Edexcel F3 2021 June — Question 6 13 marks

Exam BoardEdexcel
ModuleF3 (Further Pure Mathematics 3)
Year2021
SessionJune
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Lines & Planes
TypeLine of intersection of planes
DifficultyStandard +0.8 This is a multi-part Further Maths question on 3D vector geometry requiring finding a plane equation from two lines (using cross product of direction vectors), determining a parameter value, finding line of intersection of two planes, and calculating dihedral angle. While systematic, it requires confident manipulation of vectors in 3D, cross products, and solving simultaneous equations—more demanding than standard C4 vector work but follows established techniques without requiring novel insight.
Spec4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms4.04c Scalar product: calculate and use for angles4.04d Angles: between planes and between line and plane
The line \(l_1\) has equation $$\mathbf{r} = \mathbf{i} + \mathbf{j} + \mathbf{k} + \lambda(\mathbf{i} + 3\mathbf{k})$$ and the line \(l_2\) has equation $$\mathbf{r} = 2\mathbf{i} + s\mathbf{j} + \mu(\mathbf{i} - 2\mathbf{j} + \mathbf{k})$$ where \(s\) is a constant and \(\lambda\) and \(\mu\) are scalar parameters. Given that \(l_1\) and \(l_2\) both lie in a common plane \(\Pi_1\)
  1. show that an equation for \(\Pi_1\) is \(3x + y - z = 3\) [4]
  2. Find the value of \(s\). [1]
The plane \(\Pi_2\) has equation \(\mathbf{r} \cdot (\mathbf{i} + \mathbf{j} - 2\mathbf{k}) = 3\)
  1. Find an equation for the line of intersection of \(\Pi_1\) and \(\Pi_2\) [4]
  2. Find the acute angle between \(\Pi_1\) and \(\Pi_2\) giving your answer in degrees to 3 significant figures. [4]
Question 6:
AnswerMarks Guidance
6(b)s = −3 cao
(1)
AnswerMarks
6(c)i j k
1 1 −2 =i−5j−2k
AnswerMarks
3 1 −1Attempts cross product of normal vectors.
If the method is unclear, look for at least 2
AnswerMarks
correct components.M1
e.g. x=0,2y−2z =6,y−2z =3
AnswerMarks Guidance
⇒ y =3,z =0Any valid attempt to find a point on the
line.M1
e.g. (0,3,0)Any valid point on the line A1
r =3j+λ(i−5j−2k)Correct equation including “r =” or
y−3 z
equivalent e.g. x= =
AnswerMarks
−5 −2A1
(4)
6(c)
AnswerMarks
ALT 1r =i+j+k+λ(i+3k)+µ(i−2j+k), r.(i+j−2k)=3
⇒1+λ+µ+1−2µ−2−6λ−2µ=3
Forms equation of first plane using (1, 1, 1) and direction vectors and substitutes into the
AnswerMarks
second plane to form an equation in λ and µM1
1
⇒µ= (−5λ−3)
AnswerMarks
3Solves to obtain µ in terms of λ or λ in terms
of µM1
Correct equationA1
1
E.g. r=i+j+k+λ(i+3k)+ (−5λ−3)(i−2j+k)
3
AnswerMarks
Correct equation including “r =”A1
6(c)
AnswerMarks Guidance
ALT 23x+ y−z =3, x+ y−2z =3⇒2x+z =0 Uses the Cartesian equations of both
planes and eliminates one variableM1
1 5
z =λ⇒ x =− λ, y =3+2z−x =3+ λ
AnswerMarks
2 2Introduces parameter and expresses other
2 variables in terms of the parameterM1
Correct equationsA1
r =3j+λ(i−5j−2k)Correct equation including “r =” or
y−3 z
equivalent e.g. x= =
AnswerMarks
−5 −2A1
6(c)
AnswerMarks Guidance
ALT 33x+ y−z =3, x+ y−2z =3⇒2x+z =0 Uses the Cartesian equations of both
planes and eliminates one variableM1
3x+ y−z =3, x+ y−2z =3⇒5x+ y =3Uses the Cartesian equations of both
planes and eliminates another variableM1
z 3− y
⇒ x =− , x =
AnswerMarks
2 5Correct equations for one variable in
terms of the other 2A1
y−3 z
x= =
AnswerMarks
−5 −2Correct equation or equivalent e.g.
3− y z
x = =
AnswerMarks
5 −2A1
AnswerMarks Guidance
6(d)( 3i+j−k )  ( i+j−2k )=6 Correct value for scalar product
(3i+j−k).(i+j−2k) 6
cosθ= =
AnswerMarks Guidance
9+1+1 1+1+4 11Full scalar product attempt to reach a value
for cosθM1
For cosθ=6
11A1
θ= 42.4°Correct value. Mark their final answer. A1
(4)
6(d)
AnswerMarks Guidance
ALT( 3i+j−k )×( i+j−2k ) = 30 Correct value for magnitude of cross product
(3i+j−k).(i+j−2k) 55
sinθ= =
AnswerMarks Guidance
9+1+1 1+1+4 11Full attempt to reach a value for sinθ M1
55
For sinθ=
AnswerMarks Guidance
11A1
θ= 42.4°Correct value. Mark their final answer. A1

Total 13

Question 6:
--- 6(b) ---
6(b) | s = −3 | cao | B1
(1)
--- 6(c) ---
6(c) | i j k
1 1 −2 =i−5j−2k
3 1 −1 | Attempts cross product of normal vectors.
If the method is unclear, look for at least 2
correct components. | M1
e.g. x=0,2y−2z =6,y−2z =3
⇒ y =3,z =0 | Any valid attempt to find a point on the
line. | M1
e.g. (0,3,0) | Any valid point on the line | A1
r =3j+λ(i−5j−2k) | Correct equation including “r =” or
y−3 z
equivalent e.g. x= =
−5 −2 | A1
(4)
6(c)
ALT 1 | r =i+j+k+λ(i+3k)+µ(i−2j+k), r.(i+j−2k)=3
⇒1+λ+µ+1−2µ−2−6λ−2µ=3
Forms equation of first plane using (1, 1, 1) and direction vectors and substitutes into the
second plane to form an equation in λ and µ | M1
1
⇒µ= (−5λ−3)
3 | Solves to obtain µ in terms of λ or λ in terms
of µ | M1
Correct equation | A1
1
E.g. r=i+j+k+λ(i+3k)+ (−5λ−3)(i−2j+k)
3
Correct equation including “r =” | A1
6(c)
ALT 2 | 3x+ y−z =3, x+ y−2z =3⇒2x+z =0 | Uses the Cartesian equations of both
planes and eliminates one variable | M1
1 5
z =λ⇒ x =− λ, y =3+2z−x =3+ λ
2 2 | Introduces parameter and expresses other
2 variables in terms of the parameter | M1
Correct equations | A1
r =3j+λ(i−5j−2k) | Correct equation including “r =” or
y−3 z
equivalent e.g. x= =
−5 −2 | A1
6(c)
ALT 3 | 3x+ y−z =3, x+ y−2z =3⇒2x+z =0 | Uses the Cartesian equations of both
planes and eliminates one variable | M1
3x+ y−z =3, x+ y−2z =3⇒5x+ y =3 | Uses the Cartesian equations of both
planes and eliminates another variable | M1
z 3− y
⇒ x =− , x =
2 5 | Correct equations for one variable in
terms of the other 2 | A1
y−3 z
x= =
−5 −2 | Correct equation or equivalent e.g.
3− y z
x = =
5 −2 | A1
--- 6(d) ---
6(d) | ( 3i+j−k )  ( i+j−2k )=6 | Correct value for scalar product | B1
(3i+j−k).(i+j−2k) 6
cosθ= =
9+1+1 1+1+4 11 | Full scalar product attempt to reach a value
for cosθ | M1
For cosθ= | 6
11 | A1
θ= 42.4° | Correct value. Mark their final answer. | A1
(4)
6(d)
ALT | ( 3i+j−k )×( i+j−2k ) = 30 | Correct value for magnitude of cross product | B1
(3i+j−k).(i+j−2k) 55
sinθ= =
9+1+1 1+1+4 11 | Full attempt to reach a value for sinθ | M1
55
For sinθ=
11 | A1
θ= 42.4° | Correct value. Mark their final answer. | A1
Total 13
The line $l_1$ has equation
$$\mathbf{r} = \mathbf{i} + \mathbf{j} + \mathbf{k} + \lambda(\mathbf{i} + 3\mathbf{k})$$

and the line $l_2$ has equation
$$\mathbf{r} = 2\mathbf{i} + s\mathbf{j} + \mu(\mathbf{i} - 2\mathbf{j} + \mathbf{k})$$

where $s$ is a constant and $\lambda$ and $\mu$ are scalar parameters.

Given that $l_1$ and $l_2$ both lie in a common plane $\Pi_1$

\begin{enumerate}[label=(\alph*)]
\item show that an equation for $\Pi_1$ is $3x + y - z = 3$
[4]

\item Find the value of $s$.
[1]
\end{enumerate}

The plane $\Pi_2$ has equation $\mathbf{r} \cdot (\mathbf{i} + \mathbf{j} - 2\mathbf{k}) = 3$

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find an equation for the line of intersection of $\Pi_1$ and $\Pi_2$
[4]

\item Find the acute angle between $\Pi_1$ and $\Pi_2$ giving your answer in degrees to 3 significant figures.
[4]
\end{enumerate}

\hfill \mbox{\textit{Edexcel F3 2021 Q6 [13]}}
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