| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2013 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Equation of plane through three points |
| Difficulty | Standard +0.3 This is a standard multi-part vectors question requiring routine techniques: finding a plane equation using cross product (standard P3 skill), using ratio theorem for position vectors (straightforward), and calculating perpendicular distance using the cross product formula. All parts follow textbook methods with no novel insight required, making it slightly easier than average. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10d Vector operations: addition and scalar multiplication4.04b Plane equations: cartesian and vector forms4.04c Scalar product: calculate and use for angles |
| Answer | Marks | Guidance |
|---|---|---|
| (i) EITHER: Obtain a vector parallel to the plane, e.g. \(\overrightarrow{AB} = -2\mathbf{i} + 4\mathbf{j} - \mathbf{k}\) | B1 | |
| Use scalar product to obtain an equation in \(a, b, c\), e.g. \(-2a+4b-c=0\), \(3a-3b+3c=0\), or \(a+b+2c=0\) | M1 | |
| Obtain two correct equations in \(a, b, c\) | A1 | |
| Solve to obtain ratio \(a : b : c\) | M1 | |
| Obtain \(a : b : c = 3:1:-2\), or equivalent | A1 | |
| Obtain equation \(3x + y - 2z = 1\), or equivalent | A1 | |
| OR1: Substitute for two points, e.g. \(A\) and \(B\), and obtain \(2a-b+2c=d\) and \(3b+c=d\) | B1 | |
| Substitute for another point, e.g. \(C\), to obtain a third equation and eliminate one unknown entirely from the three equations | M1 | |
| Obtain two correct equations in three unknowns, e.g. in \(a, b, c\) | A1 | |
| Solve to obtain their ratio, e.g. \(a : b : c\) | M1 | |
| Obtain \(a : b : c = 3:1:-2\), \(a:c:d = 3:-2:1\) or \(b:c:d = 1:-2:1\), or equivalent | A1 | |
| Obtain equation \(3x + y - 2z = 1\), or equivalent | A1 | |
| OR2: Obtain a vector parallel to the plane, e.g. \(\overrightarrow{AC} = \mathbf{i} + \mathbf{j} + 2\mathbf{k}\) | B1 | |
| Obtain a second such vector and calculate their vector product e.g. \((-2\mathbf{i}+4\mathbf{j}-\mathbf{k})\times(3\mathbf{i}-3\mathbf{j}+3\mathbf{k})\) | M1 | |
| Obtain two correct components of the product | A1 | |
| Obtain correct answer, e.g. \(9\mathbf{i} + 3\mathbf{j} - 6\mathbf{k}\) | A1 | |
| Substitute in \(9x + 3y - 6z = d\) to find \(d\) | M1 | |
| Obtain equation \(3x + y - 2z = 1\), or equivalent | A1 | |
| OR3: Obtain a vector parallel to the plane, e.g. \(\overrightarrow{AC} = \mathbf{i} + \mathbf{j} + 2\mathbf{k}\) | B1 | |
| Obtain a second such vector and form correctly a 2-parameter equation for the plane | M1 | |
| Obtain a correct equation, e.g. \(\mathbf{r} = 3\mathbf{i} + 4\mathbf{k} + \lambda(-2\mathbf{i}+4\mathbf{j}-\mathbf{k}) + \mu(\mathbf{i}+\mathbf{j}+2\mathbf{k})\) | A1 | |
| State three correct equations in \(x, y, z, \lambda, \mu\) | A1 | |
| Eliminate \(\lambda\) and \(\mu\) | M1 | |
| Obtain equation \(3x + y - 2z = 1\), or equivalent | A1 | [6] |
| (ii) Obtain answer \(\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}\), or equivalent | B1 | [1] |
| (iii) EITHER: Use \(\frac{\overrightarrow{OA}.\overrightarrow{OD}}{ | \overrightarrow{OD} | }\) to find projection \(ON\) of \(OA\) onto \(OD\) |
| Obtain \(ON = \frac{4}{3}\) | A1 | |
| Use Pythagoras in triangle \(OAN\) to find \(AN\) | M1 | |
| Obtain the given answer | A1 | |
| OR1: Calculate the vector product of \(\overrightarrow{OA}\) and \(\overrightarrow{OD}\) | M1 | |
| Obtain answer \(6\mathbf{i} + 2\mathbf{j} - 5\mathbf{k}\) | A1 | |
| Divide the modulus of the vector product by the modulus of \(\overrightarrow{OD}\) | M1 | |
| Obtain the given answer | A1 | |
| OR2: Taking general point \(P\) of \(OD\) to have position vector \(\lambda(2\mathbf{i}+2\mathbf{j}+2\mathbf{k})\), form an equation in \(\lambda\) by either equating the scalar product of \(\overrightarrow{AP}\) and \(\overrightarrow{OP}\) to zero, or using Pythagoras in triangle \(OPA\), or setting the derivative of \( | \overrightarrow{AP} | \) to zero |
| Solve and obtain \(\lambda = \frac{4}{9}\) | A1 | |
| Carry out method to calculate \(AP\) when \(\lambda = \frac{4}{9}\) | M1 | |
| Obtain the given answer | A1 | |
| OR3: Use a relevant scalar product to find the cosine of \(AOD\) or \(ADO\) | M1 | |
| Obtain \(\cos AOD = \frac{4}{9}\) or \(\cos ADO = \frac{5}{3\sqrt{10}}\), or equivalent | A1 | |
| Use trig to find the length of the perpendicular | M1 | |
| Obtain the given answer | A1 | |
| OR4: Use the cosine formula in triangle \(AOD\) to find \(\cos AOD\) or \(\cos ADO\) | M1 | |
| Obtain \(\cos AOD = \frac{8}{18}\) or \(\cos ADO = \frac{10}{6\sqrt{10}}\), or equivalent | A1 | |
| Use trig to find the length of the perpendicular | M1 | |
| Obtain the given answer | A1 | [4] |
**(i)** **EITHER:** Obtain a vector parallel to the plane, e.g. $\overrightarrow{AB} = -2\mathbf{i} + 4\mathbf{j} - \mathbf{k}$ | B1 |
Use scalar product to obtain an equation in $a, b, c$, e.g. $-2a+4b-c=0$, $3a-3b+3c=0$, or $a+b+2c=0$ | M1 |
Obtain two correct equations in $a, b, c$ | A1 |
Solve to obtain ratio $a : b : c$ | M1 |
Obtain $a : b : c = 3:1:-2$, or equivalent | A1 |
Obtain equation $3x + y - 2z = 1$, or equivalent | A1 |
**OR1:** Substitute for two points, e.g. $A$ and $B$, and obtain $2a-b+2c=d$ and $3b+c=d$ | B1 |
Substitute for another point, e.g. $C$, to obtain a third equation and eliminate one unknown entirely from the three equations | M1 |
Obtain two correct equations in three unknowns, e.g. in $a, b, c$ | A1 |
Solve to obtain their ratio, e.g. $a : b : c$ | M1 |
Obtain $a : b : c = 3:1:-2$, $a:c:d = 3:-2:1$ or $b:c:d = 1:-2:1$, or equivalent | A1 |
Obtain equation $3x + y - 2z = 1$, or equivalent | A1 |
**OR2:** Obtain a vector parallel to the plane, e.g. $\overrightarrow{AC} = \mathbf{i} + \mathbf{j} + 2\mathbf{k}$ | B1 |
Obtain a second such vector and calculate their vector product e.g. $(-2\mathbf{i}+4\mathbf{j}-\mathbf{k})\times(3\mathbf{i}-3\mathbf{j}+3\mathbf{k})$ | M1 |
Obtain two correct components of the product | A1 |
Obtain correct answer, e.g. $9\mathbf{i} + 3\mathbf{j} - 6\mathbf{k}$ | A1 |
Substitute in $9x + 3y - 6z = d$ to find $d$ | M1 |
Obtain equation $3x + y - 2z = 1$, or equivalent | A1 |
**OR3:** Obtain a vector parallel to the plane, e.g. $\overrightarrow{AC} = \mathbf{i} + \mathbf{j} + 2\mathbf{k}$ | B1 |
Obtain a second such vector and form correctly a 2-parameter equation for the plane | M1 |
Obtain a correct equation, e.g. $\mathbf{r} = 3\mathbf{i} + 4\mathbf{k} + \lambda(-2\mathbf{i}+4\mathbf{j}-\mathbf{k}) + \mu(\mathbf{i}+\mathbf{j}+2\mathbf{k})$ | A1 |
State three correct equations in $x, y, z, \lambda, \mu$ | A1 |
Eliminate $\lambda$ and $\mu$ | M1 |
Obtain equation $3x + y - 2z = 1$, or equivalent | A1 | [6]
**(ii)** Obtain answer $\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}$, or equivalent | B1 | [1]
**(iii)** **EITHER:** Use $\frac{\overrightarrow{OA}.\overrightarrow{OD}}{|\overrightarrow{OD}|}$ to find projection $ON$ of $OA$ onto $OD$ | M1 |
Obtain $ON = \frac{4}{3}$ | A1 |
Use Pythagoras in triangle $OAN$ to find $AN$ | M1 |
Obtain the given answer | A1 |
**OR1:** Calculate the vector product of $\overrightarrow{OA}$ and $\overrightarrow{OD}$ | M1 |
Obtain answer $6\mathbf{i} + 2\mathbf{j} - 5\mathbf{k}$ | A1 |
Divide the modulus of the vector product by the modulus of $\overrightarrow{OD}$ | M1 |
Obtain the given answer | A1 |
**OR2:** Taking general point $P$ of $OD$ to have position vector $\lambda(2\mathbf{i}+2\mathbf{j}+2\mathbf{k})$, form an equation in $\lambda$ by either equating the scalar product of $\overrightarrow{AP}$ and $\overrightarrow{OP}$ to zero, or using Pythagoras in triangle $OPA$, or setting the derivative of $|\overrightarrow{AP}|$ to zero | M1 |
Solve and obtain $\lambda = \frac{4}{9}$ | A1 |
Carry out method to calculate $AP$ when $\lambda = \frac{4}{9}$ | M1 |
Obtain the given answer | A1 |
**OR3:** Use a relevant scalar product to find the cosine of $AOD$ or $ADO$ | M1 |
Obtain $\cos AOD = \frac{4}{9}$ or $\cos ADO = \frac{5}{3\sqrt{10}}$, or equivalent | A1 |
Use trig to find the length of the perpendicular | M1 |
Obtain the given answer | A1 |
**OR4:** Use the cosine formula in triangle $AOD$ to find $\cos AOD$ or $\cos ADO$ | M1 |
Obtain $\cos AOD = \frac{8}{18}$ or $\cos ADO = \frac{10}{6\sqrt{10}}$, or equivalent | A1 |
Use trig to find the length of the perpendicular | M1 |
Obtain the given answer | A1 | [4]
\includegraphics{figure_9}
The diagram shows three points $A$, $B$ and $C$ whose position vectors with respect to the origin $O$ are given by $\overrightarrow{OA} = \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}$, $\overrightarrow{OB} = \begin{pmatrix} 0 \\ 3 \\ 1 \end{pmatrix}$ and $\overrightarrow{OC} = \begin{pmatrix} 3 \\ 0 \\ 4 \end{pmatrix}$. The point $D$ lies on $BC$, between $B$ and $C$, and is such that $CD = 2DB$.
\begin{enumerate}[label=(\roman*)]
\item Find the equation of the plane $ABC$, giving your answer in the form $ax + by + cz = d$. [6]
\item Find the position vector of $D$. [1]
\item Show that the length of the perpendicular from $A$ to $OD$ is $\frac{1}{3}\sqrt{(65)}$. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2013 Q9 [11]}}