
$\mathbf { a } = 2 \mathbf { i } + 3 \mathbf { j } - \mathbf { k }$,\\
$\mathbf { b } = 4 \mathbf { j } + 5 \mathbf { k }$,\\
$\mathbf { c } = 7 \mathbf { i } - 3 \mathbf { k }$,\\
$\mathbf { d } = - 3 \mathbf { i } - \mathbf { j } - 5 \mathbf { k }$,\\
respectively.
\begin{enumerate}[label=(\alph*)]
\item Find the vector equations of the lines $A B$ and $C D$.
\item Determine whether or not the lines $A B$ and $C D$ are perpendicular.
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 1 2019 Q2 [7]}}