OCR MEI C4 — Question 4 6 marks

Exam BoardOCR MEI
ModuleC4 (Core Mathematics 4)
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors 3D & Lines
TypePerpendicularity conditions
DifficultyModerate -0.8 This is a straightforward application of the scalar product formula requiring direct recall and basic calculation. Part (i) involves computing a·b, finding magnitudes, and using the cosine formula—all standard steps. Part (ii) is even simpler, setting the scalar product to zero and solving for k. Both parts are routine textbook exercises with no problem-solving insight required, making this easier than average.
Spec1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication4.04c Scalar product: calculate and use for angles

4 You are given that \(\mathbf { a } = \left( \begin{array} { c } 1 \\ 2 \\ - 1 \end{array} \right)\) and \(\mathbf { b } = \left( \begin{array} { c } 3 \\ - 1 \\ k \end{array} \right)\).
  1. Find the angle between \(\mathbf { a }\) and \(\mathbf { b }\) when \(k = 2\).
  2. Find the value of \(k\) such that \(\mathbf { a }\) and \(\mathbf { b }\) are perpendicular.

Question 4(i):
AnswerMarks Guidance
AnswerMarks Guidance
\(\begin{pmatrix}1\\2\\-1\end{pmatrix}\cdot\begin{pmatrix}3\\-1\\2\end{pmatrix}=3-2-2=-1\)M1
\(n_1 n_2=\sqrt{6}\sqrt{14}\cos\theta \Rightarrow \cos\theta=\frac{-1}{\sqrt{84}}\approx -0.1091\)M1, A1
\(\Rightarrow \theta=96.3°\)A1
Total: 4 marks
Question 4(ii):
AnswerMarks Guidance
AnswerMarks Guidance
\(\mathbf{a}\cdot\mathbf{b}=0 \Rightarrow 3-2-k=0 \Rightarrow k=1\)M1, A1
Total: 2 marks
## Question 4(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}1\\2\\-1\end{pmatrix}\cdot\begin{pmatrix}3\\-1\\2\end{pmatrix}=3-2-2=-1$ | M1 | |
| $n_1 n_2=\sqrt{6}\sqrt{14}\cos\theta \Rightarrow \cos\theta=\frac{-1}{\sqrt{84}}\approx -0.1091$ | M1, A1 | |
| $\Rightarrow \theta=96.3°$ | A1 | |

**Total: 4 marks**

## Question 4(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{a}\cdot\mathbf{b}=0 \Rightarrow 3-2-k=0 \Rightarrow k=1$ | M1, A1 | |

**Total: 2 marks**

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4 You are given that $\mathbf { a } = \left( \begin{array} { c } 1 \\ 2 \\ - 1 \end{array} \right)$ and $\mathbf { b } = \left( \begin{array} { c } 3 \\ - 1 \\ k \end{array} \right)$.\\
(i) Find the angle between $\mathbf { a }$ and $\mathbf { b }$ when $k = 2$.\\
(ii) Find the value of $k$ such that $\mathbf { a }$ and $\mathbf { b }$ are perpendicular.

\hfill \mbox{\textit{OCR MEI C4  Q4 [6]}}