CAIE P1 2019 March — Question 5 7 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2019
SessionMarch
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors 3D & Lines
TypePerpendicularity conditions
DifficultyModerate -0.8 This is a straightforward two-part question on basic 3D vector operations. Part (i) requires setting the dot product to zero and solving a quadratic equation. Part (ii) involves computing magnitudes and using the standard angle formula. Both parts are routine applications of standard formulas with no problem-solving insight required, making it easier than average.
Spec1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication
Two vectors, \(\mathbf{u}\) and \(\mathbf{v}\), are such that $$\mathbf{u} = \begin{pmatrix} q \\ 1 \\ 6 \end{pmatrix} \quad \text{and} \quad \mathbf{v} = \begin{pmatrix} 8 \\ q - 1 \\ q^2 - 7 \end{pmatrix},$$ where \(q\) is a constant.
  1. Find the values of \(q\) for which \(\mathbf{u}\) is perpendicular to \(\mathbf{v}\). [3]
  2. Find the angle between \(\mathbf{u}\) and \(\mathbf{v}\) when \(q = 0\). [4]
Question 5:
AnswerMarks Guidance
5(i)u.v = 8q+2q−2+6q2 −42 B1
6q2 +10q−44=0 oeM1 Simplify, set to zero and attempt to solve
q = 2, ‒11/3A1 Both required. Accept ‒3.67
3
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
5(ii)0  8 
   
u = 2 v = −1 u.v = ‒2 ‒ 42
   
   
AnswerMarks Guidance
6 −7M1 Correct method for scalar product
u× v
−44 −44 −4
cosθ= = =
AnswerMarks Guidance
40× 114 4 285 √11M1 All linked correctly and inverse cos used correctly
θ=130.7° or 2.28(05) radsA1 No other angles between 0° and 180°
4
AnswerMarks Guidance
QuestionAnswer Marks 
Question 5:
--- 5(i) ---
5(i) | u.v = 8q+2q−2+6q2 −42 | B1 | May be unsimplified
6q2 +10q−44=0 oe | M1 | Simplify, set to zero and attempt to solve
q = 2, ‒11/3 | A1 | Both required. Accept ‒3.67
3
Question | Answer | Marks | Guidance
--- 5(ii) ---
5(ii) | 0  8 
   
u = 2 v = −1 u.v = ‒2 ‒ 42
   
   
6 −7 | M1 | Correct method for scalar product
|u| × |v| = 22 +62 × 82 +12 +72 | M1 | Prod of mods. At least one methodically correct.
−44 −44 −4
cosθ= = =
40× 114 4 285 √11 | M1 | All linked correctly and inverse cos used correctly
θ=130.7° or 2.28(05) rads | A1 | No other angles between 0° and 180°
4
Question | Answer | Marks | Guidance
Two vectors, $\mathbf{u}$ and $\mathbf{v}$, are such that
$$\mathbf{u} = \begin{pmatrix} q \\ 1 \\ 6 \end{pmatrix} \quad \text{and} \quad \mathbf{v} = \begin{pmatrix} 8 \\ q - 1 \\ q^2 - 7 \end{pmatrix},$$
where $q$ is a constant.

\begin{enumerate}[label=(\roman*)]
\item Find the values of $q$ for which $\mathbf{u}$ is perpendicular to $\mathbf{v}$. [3]
\item Find the angle between $\mathbf{u}$ and $\mathbf{v}$ when $q = 0$. [4]
\end{enumerate}

\hfill \mbox{\textit{CAIE P1 2019 Q5 [7]}}
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