| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2010 |
| Session | June |
| Marks | 10 |
| Topic | Vectors: Cross Product & Distances |
| Type | Equation of plane through three points |
| Difficulty | Challenging +1.2 This is a structured Further Maths vectors question requiring cross product manipulation and geometric interpretation. Part (i)(a) needs proof using vector identities, (i)(b) tests conceptual understanding of normals, while (ii) applies these to a specific symmetric case. The question guides students through each step and uses standard techniques, making it moderately above average difficulty but not requiring exceptional insight. |
| Spec | 4.04d Angles: between planes and between line and plane4.04g Vector product: a x b perpendicular vector |
| Answer | Marks | Guidance |
|---|---|---|
| \(\triangle ABC = \frac{1}{2} | (\mathbf{b}-\mathbf{a})\times(\mathbf{c}-\mathbf{a}) | \) M1 (*) |
| \(= \frac{1}{2} | \mathbf{b}\times\mathbf{c}-\mathbf{a}\times\mathbf{c}-\mathbf{b}\times\mathbf{a}+\mathbf{a}\times\mathbf{a} | \) M1 |
| \(= \frac{1}{2} | \mathbf{a}\times\mathbf{b}+\mathbf{b}\times\mathbf{c}+\mathbf{c}\times\mathbf{a} | \) A1 With explanations since \(-\mathbf{a}\times\mathbf{c}=\mathbf{c}\times\mathbf{a}\), \(-\mathbf{b}\times\mathbf{a}=\mathbf{a}\times\mathbf{b}\) and \(\mathbf{a}\times\mathbf{a}=\mathbf{0}\) [3] |
| Answer | Marks | Guidance |
|---|---|---|
| (b) Sh. Dist. \(= \frac{d}{ | \mathbf{n} | }\) M1 \(= \frac{abc}{\sqrt{a^2b^2+b^2c^2+c^2a^2}}\) A1 (ignore mod. top) [2] |
**(i)(a)**
$\triangle ABC = \frac{1}{2}|(\mathbf{b}-\mathbf{a})\times(\mathbf{c}-\mathbf{a})|$ **M1** (*)
$= \frac{1}{2}|\mathbf{b}\times\mathbf{c}-\mathbf{a}\times\mathbf{c}-\mathbf{b}\times\mathbf{a}+\mathbf{a}\times\mathbf{a}|$ **M1**
$= \frac{1}{2}|\mathbf{a}\times\mathbf{b}+\mathbf{b}\times\mathbf{c}+\mathbf{c}\times\mathbf{a}|$ **A1** With explanations since $-\mathbf{a}\times\mathbf{c}=\mathbf{c}\times\mathbf{a}$, $-\mathbf{b}\times\mathbf{a}=\mathbf{a}\times\mathbf{b}$ and $\mathbf{a}\times\mathbf{a}=\mathbf{0}$ [3]
**(b)** $\mathbf{a}\times\mathbf{b}+\mathbf{b}\times\mathbf{c}+\mathbf{c}\times\mathbf{a}$ is a normal vector to the plane **B1** [1]
**(ii)(a)**
$\mathbf{n} = \mathbf{a}\times\mathbf{b}+\mathbf{b}\times\mathbf{c}+\mathbf{c}\times\mathbf{a} = \begin{pmatrix}0\\0\\ab\end{pmatrix}+\begin{pmatrix}bc\\0\\0\end{pmatrix}+\begin{pmatrix}0\\ca\\0\end{pmatrix} = \begin{pmatrix}bc\\ca\\ab\end{pmatrix}$ **M1 A1** or via (*)'s form
$d = \begin{pmatrix}bc\\ca\\ab\end{pmatrix}\cdot\begin{pmatrix}a\\0\\0\end{pmatrix} = abc$ **M1 A1**
ALTERNATIVELY: Plane through $(a,0,0)$, $(0,b,0)$, $(0,0,c)$ is $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ **M1 A1**
**M1 A1** for turning this into a correct vector equation. [4]
**(b)** Sh. Dist. $= \frac{d}{|\mathbf{n}|}$ **M1** $= \frac{abc}{\sqrt{a^2b^2+b^2c^2+c^2a^2}}$ **A1** (ignore mod. top) [2]9 Three non-collinear points $A , B$ and $C$ have position vectors $\mathbf { a } , \mathbf { b }$ and $\mathbf { c }$ respectively, relative to the origin $O$. The plane through $A , B$ and $C$ is denoted by $\Pi$.
\begin{enumerate}[label=(\roman*)]
\item (a) Prove that the area of triangle $A B C$ is $\frac { 1 } { 2 } | \mathbf { a } \times \mathbf { b } + \mathbf { b } \times \mathbf { c } + \mathbf { c } \times \mathbf { a } |$.\\
(b) Describe the significance of the vector $\mathbf { a } \times \mathbf { b } + \mathbf { b } \times \mathbf { c } + \mathbf { c } \times \mathbf { a }$ in relation to $\Pi$.
\item (a) In the case when $\mathbf { a } = a \mathbf { i } , \mathbf { b } = b \mathbf { j }$ and $\mathbf { c } = c \mathbf { k }$, where $a , b$ and $c$ are positive scalar constants, determine the equation of $\Pi$ in the form r.n $= d$, where the components of $\mathbf { n }$ and the value of the scalar constant $d$ are to be given in terms of $a , b$ and $c$.\\
(b) Deduce the shortest distance from the origin $O$ to $\Pi$ in this case.
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2010 Q9 [10]}}