CAIE P3 2018 November — Question 10 11 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2018
SessionNovember
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Cross Product & Distances
TypeIntersection of line with plane
DifficultyStandard +0.8 This is a multi-part 3D vector geometry question requiring: (i) line-plane intersection (substitution into plane equation), (ii) angle calculation using dot product with correct formula for line-plane angle, and (iii) finding a perpendicular line in a plane—requiring cross product to find direction and verification of both perpendicularity and plane containment. Part (iii) requires synthesis of multiple concepts and careful algebraic manipulation, elevating this above routine exercises.
Spec4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms4.04e Line intersections: parallel, skew, or intersecting4.04f Line-plane intersection: find point

The line \(l\) has equation \(\mathbf{r} = 5\mathbf{i} - 3\mathbf{j} - \mathbf{k} + \lambda(\mathbf{i} - 2\mathbf{j} + \mathbf{k})\). The plane \(p\) has equation $$(\mathbf{r} - \mathbf{i} - 2\mathbf{j}) \cdot (3\mathbf{i} + \mathbf{j} + \mathbf{k}) = 0.$$ The line \(l\) intersects the plane \(p\) at the point \(A\).
  1. Find the position vector of \(A\). [3]
  2. Calculate the acute angle between \(l\) and \(p\). [4]
  3. Find the equation of the line which lies in \(p\) and intersects \(l\) at right angles. [4]

Question 10:

AnswerMarks Guidance
10(i)Substitute for r and expand the scalar product to obtain an equation
in λM1* e.g. 3 ( 5+λ )+( −3−2λ )+( −1+λ )=5 ( 2λ=5−11 )
or 3 ( 4+λ )+1 ( −5−2λ )+( −1+λ )=0
Must attempt to deal with i + 2j
AnswerMarks
Solve a linear equation for λM1(dep*)
Obtain λ = – 3 and position vector r = 2i +3j – 4k for A
AnswerMarks Guidance
AA1 Accept coordinates
3

AnswerMarks Guidance
10(ii)State or imply a normal vector of p is 3i + j + k, or equivalent B1
Use correct method to evaluate a scalar product of relevant vectors
AnswerMarks
e.g. (i – 2j + k).(3i + j + k)M1
Using the correct process for calculating the moduli, divide the
scalar product by the product of the moduli and evaluate the inverse
AnswerMarks Guidance
sine or cosine of the resultM1 2
cosθ=
6 11
Second M1 available if working with the wrong vectors
AnswerMarks Guidance
Obtain answer 14.3° or 0.249 radiansA1 Or better
QuestionAnswer Marks
10(ii)Alternative 1
Use of a point on l and Cartesian equation 3x+ y+z=5 to find
( )
distance of point from plane e.g.B 5,−3,−1
d = 3×5−3−1−5
AnswerMarks
9+1+1M1
= 6 (=1.809... )
AnswerMarks
11A1
d
Complete method to find angle e.g. sinθ=
AnswerMarks
ABM1
 6 
θ=sin−1 =0.249
 
AnswerMarks Guidance
 11 54A1 Or better
Alternative 2
AnswerMarks
State or imply a normal vector of p is 3i + j + k, or equivalentB1
Use correct method to evaluate a vector product of relevant vectors
AnswerMarks Guidance
e.g. (i – 2j + k)x(3i + j + k)M1 3i−2j+7k
Using the correct process for calculating the moduli, divide the
vector product by the product of the moduli and evaluate the inverse
AnswerMarks Guidance
sine or cosine of the resultM1 32 +22 +72
sinθ= .
11 6
Second M1 available if working with the wrong vectors
AnswerMarks Guidance
Obtain answer 14.3° or 0.249 radiansA1 Or better
4
AnswerMarks Guidance
QuestionAnswer Marks

AnswerMarks
10(iii)Taking the direction vector of the line to be ai + bj + ck , state a
relevant equation in a, b, c, e.g. 3a + b + c = 0B1
State a second relevant equation, e.g. a – 2b + c = 0, and solve for
AnswerMarks Guidance
one ratio, e.g. a : bM1
Obtain a : b : c = 3 : – 2 : – 7, or equivalentA1
State answer r = 2i + 3j – 4k + µ (3i – 2j – 7k)A1ft Or equivalent. The f.t. is on r Requires ‘r = ….’
A
Alternative
Attempt to calculate the vector product of relevant vectors, e.g.
AnswerMarks Guidance
(3i + j + k) × (i – 2j + k)M1
Obtain two correct components of the productA1
Obtain correct product, e.g. 3i – 2j – 7kA1
State answer r = 2i + 3j – 4k + µ ( 3i – 2j – 7k)A1ft Or equivalent. The f.t. is on r Requires “ r = ….”
A.
4
Question 10:
--- 10(i) ---
10(i) | Substitute for r and expand the scalar product to obtain an equation
in λ | M1* | e.g. 3 ( 5+λ )+( −3−2λ )+( −1+λ )=5 ( 2λ=5−11 )
or 3 ( 4+λ )+1 ( −5−2λ )+( −1+λ )=0
Must attempt to deal with i + 2j
Solve a linear equation for λ | M1(dep*)
Obtain λ = – 3 and position vector r = 2i +3j – 4k for A
A | A1 | Accept coordinates
3
--- 10(ii) ---
10(ii) | State or imply a normal vector of p is 3i + j + k, or equivalent | B1
Use correct method to evaluate a scalar product of relevant vectors
e.g. (i – 2j + k).(3i + j + k) | M1
Using the correct process for calculating the moduli, divide the
scalar product by the product of the moduli and evaluate the inverse
sine or cosine of the result | M1 | 2
cosθ=
6 11
Second M1 available if working with the wrong vectors
Obtain answer 14.3° or 0.249 radians | A1 | Or better
Question | Answer | Marks | Guidance
10(ii) | Alternative 1
Use of a point on l and Cartesian equation 3x+ y+z=5 to find
( )
distance of point from plane e.g.B 5,−3,−1
d = 3×5−3−1−5
9+1+1 | M1
= 6 (=1.809... )
11 | A1
d
Complete method to find angle e.g. sinθ=
AB | M1
 6 
θ=sin−1 =0.249
 
 11 54 | A1 | Or better
Alternative 2
State or imply a normal vector of p is 3i + j + k, or equivalent | B1
Use correct method to evaluate a vector product of relevant vectors
e.g. (i – 2j + k)x(3i + j + k) | M1 | 3i−2j+7k
Using the correct process for calculating the moduli, divide the
vector product by the product of the moduli and evaluate the inverse
sine or cosine of the result | M1 | 32 +22 +72
sinθ= .
11 6
Second M1 available if working with the wrong vectors
Obtain answer 14.3° or 0.249 radians | A1 | Or better
4
Question | Answer | Marks | Guidance
--- 10(iii) ---
10(iii) | Taking the direction vector of the line to be ai + bj + ck , state a
relevant equation in a, b, c, e.g. 3a + b + c = 0 | B1
State a second relevant equation, e.g. a – 2b + c = 0, and solve for
one ratio, e.g. a : b | M1
Obtain a : b : c = 3 : – 2 : – 7, or equivalent | A1
State answer r = 2i + 3j – 4k + µ (3i – 2j – 7k) | A1ft | Or equivalent. The f.t. is on r Requires ‘r = ….’
A
Alternative
Attempt to calculate the vector product of relevant vectors, e.g.
(3i + j + k) × (i – 2j + k) | M1
Obtain two correct components of the product | A1
Obtain correct product, e.g. 3i – 2j – 7k | A1
State answer r = 2i + 3j – 4k + µ ( 3i – 2j – 7k) | A1ft | Or equivalent. The f.t. is on r Requires “ r = ….”
A.
4
The line $l$ has equation $\mathbf{r} = 5\mathbf{i} - 3\mathbf{j} - \mathbf{k} + \lambda(\mathbf{i} - 2\mathbf{j} + \mathbf{k})$. The plane $p$ has equation
$$(\mathbf{r} - \mathbf{i} - 2\mathbf{j}) \cdot (3\mathbf{i} + \mathbf{j} + \mathbf{k}) = 0.$$
The line $l$ intersects the plane $p$ at the point $A$.

\begin{enumerate}[label=(\roman*)]
\item Find the position vector of $A$. [3]

\item Calculate the acute angle between $l$ and $p$. [4]

\item Find the equation of the line which lies in $p$ and intersects $l$ at right angles. [4]
\end{enumerate}

\hfill \mbox{\textit{CAIE P3 2018 Q10 [11]}}