SPS SPS SM Pure 2024 February — Question 4 5 marks

Exam BoardSPS
ModuleSPS SM Pure (SPS SM Pure)
Year2024
SessionFebruary
Marks5
TopicVectors 3D & Lines
TypeEqual length conditions
DifficultyModerate -0.3 This is a straightforward vectors question with standard techniques: part (a) requires simple vector addition using the given condition, and part (b) involves finding a magnitude and solving a quadratic equation. Both parts are routine applications of basic vector operations with no conceptual challenges or novel problem-solving required.
Spec1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10f Distance between points: using position vectors

4. Relative to a fixed origin \(O\),
the point \(A\) has position vector \(( 2 \mathbf { i } + 3 \mathbf { j } - 4 \mathbf { k } )\),
the point \(B\) has position vector \(( 4 \mathbf { i } - 2 \mathbf { j } + 3 \mathbf { k } )\),
and the point \(C\) has position vector \(( a \mathbf { i } + 5 \mathbf { j } - 2 \mathbf { k } )\), where \(a\) is a constant and \(a < 0\) \(D\) is the point such that \(\overrightarrow { A B } = \overrightarrow { B D }\).
  1. Find the position vector of \(D\).
    (2) Given \(| \overrightarrow { A C } | = 4\)
  2. find the value of \(a\).
    (3)

4.

Relative to a fixed origin $O$,\\
the point $A$ has position vector $( 2 \mathbf { i } + 3 \mathbf { j } - 4 \mathbf { k } )$,\\
the point $B$ has position vector $( 4 \mathbf { i } - 2 \mathbf { j } + 3 \mathbf { k } )$,\\
and the point $C$ has position vector $( a \mathbf { i } + 5 \mathbf { j } - 2 \mathbf { k } )$, where $a$ is a constant and $a < 0$\\
$D$ is the point such that $\overrightarrow { A B } = \overrightarrow { B D }$.
\begin{enumerate}[label=(\alph*)]
\item Find the position vector of $D$.\\
(2)

Given $| \overrightarrow { A C } | = 4$
\item find the value of $a$.\\
(3)
\end{enumerate}

\hfill \mbox{\textit{SPS SPS SM Pure 2024 Q4 [5]}}