OCR Further Additional Pure 2021 November — Question 3 6 marks

Exam BoardOCR
ModuleFurther Additional Pure (Further Additional Pure)
Year2021
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVector Product and Surfaces
TypeVolume of tetrahedron using scalar triple product
DifficultyStandard +0.8 This is a Further Maths question requiring vector product calculation and scalar triple product for volume. Part (a) involves computing a cross product and applying parallelism conditions, while part (b) requires the formula V = (1/6)|[p,q,r]| and solving the resulting equation. These are standard Further Maths techniques but require multiple steps and careful algebraic manipulation, placing it moderately above average difficulty.
Spec1.10b Vectors in 3D: i,j,k notation4.04g Vector product: a x b perpendicular vector
3 The points \(P , Q\) and \(R\) have position vectors \(\mathbf { p } = 2 \mathbf { i } + \mathbf { j } + 5 \mathbf { k } , \mathbf { q } = \mathbf { i } - \mathbf { j } + \mathbf { k }\) and \(\mathbf { r } = 2 \mathbf { i } + \mathbf { j } + t \mathbf { k }\) respectively, relative to the origin \(O\). Determine the value(s) of \(t\) in each of the following cases.
  1. The line \(O R\) is parallel to \(\mathbf { p } \times \mathbf { q }\).
  2. The volume of tetrahedron \(O P Q R\) is 13 .
Question 3:
AnswerMarks Guidance
3(a) i j k  6 
p  q = 2 1 5 = 3
1 −1 1 − 3
 2
 
= 3 1  t = –1
 
−1
AnswerMarks
M1
A1
AnswerMarks
[2]3.1a
2.2aPossibly BC
Correct vector product; t correct
AnswerMarks
(b) 6 2
   
AnswerMarks Guidance
Vol. OABC = 16p  q . r = 1  3•1
6
   
 −3 t
AnswerMarks
= 1612 + 3 – 3t
Solving 5 – t = 26 and/or t – 5 = 26
AnswerMarks
 t = –21 or 31B1
M1
M1
A1
AnswerMarks
[4]1.1
1.1
3.1a
AnswerMarks
2.2aA correct scalar triple product involving t
Correct formula attempted
Method may be implied by one correct t
Question 3:
3 | (a) | i j k  6 
p  q = 2 1 5 = 3
1 −1 1 − 3
 2
 
= 3 1  t = –1
 
−1
 | M1
A1
[2] | 3.1a
2.2a | Possibly BC
Correct vector product; t correct
(b) |  6 2
   
Vol. OABC = 16 | p  q . r | = 1  3•1
6
   
 −3 t
= 16 | 12 + 3 – 3t |
Solving 5 – t = 26 and/or t – 5 = 26
 t = –21 or 31 | B1
M1
M1
A1
[4] | 1.1
1.1
3.1a
2.2a | A correct scalar triple product involving t
Correct formula attempted
Method may be implied by one correct t
3 The points $P , Q$ and $R$ have position vectors $\mathbf { p } = 2 \mathbf { i } + \mathbf { j } + 5 \mathbf { k } , \mathbf { q } = \mathbf { i } - \mathbf { j } + \mathbf { k }$ and $\mathbf { r } = 2 \mathbf { i } + \mathbf { j } + t \mathbf { k }$ respectively, relative to the origin $O$.

Determine the value(s) of $t$ in each of the following cases.
\begin{enumerate}[label=(\alph*)]
\item The line $O R$ is parallel to $\mathbf { p } \times \mathbf { q }$.
\item The volume of tetrahedron $O P Q R$ is 13 .
\end{enumerate}

\hfill \mbox{\textit{OCR Further Additional Pure 2021 Q3 [6]}}
This paper (10 questions)
View full paper
Q1 3 Q2 5 Q3 6 Q4 6 Q5 8 Q6 10 Q7 8 Q8 12 Q9 10 Q10 7