| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2017 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Geometric configuration of planes |
| Difficulty | Standard +0.8 This is a multi-part vectors question requiring: (i) angle between planes using normal vectors (standard technique), and (ii) finding a point on the line of intersection, then constructing a plane perpendicular to both given planes using the cross product of normals. Part (ii) requires systematic problem-solving across multiple steps—finding the intersection line, locating point A, computing the cross product, and forming the plane equation. This is above-average difficulty for A-level but uses standard Further Maths techniques without requiring novel insight. |
| Spec | 4.04b Plane equations: cartesian and vector forms4.04c Scalar product: calculate and use for angles4.04d Angles: between planes and between line and plane |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| State or imply a correct normal vector to either plane, e.g. \(\mathbf{i}+\mathbf{j}+3\mathbf{k}\) or \(2\mathbf{i}-2\mathbf{j}+\mathbf{k}\) | B1 | |
| Carry out correct process for evaluating the scalar product of two normal vectors | M1 | |
| Using the correct process for the moduli, divide the scalar product of the two normals by the product of their moduli and evaluate the inverse cosine of the result | M1 | |
| Obtain final answer \(72.5°\) or \(1.26\) radians | A1 | |
| Total | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| *EITHER:* Substitute \(y=2\) in both plane equations and solve for \(x\) or for \(z\) | (M1 | |
| Obtain \(x=3\) and \(z=1\) | A1) | |
| *OR:* Find the equation of the line of intersection of the planes | ||
| Substitute \(y=2\) in line equation and solve for \(x\) or for \(z\) | (M1 | |
| Obtain \(x=3\) and \(z=1\) | A1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| *EITHER:* Use scalar product to obtain an equation in \(a\), \(b\) and \(c\), e.g. \(a+b+3c=0\) | (B1 | |
| Form a second relevant equation, e.g. \(2a-2b+c=0\), and solve for one ratio, e.g. \(a:b\) | *M1 | |
| Obtain final answer \(a:b:c = 7:5:-4\) | A1 | |
| Use coordinates of \(A\) and values of \(a\), \(b\) and \(c\) in general equation and find \(d\) | DM1 | |
| Obtain answer \(7x+5y-4z=27\), or equivalent | A1 FT) | |
| *OR1:* Calculate the vector product of relevant vectors, e.g. \((\mathbf{i}+\mathbf{j}+3\mathbf{k})\times(2\mathbf{i}-2\mathbf{j}+\mathbf{k})\) | (*M1 | |
| Obtain two correct components | A1 | |
| Obtain correct answer, e.g. \(7\mathbf{i}+5\mathbf{j}-4\mathbf{k}\) | A1 | |
| Substitute coordinates of \(A\) in plane equation with their normal and find \(d\) | DM1 | |
| Obtain answer \(7x+5y-4z=27\), or equivalent | A1 FT) | |
| *OR2:* Using relevant vectors, form a two-parameter equation for the plane | (*M1 | |
| State a correct equation, e.g. \(\mathbf{r}=3\mathbf{i}+2\mathbf{j}+\mathbf{k}+\lambda(\mathbf{i}+\mathbf{j}+3\mathbf{k})+\mu(2\mathbf{i}-2\mathbf{j}+\mathbf{k})\) | A1 FT | |
| State 3 correct equations in \(x\), \(y\), \(z\), \(\lambda\) and \(\mu\) | A1 FT | |
| Eliminate \(\lambda\) and \(\mu\) | DM1 | |
| Obtain answer \(7x+5y-4z=27\), or equivalent | A1 FT) | |
| *OR3:* Use the direction vector of the line of intersection of the two planes as normal vector to the plane | (*M1 | |
| Two correct components | A1 | |
| Three correct components | A1 | |
| Substitute coordinates of \(A\) in plane equation with their normal and find \(d\) | DM1 | |
| Obtain answer \(7x+5y-4z=27\), or equivalent | A1 FT) | |
| Total | 7 |
# Question 10(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| State or imply a correct normal vector to either plane, e.g. $\mathbf{i}+\mathbf{j}+3\mathbf{k}$ or $2\mathbf{i}-2\mathbf{j}+\mathbf{k}$ | B1 | |
| Carry out correct process for evaluating the scalar product of two normal vectors | M1 | |
| Using the correct process for the moduli, divide the scalar product of the two normals by the product of their moduli and evaluate the inverse cosine of the result | M1 | |
| Obtain final answer $72.5°$ or $1.26$ radians | A1 | |
| **Total** | **4** | |
---
# Question 10(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| *EITHER:* Substitute $y=2$ in both plane equations and solve for $x$ or for $z$ | (M1 | |
| Obtain $x=3$ and $z=1$ | A1) | |
| *OR:* Find the equation of the line of intersection of the planes | | |
| Substitute $y=2$ in line equation and solve for $x$ or for $z$ | (M1 | |
| Obtain $x=3$ and $z=1$ | A1) | |
---
# Question 10(iii) [continued]:
| Answer | Mark | Guidance |
|--------|------|----------|
| *EITHER:* Use scalar product to obtain an equation in $a$, $b$ and $c$, e.g. $a+b+3c=0$ | (B1 | |
| Form a second relevant equation, e.g. $2a-2b+c=0$, and solve for one ratio, e.g. $a:b$ | *M1 | |
| Obtain final answer $a:b:c = 7:5:-4$ | A1 | |
| Use coordinates of $A$ and values of $a$, $b$ and $c$ in general equation and find $d$ | DM1 | |
| Obtain answer $7x+5y-4z=27$, or equivalent | A1 FT) | |
| *OR1:* Calculate the vector product of relevant vectors, e.g. $(\mathbf{i}+\mathbf{j}+3\mathbf{k})\times(2\mathbf{i}-2\mathbf{j}+\mathbf{k})$ | (*M1 | |
| Obtain two correct components | A1 | |
| Obtain correct answer, e.g. $7\mathbf{i}+5\mathbf{j}-4\mathbf{k}$ | A1 | |
| Substitute coordinates of $A$ in plane equation with their normal and find $d$ | DM1 | |
| Obtain answer $7x+5y-4z=27$, or equivalent | A1 FT) | |
| *OR2:* Using relevant vectors, form a two-parameter equation for the plane | (*M1 | |
| State a correct equation, e.g. $\mathbf{r}=3\mathbf{i}+2\mathbf{j}+\mathbf{k}+\lambda(\mathbf{i}+\mathbf{j}+3\mathbf{k})+\mu(2\mathbf{i}-2\mathbf{j}+\mathbf{k})$ | A1 FT | |
| State 3 correct equations in $x$, $y$, $z$, $\lambda$ and $\mu$ | A1 FT | |
| Eliminate $\lambda$ and $\mu$ | DM1 | |
| Obtain answer $7x+5y-4z=27$, or equivalent | A1 FT) | |
| *OR3:* Use the direction vector of the line of intersection of the two planes as normal vector to the plane | (*M1 | |
| Two correct components | A1 | |
| Three correct components | A1 | |
| Substitute coordinates of $A$ in plane equation with their normal and find $d$ | DM1 | |
| Obtain answer $7x+5y-4z=27$, or equivalent | A1 FT) | |
| **Total** | **7** | |10 Two planes $p$ and $q$ have equations $x + y + 3 z = 8$ and $2 x - 2 y + z = 3$ respectively.\\
(i) Calculate the acute angle between the planes $p$ and $q$.\\
(ii) The point $A$ on the line of intersection of $p$ and $q$ has $y$-coordinate equal to 2 . Find the equation of the plane which contains the point $A$ and is perpendicular to both the planes $p$ and $q$. Give your answer in the form $a x + b y + c z = d$.\\
\hfill \mbox{\textit{CAIE P3 2017 Q10 [11]}}