| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2017 |
| Session | June |
| Marks | 11 |
| Topic | Vectors: Lines & Planes |
| Type | Perpendicular distance point to plane |
| Difficulty | Standard +0.3 This is a standard multi-part vectors question testing routine techniques: dot product for angles, substitution for intersection, perpendicular distance formula for point-to-plane distance, and recognizing parallel planes. All methods are textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 4.04c Scalar product: calculate and use for angles4.04f Line-plane intersection: find point4.04j Shortest distance: between a point and a plane |
| Answer | Marks | Guidance |
|---|---|---|
| Quote formula: \(\text{SD} = \left | \dfrac{d}{\ | \mathbf{n}\ |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{SD} = \dfrac{\left | 12\!\left(\tfrac{7}{4}\right)-4\!\left(-\tfrac{7}{2}\right)+6\!\left(\tfrac{7}{2}\right)+21\right | }{\sqrt{12^2+4^2+6^2}} = \dfrac{91}{14} = \dfrac{13}{2}\) A1A1 |
**Question 8(i)(a)**
$\cos\theta = \dfrac{12+2+6}{3\times 7} = \dfrac{20}{21}$ **M1A2** A1 scalar product; A1 both moduli. Give B1s for correct scalar product; both moduli if $\sin\theta = \ldots$ used.
**Question 8(i)(b)**
Substituting $(2\lambda, -\lambda, 2\lambda)$ into $6x - 2y + 3z = 35$ **M1**
$\Rightarrow \lambda = \dfrac{7}{4} \Rightarrow \mathbf{p} = \dfrac{7}{4}\begin{pmatrix}2\\-1\\2\end{pmatrix}$ **A1A1** Second A1 is FT
**Question 8(i)(c)**
SD $O$ to $\Pi_1 = OP\cos\theta = \dfrac{7}{4}\times 3\times\dfrac{20}{21} = 5$ **M1A1** A1FT
**Alternative I:**
$(6\lambda, -2\lambda, 3\lambda)$ in plane $\Rightarrow 36\lambda + 4\lambda + 9\lambda = 35$ **M1** $\Rightarrow \lambda = \dfrac{5}{7}$
$\Rightarrow \text{SD} = \lambda\sqrt{6^2+2^2+3^2} = 5$ cao **A1**
**Alternative II:**
Quote formula: $\text{SD} = \left|\dfrac{d}{\|\mathbf{n}\|}\right| = \dfrac{35}{\sqrt{6^2+2^2+3^2}} = 5$ cao **M1A1**
**Question 8(ii)**
Similar working gives $\lambda_1 = -\dfrac{21}{40}$ **B1**
Planes parallel, and on opposite sides of $O$, so total distance is $3\!\left(\dfrac{7}{4}+\dfrac{21}{40}\right)\cos\theta = \dfrac{13}{2}$ **M1A1**
**Alternative I:**
$\Pi_2$ has equation $\mathbf{r}\cdot\begin{pmatrix}6\\-2\\3\end{pmatrix} = -\dfrac{21}{2}$ **B1**
$\Rightarrow$ SD to $\Pi_2$ is $-\dfrac{3}{2}$ **B1**
Planes parallel, and on opposite sides of $O$, so distance between them is $5 - (-\tfrac{3}{2}) = \dfrac{13}{2}$ **B1** FT
**Alternative II:**
Quote Sh. Dist. formula for $P\!\left(\tfrac{7}{4},-\tfrac{7}{2},\tfrac{7}{2}\right)$ to $\Pi_2$ **M1**
$\text{SD} = \dfrac{\left|12\!\left(\tfrac{7}{4}\right)-4\!\left(-\tfrac{7}{2}\right)+6\!\left(\tfrac{7}{2}\right)+21\right|}{\sqrt{12^2+4^2+6^2}} = \dfrac{91}{14} = \dfrac{13}{2}$ **A1A1**
**Total: 11 marks**8 The line $l$ has equation $\mathbf { r } = \lambda \mathbf { d }$ and the plane $\Pi _ { 1 }$ has equation $\mathbf { r } . \mathbf { n } = 35$, where
$$\mathbf { d } = \left( \begin{array} { r }
2 \\
- 1 \\
2
\end{array} \right) \quad \text { and } \quad \mathbf { n } = \left( \begin{array} { r }
6 \\
- 2 \\
3
\end{array} \right) .$$
\begin{enumerate}[label=(\roman*)]
\item (a) Determine the exact value of $\cos \theta$, where $\theta$ is the angle between $\mathbf { d }$ and $\mathbf { n }$.\\
(b) Determine the position vector of the point of intersection of $l$ and $\Pi _ { 1 }$.\\
(c) Determine the shortest distance from $O$ to $\Pi _ { 1 }$.
\item The plane $\Pi _ { 2 }$ has cartesian equation $12 x - 4 y + 6 z + 21 = 0$. Determine the distance between $\Pi _ { 1 }$ and $\Pi _ { 2 }$.
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2017 Q8 [11]}}