OCR MEI Further Pure Core AS 2018 June — Question 2 3 marks

Exam BoardOCR MEI
ModuleFurther Pure Core AS (Further Pure Core AS)
Year2018
SessionJune
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors 3D & Lines
TypeAngle between two vectors/lines (direct)
DifficultyModerate -0.8 This is a straightforward application of the standard dot product formula for finding angles between vectors (cos θ = a·b/|a||b|). It requires only direct substitution into a memorized formula with basic arithmetic and calculator work, making it easier than average even for Further Maths students.
Spec4.04c Scalar product: calculate and use for angles

Find, to the nearest degree, the angle between the vectors \(\begin{pmatrix} 1 \\ 0 \\ -2 \end{pmatrix}\) and \(\begin{pmatrix} -2 \\ 3 \\ -3 \end{pmatrix}\). [3]

Question 2:
AnswerMarks
21(2)03(2)(3)
cos 0.381K
5 22
AnswerMarks
 = 68 to nearest degreeM1
A1
A1
AnswerMarks
[3]1.1a,
1.1
AnswerMarks
1.1a.b
Use of cos
a b
4
0.381… or
5 22
AnswerMarks
must be nearest degreeallow unsupported correct
answers
Question 2:
2 | 1(2)03(2)(3)
cos 0.381K
5 22
 = 68 to nearest degree | M1
A1
A1
[3] | 1.1a,
1.1
1.1 | a.b
Use of cos
a b
4
0.381… or
5 22
must be nearest degree | allow unsupported correct
answers
Find, to the nearest degree, the angle between the vectors $\begin{pmatrix} 1 \\ 0 \\ -2 \end{pmatrix}$ and $\begin{pmatrix} -2 \\ 3 \\ -3 \end{pmatrix}$. [3]

\hfill \mbox{\textit{OCR MEI Further Pure Core AS 2018 Q2 [3]}}