OCR PURE — Question 4 7 marks

Exam BoardOCR
ModulePURE
Marks7
PaperDownload PDF ↗
TopicVectors Introduction & 2D
TypeParallel or perpendicular vectors condition
DifficultyModerate -0.8 Part (a) is straightforward vector arithmetic requiring only scalar multiplication and addition. Part (b) requires understanding that parallel to y-axis means x-component equals zero, then solving a simple linear equation followed by using magnitude formula. This is a standard introductory vectors question with routine techniques and no problem-solving insight required.
Spec1.10a Vectors in 2D: i,j notation and column vectors1.10d Vector operations: addition and scalar multiplication

4
  1. Simplify \(2 \binom { 6 } { - 3 } - 3 \binom { - 1 } { 2 }\).
  2. The vector \(\mathbf { a }\) is defined by \(\mathbf { a } = r \binom { 6 } { - 3 } + s \binom { - 1 } { 2 }\), where \(r\) and \(s\) are constants. Determine two pairs of values of \(r\) and \(s\) such that \(\mathbf { a }\) is parallel to the \(y\)-axis and \(| \mathbf { a } | = 3\).

Question 4:
Part (a):
AnswerMarks Guidance
\(\begin{pmatrix}15\\-12\end{pmatrix}\) or \(15\mathbf{i} - 12\mathbf{j}\)B1, B1 B1 for each element. Allow i, j notation without "squiggles"
[1]
Part (b):
AnswerMarks Guidance
\(6r - s = 0\), \(s = 6r\)M1
\(0^2 + (9r)^2 = 9\), \(81r^2 = 9\)M1, A1 Attempt \(
\(r = \pm\frac{1}{3}\)A1 Allow just \(r = \frac{1}{3}\)
\(r = \frac{1}{3}\) and \(s=2\) or \(r = -\frac{1}{3}\) and \(s=-2\)A1 Correctly paired
[5]
# Question 4:

## Part (a):
| $\begin{pmatrix}15\\-12\end{pmatrix}$ or $15\mathbf{i} - 12\mathbf{j}$ | B1, B1 | B1 for each element. Allow **i**, **j** notation without "squiggles" |
|---|---|---|
| | [1] | |

## Part (b):
| $6r - s = 0$, $s = 6r$ | M1 | |
|---|---|---|
| $0^2 + (9r)^2 = 9$, $81r^2 = 9$ | M1, A1 | Attempt $|\mathbf{a}|^2 = 9$, or $|\mathbf{a}|=3$; allow in terms of both $r$ and $s$ |
| $r = \pm\frac{1}{3}$ | A1 | Allow just $r = \frac{1}{3}$ |
| $r = \frac{1}{3}$ and $s=2$ or $r = -\frac{1}{3}$ and $s=-2$ | A1 | Correctly paired |
| | [5] | |

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4
\begin{enumerate}[label=(\alph*)]
\item Simplify $2 \binom { 6 } { - 3 } - 3 \binom { - 1 } { 2 }$.
\item The vector $\mathbf { a }$ is defined by $\mathbf { a } = r \binom { 6 } { - 3 } + s \binom { - 1 } { 2 }$, where $r$ and $s$ are constants.

Determine two pairs of values of $r$ and $s$ such that $\mathbf { a }$ is parallel to the $y$-axis and $| \mathbf { a } | = 3$.
\end{enumerate}

\hfill \mbox{\textit{OCR PURE  Q4 [7]}}