Edexcel FP3 2014 June — Question 8 8 marks

Exam BoardEdexcel
ModuleFP3 (Further Pure Mathematics 3)
Year2014
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Cross Product & Distances
TypeArea of triangle using cross product
DifficultyStandard +0.3 This is a straightforward Further Maths FP3 vectors question testing standard techniques: computing a cross product for triangle area, evaluating a scalar triple product, and interpreting the geometric meaning of zero scalar triple product (coplanarity). All parts follow routine procedures with no novel insight required, making it slightly easier than average even for Further Maths.
Spec1.10g Problem solving with vectors: in geometry4.04g Vector product: a x b perpendicular vector

The position vectors of the points \(A\), \(B\) and \(C\) from a fixed origin \(O\) are $$\mathbf{a} = \mathbf{i} - \mathbf{j}, \quad \mathbf{b} = \mathbf{i} + \mathbf{j} + \mathbf{k}, \quad \mathbf{c} = 2\mathbf{j} + \mathbf{k}$$ respectively.
  1. Using vector products, find the area of the triangle \(ABC\). [4]
  2. Show that \(\frac{1}{6}\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = 0\) [3]
  3. Hence or otherwise, state what can be deduced about the vectors \(\mathbf{a}\), \(\mathbf{b}\) and \(\mathbf{c}\). [1]

The position vectors of the points $A$, $B$ and $C$ from a fixed origin $O$ are
$$\mathbf{a} = \mathbf{i} - \mathbf{j}, \quad \mathbf{b} = \mathbf{i} + \mathbf{j} + \mathbf{k}, \quad \mathbf{c} = 2\mathbf{j} + \mathbf{k}$$
respectively.

\begin{enumerate}[label=(\alph*)]
\item Using vector products, find the area of the triangle $ABC$. [4]
\item Show that $\frac{1}{6}\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = 0$ [3]
\item Hence or otherwise, state what can be deduced about the vectors $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$. [1]
\end{enumerate}

\hfill \mbox{\textit{Edexcel FP3 2014 Q8 [8]}}