| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2017 |
| Session | March |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Parallel and perpendicular planes |
| Difficulty | Standard +0.8 This is a multi-step Further Maths vectors question requiring: (i) verification that a line lies in a plane (checking point and direction vector), and (ii) finding a plane given three constraints (parallel to a line, perpendicular to another plane, through a point). Part (ii) requires finding a normal vector via cross product of two direction vectors, which is non-trivial but systematic. Slightly above average difficulty due to the conceptual understanding needed and multiple techniques, but follows standard Further Maths procedures. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms4.04f Line-plane intersection: find point |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Verify that the point with position vector \(\mathbf{i} + 2\mathbf{j} - 3\mathbf{k}\) lies in the plane | B1 | |
| EITHER: Find a second point on \(l\) and substitute its coordinates in the equation of \(p\) | (M1 | |
| Verify that the second point, e.g. \((3, 1, -2)\), lies in the plane | A1) | |
| OR: Expand scalar product of a normal to \(p\) and the direction vector of \(l\) | (M1 | |
| Verify scalar product is zero | A1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use scalar product to obtain relevant equation in \(a\), \(b\) and \(c\), e.g. \(2a - b + c = 0\) | B1 | |
| Obtain second equation, e.g. \(3a + b - 5c = 0\), and solve for one ratio e.g. \(a:b\) | M1 | |
| Obtain \(a:b:c = 4:13:5\), or equivalent | A1 | |
| Substitute \((3,-1,2)\) and values of \(a\), \(b\), \(c\) in general equation to find \(d\) | M1 | |
| Obtain \(4x + 13y + 5z = 9\), or equivalent | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Attempt to calculate vector product of relevant vectors, e.g. \((2\mathbf{i}-\mathbf{j}+\mathbf{k})\times(3\mathbf{i}+\mathbf{j}-5\mathbf{k})\) | M1 | |
| Obtain two correct components | A1 | |
| Obtain correct answer, e.g. \(4\mathbf{i}+13\mathbf{j}+5\mathbf{k}\) | A1 | |
| Substitute \((3,-1,2)\) in \(4x+13y+5z=d\), and find \(d\) | M1 | |
| Obtain \(4x+13y+5z=9\), or equivalent | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Using relevant point and vectors form a 2-parameter equation for the plane | M1 | |
| State correct equation, e.g. \(\mathbf{r} = 3\mathbf{i}-\mathbf{j}+2\mathbf{k}+\lambda(2\mathbf{i}-\mathbf{j}+\mathbf{k})+\mu(3\mathbf{i}+\mathbf{j}-5\mathbf{k})\) | A1 | |
| State three correct equations in \(x, y, z, \lambda\) and \(\mu\) | A1 | |
| Eliminate \(\lambda\) and \(\mu\) | M1 | |
| Obtain \(4x+13y+5z=9\), or equivalent | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Using relevant point and vectors form a determinant equation for the plane | M1 | |
| State correct equation, e.g. \(\begin{vmatrix} x-3 & y+1 & z-2 \\ 2 & -1 & 1 \\ 3 & 1 & -5 \end{vmatrix} = 0\) | A1 | |
| Attempt to expand the determinant | M1 | |
| Obtain or imply two correct cofactors | A1 | |
| Obtain \(4x+13y+5z=9\), or equivalent | A1 |
## Question 6(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Verify that the point with position vector $\mathbf{i} + 2\mathbf{j} - 3\mathbf{k}$ lies in the plane | B1 | |
| **EITHER:** Find a second point on $l$ and substitute its coordinates in the equation of $p$ | (M1 | |
| Verify that the second point, e.g. $(3, 1, -2)$, lies in the plane | A1) | |
| **OR:** Expand scalar product of a normal to $p$ and the direction vector of $l$ | (M1 | |
| Verify scalar product is zero | A1) | |
**Total: 3**
## Question 6(ii):
**EITHER method:**
| Answer | Mark | Guidance |
|--------|------|----------|
| Use scalar product to obtain relevant equation in $a$, $b$ and $c$, e.g. $2a - b + c = 0$ | B1 | |
| Obtain second equation, e.g. $3a + b - 5c = 0$, and solve for one ratio e.g. $a:b$ | M1 | |
| Obtain $a:b:c = 4:13:5$, or equivalent | A1 | |
| Substitute $(3,-1,2)$ and values of $a$, $b$, $c$ in general equation to find $d$ | M1 | |
| Obtain $4x + 13y + 5z = 9$, or equivalent | A1 | |
**OR1:**
| Answer | Mark | Guidance |
|--------|------|----------|
| Attempt to calculate vector product of relevant vectors, e.g. $(2\mathbf{i}-\mathbf{j}+\mathbf{k})\times(3\mathbf{i}+\mathbf{j}-5\mathbf{k})$ | M1 | |
| Obtain two correct components | A1 | |
| Obtain correct answer, e.g. $4\mathbf{i}+13\mathbf{j}+5\mathbf{k}$ | A1 | |
| Substitute $(3,-1,2)$ in $4x+13y+5z=d$, and find $d$ | M1 | |
| Obtain $4x+13y+5z=9$, or equivalent | A1 | |
**OR2:**
| Answer | Mark | Guidance |
|--------|------|----------|
| Using relevant point and vectors form a 2-parameter equation for the plane | M1 | |
| State correct equation, e.g. $\mathbf{r} = 3\mathbf{i}-\mathbf{j}+2\mathbf{k}+\lambda(2\mathbf{i}-\mathbf{j}+\mathbf{k})+\mu(3\mathbf{i}+\mathbf{j}-5\mathbf{k})$ | A1 | |
| State three correct equations in $x, y, z, \lambda$ and $\mu$ | A1 | |
| Eliminate $\lambda$ and $\mu$ | M1 | |
| Obtain $4x+13y+5z=9$, or equivalent | A1 | |
**OR3:**
| Answer | Mark | Guidance |
|--------|------|----------|
| Using relevant point and vectors form a determinant equation for the plane | M1 | |
| State correct equation, e.g. $\begin{vmatrix} x-3 & y+1 & z-2 \\ 2 & -1 & 1 \\ 3 & 1 & -5 \end{vmatrix} = 0$ | A1 | |
| Attempt to expand the determinant | M1 | |
| Obtain or imply two correct cofactors | A1 | |
| Obtain $4x+13y+5z=9$, or equivalent | A1 | |
**Total: 5**
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6 The line $l$ has equation $\mathbf { r } = \mathbf { i } + 2 \mathbf { j } - 3 \mathbf { k } + \lambda ( 2 \mathbf { i } - \mathbf { j } + \mathbf { k } )$. The plane $p$ has equation $3 x + y - 5 z = 20$.\\
(i) Show that the line $l$ lies in the plane $p$.\\
(ii) A second plane is parallel to $l$, perpendicular to $p$ and contains the point with position vector $3 \mathbf { i } - \mathbf { j } + 2 \mathbf { k }$. Find the equation of this plane, giving your answer in the form $a x + b y + c z = d$. [5]\\
\hfill \mbox{\textit{CAIE P3 2017 Q6 [8]}}