| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2018 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Intersection of line with plane |
| Difficulty | Standard +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. |
| Spec | 4.04a 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 |
| Answer | Marks | Guidance |
|---|---|---|
| 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 ) |
| Answer | Marks |
|---|---|
| Solve a linear equation for λ | M1(dep*) |
| Answer | Marks | Guidance |
|---|---|---|
| A | A1 | Accept coordinates |
| Answer | Marks | Guidance |
|---|---|---|
| 10(ii) | State or imply a normal vector of p is 3i + j + k, or equivalent | B1 |
| Answer | Marks |
|---|---|
| e.g. (i – 2j + k).(3i + j + k) | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| sine or cosine of the result | M1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Obtain answer 14.3° or 0.249 radians | A1 | Or better |
| Question | Answer | Marks |
| 10(ii) | Alternative 1 |
| Answer | Marks |
|---|---|
| 9+1+1 | M1 |
| Answer | Marks |
|---|---|
| 11 | A1 |
| Answer | Marks |
|---|---|
| AB | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 11 54 | A1 | Or better |
| Answer | Marks |
|---|---|
| State or imply a normal vector of p is 3i + j + k, or equivalent | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| e.g. (i – 2j + k)x(3i + j + k) | M1 | 3i−2j+7k |
| Answer | Marks | Guidance |
|---|---|---|
| sine or cosine of the result | M1 | 32 +22 +72 |
| Answer | Marks | Guidance |
|---|---|---|
| Obtain answer 14.3° or 0.249 radians | A1 | Or better |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks |
|---|---|
| 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 |
| Answer | Marks | Guidance |
|---|---|---|
| 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 = ….’ |
| Answer | Marks | Guidance |
|---|---|---|
| (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 = ….” |
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]}}