OCR H240/02 2023 June — Question 2 5 marks

Exam BoardOCR
ModuleH240/02 (Pure Mathematics and Statistics)
Year2023
SessionJune
Marks5
PaperDownload PDF ↗
TopicVectors 3D & Lines
TypeFoot of perpendicular from origin to line
DifficultyModerate -0.3 Part (a) is straightforward vector magnitude calculation using scalar multiplication. Part (b) requires setting up a dot product equation using perpendicularity, which is a standard A-level technique, though the algebra involves expanding and solving a quadratic. This is a routine multi-step vectors question with no novel insight required, making it slightly easier than average.
Spec1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10f Distance between points: using position vectors

The points \(O\) and \(A\) have position vectors \(\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\) and \(\begin{pmatrix} 6 \\ 0 \\ 8 \end{pmatrix}\) respectively. The point \(P\) is such that \(\overrightarrow{OP} = k\overrightarrow{OA}\), where \(k\) is a non-zero constant.
  1. Find, in terms of \(k\), the length of \(OP\). [1] Point \(B\) has position vector \(\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}\) and angle \(OPB\) is a right angle.
  2. Determine the value of \(k\). [4]

The points $O$ and $A$ have position vectors $\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} 6 \\ 0 \\ 8 \end{pmatrix}$ respectively. The point $P$ is such that $\overrightarrow{OP} = k\overrightarrow{OA}$, where $k$ is a non-zero constant.

\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $k$, the length of $OP$. [1]

Point $B$ has position vector $\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$ and angle $OPB$ is a right angle.

\item Determine the value of $k$. [4]
\end{enumerate}

\hfill \mbox{\textit{OCR H240/02 2023 Q2 [5]}}