CAIE P3 2004 June — Question 11 12 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2004
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Lines & Planes
TypeCartesian equation of a plane
DifficultyStandard +0.3 This is a standard two-part vectors question requiring finding a plane equation via cross product of two vectors in the plane, then finding the foot of perpendicular using the parametric form of a line. Both are routine A-level Further Maths techniques with straightforward arithmetic, making it slightly easier than average.
Spec4.04b Plane equations: cartesian and vector forms4.04j Shortest distance: between a point and a plane

11 With respect to the origin \(O\), the points \(P , Q , R , S\) have position vectors given by $$\overrightarrow { O P } = \mathbf { i } - \mathbf { k } , \quad \overrightarrow { O Q } = - 2 \mathbf { i } + 4 \mathbf { j } , \quad \overrightarrow { O R } = 4 \mathbf { i } + 2 \mathbf { j } + \mathbf { k } , \quad \overrightarrow { O S } = 3 \mathbf { i } + 5 \mathbf { j } - 6 \mathbf { k } .$$
  1. Find the equation of the plane containing \(P , Q\) and \(R\), giving your answer in the form \(a x + b y + c z = d\).
  2. The point \(N\) is the foot of the perpendicular from \(S\) to this plane. Find the position vector of \(N\) and show that the length of \(S N\) is 7 .

Question 11:
AnswerMarks Guidance
Answer/WorkingMark Guidance
(i) EITHER: Obtain a vector in the plane e.g. \(\overrightarrow{PQ} = -3\mathbf{i} + 4\mathbf{j} + \mathbf{k}\)B1
Use scalar product to obtain relevant equation in \(a, b, c\) e.g. \(-3a + 4b + c = 0\) or \(6a - 2b + c = 0\) or \(3a + 2b + 2c = 0\)M1
State two correct equations in \(a, b, c\)A1
Solve simultaneous equations to obtain one ratio e.g. \(a : b\)M1
Obtain \(a : b : c = 2 : 3 : -6\) or equivalentA1
Obtain equation \(2x + 3y - 6z = 8\) or equivalentA1
OR: Substitute for \(P, Q, R\) in equation of plane and state 3 equations in \(a, b, c, d\)B1
Eliminate one unknown, e.g. \(d\), entirelyM1
Obtain 2 equations in 3 unknownsA1
Solve to obtain one ratio e.g. \(a : b\)M1
Obtain \(a : b : c = 2 : 3 : -6\) or equivalentA1
Obtain equation \(2x + 3y - 6z = 8\) or equivalentA1
OR: Obtain a vector in plane e.g. \(\overrightarrow{QR} = 6\mathbf{i} - 2\mathbf{j} + \mathbf{k}\)B1
Find second vector in plane and form correctly a 2-parameter equation for the planeM1
Obtain equation in any correct form e.g. \(\mathbf{r} = \lambda(-3\mathbf{i}+4\mathbf{j}+\mathbf{k}) + \mu(6\mathbf{i}-2\mathbf{j}+\mathbf{k}) + \mathbf{i} - \mathbf{k}\)A1
State 3 equations in \(x, y, z, \lambda\), and \(\mu\)A1
Eliminate \(\lambda\) and \(\mu\)M1
Obtain equation \(2x + 3y - 6z = 8\) or equivalentA1
OR: Obtain a vector in plane e.g. \(\overrightarrow{PR} = 3\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}\)B1
Obtain second vector in plane and calculate vector product of the two vectors, e.g. \((-3\mathbf{i}+4\mathbf{j}+\mathbf{k}) \times (3\mathbf{i}+2\mathbf{j}+2\mathbf{k})\)M1
Obtain 2 correct components of the productA1
Obtain correct product e.g. \(6\mathbf{i} + 9\mathbf{j} - 18\mathbf{k}\) or equivalentA1
Substitute in \(2x + 3y - 6z = d\) and find \(d\) or equivalentM1
Obtain equation \(2x + 3y - 6z = 8\) or equivalentA1 Total: 6
Question (ii) EITHER:
AnswerMarks Guidance
Answer/WorkingMark Guidance Notes
State equation of \(SN\) is \(\mathbf{r} = 3\mathbf{i} + 5\mathbf{j} - 6\mathbf{k} + \lambda(2\mathbf{i} + 3\mathbf{j} - 6\mathbf{k})\) or equivalent\(B1\sqrt{}\)
Express \(x, y, z\) in terms of \(\lambda\) e.g. \((3 + 2\lambda,\ 5 + 3\lambda,\ -6 - 6\lambda)\)\(B1\sqrt{}\)
Substitute in the equation of the plane and solve for \(\lambda\)\(M1\)
Obtain \(\overrightarrow{ON} = \mathbf{i} + 2\mathbf{j}\), or equivalent\(A1\)
Carry out method for finding \(SN\)\(M1\)
Show that \(SN = 7\) correctly\(A1\)
Question (ii) OR (Method 2):
AnswerMarks Guidance
Answer/WorkingMark Guidance Notes
Letting \(\overrightarrow{ON} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}\), obtain two equations in \(x, y, z\) by equating scalar product of \(\overrightarrow{NS}\) with two of \(\overrightarrow{PQ}, \overrightarrow{QR}, \overrightarrow{RP}\) to zero\(B1\sqrt{} + B1\sqrt{}\)
Using the plane equation as third equation, solve for \(x\), \(y\), and \(z\)\(M1\)
Obtain \(\overrightarrow{ON} = \mathbf{i} + 2\mathbf{j}\), or equivalent\(A1\)
Carry out method for finding \(SN\)\(M1\)
Show that \(SN = 7\) correctly\(A1\)
Question (ii) OR (Method 3):
AnswerMarks Guidance
Answer/WorkingMark Guidance Notes
Use Cartesian formula or scalar product of \(\overrightarrow{PS}\) with a normal vector to find \(SN\)\(M1\)
Obtain \(SN = 7\)\(A1\)
State a unit normal \(\hat{\mathbf{n}}\) to the plane\(B1\sqrt{}\)
Use \(\overrightarrow{ON} = \overrightarrow{OS} \pm 7\hat{\mathbf{n}}\)\(M1\)
Obtain an unsimplified expression e.g. \(3\mathbf{i} + 5\mathbf{j} - 6\mathbf{k} \pm 7\left(\frac{2}{7}\mathbf{i} + \frac{3}{7}\mathbf{j} - \frac{6}{7}\mathbf{k}\right)\)\(A1\sqrt{}\)
Obtain \(\overrightarrow{ON} = \mathbf{i} + 2\mathbf{j}\), or equivalent, only\(A1\) [6]
## Question 11:

| Answer/Working | Mark | Guidance |
|---|---|---|
| **(i) EITHER:** Obtain a vector in the plane e.g. $\overrightarrow{PQ} = -3\mathbf{i} + 4\mathbf{j} + \mathbf{k}$ | B1 | |
| Use scalar product to obtain relevant equation in $a, b, c$ e.g. $-3a + 4b + c = 0$ or $6a - 2b + c = 0$ or $3a + 2b + 2c = 0$ | M1 | |
| State two correct equations in $a, b, c$ | A1 | |
| Solve simultaneous equations to obtain one ratio e.g. $a : b$ | M1 | |
| Obtain $a : b : c = 2 : 3 : -6$ or equivalent | A1 | |
| Obtain equation $2x + 3y - 6z = 8$ or equivalent | A1 | |
| **OR:** Substitute for $P, Q, R$ in equation of plane and state 3 equations in $a, b, c, d$ | B1 | |
| Eliminate one unknown, e.g. $d$, entirely | M1 | |
| Obtain 2 equations in 3 unknowns | A1 | |
| Solve to obtain one ratio e.g. $a : b$ | M1 | |
| Obtain $a : b : c = 2 : 3 : -6$ or equivalent | A1 | |
| Obtain equation $2x + 3y - 6z = 8$ or equivalent | A1 | |
| **OR:** Obtain a vector in plane e.g. $\overrightarrow{QR} = 6\mathbf{i} - 2\mathbf{j} + \mathbf{k}$ | B1 | |
| Find second vector in plane and form correctly a 2-parameter equation for the plane | M1 | |
| Obtain equation in any correct form e.g. $\mathbf{r} = \lambda(-3\mathbf{i}+4\mathbf{j}+\mathbf{k}) + \mu(6\mathbf{i}-2\mathbf{j}+\mathbf{k}) + \mathbf{i} - \mathbf{k}$ | A1 | |
| State 3 equations in $x, y, z, \lambda$, and $\mu$ | A1 | |
| Eliminate $\lambda$ and $\mu$ | M1 | |
| Obtain equation $2x + 3y - 6z = 8$ or equivalent | A1 | |
| **OR:** Obtain a vector in plane e.g. $\overrightarrow{PR} = 3\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}$ | B1 | |
| Obtain second vector in plane and calculate vector product of the two vectors, e.g. $(-3\mathbf{i}+4\mathbf{j}+\mathbf{k}) \times (3\mathbf{i}+2\mathbf{j}+2\mathbf{k})$ | M1 | |
| Obtain 2 correct components of the product | A1 | |
| Obtain correct product e.g. $6\mathbf{i} + 9\mathbf{j} - 18\mathbf{k}$ or equivalent | A1 | |
| Substitute in $2x + 3y - 6z = d$ and find $d$ or equivalent | M1 | |
| Obtain equation $2x + 3y - 6z = 8$ or equivalent | A1 | **Total: 6** |

## Question (ii) EITHER:

| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| State equation of $SN$ is $\mathbf{r} = 3\mathbf{i} + 5\mathbf{j} - 6\mathbf{k} + \lambda(2\mathbf{i} + 3\mathbf{j} - 6\mathbf{k})$ or equivalent | $B1\sqrt{}$ | |
| Express $x, y, z$ in terms of $\lambda$ e.g. $(3 + 2\lambda,\ 5 + 3\lambda,\ -6 - 6\lambda)$ | $B1\sqrt{}$ | |
| Substitute in the equation of the plane and solve for $\lambda$ | $M1$ | |
| Obtain $\overrightarrow{ON} = \mathbf{i} + 2\mathbf{j}$, or equivalent | $A1$ | |
| Carry out method for finding $SN$ | $M1$ | |
| Show that $SN = 7$ correctly | $A1$ | |

## Question (ii) OR (Method 2):

| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| Letting $\overrightarrow{ON} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$, obtain two equations in $x, y, z$ by equating scalar product of $\overrightarrow{NS}$ with two of $\overrightarrow{PQ}, \overrightarrow{QR}, \overrightarrow{RP}$ to zero | $B1\sqrt{} + B1\sqrt{}$ | |
| Using the plane equation as third equation, solve for $x$, $y$, and $z$ | $M1$ | |
| Obtain $\overrightarrow{ON} = \mathbf{i} + 2\mathbf{j}$, or equivalent | $A1$ | |
| Carry out method for finding $SN$ | $M1$ | |
| Show that $SN = 7$ correctly | $A1$ | |

## Question (ii) OR (Method 3):

| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| Use Cartesian formula or scalar product of $\overrightarrow{PS}$ with a normal vector to find $SN$ | $M1$ | |
| Obtain $SN = 7$ | $A1$ | |
| State a unit normal $\hat{\mathbf{n}}$ to the plane | $B1\sqrt{}$ | |
| Use $\overrightarrow{ON} = \overrightarrow{OS} \pm 7\hat{\mathbf{n}}$ | $M1$ | |
| Obtain an unsimplified expression e.g. $3\mathbf{i} + 5\mathbf{j} - 6\mathbf{k} \pm 7\left(\frac{2}{7}\mathbf{i} + \frac{3}{7}\mathbf{j} - \frac{6}{7}\mathbf{k}\right)$ | $A1\sqrt{}$ | |
| Obtain $\overrightarrow{ON} = \mathbf{i} + 2\mathbf{j}$, or equivalent, only | $A1$ | **[6]** |
11 With respect to the origin $O$, the points $P , Q , R , S$ have position vectors given by

$$\overrightarrow { O P } = \mathbf { i } - \mathbf { k } , \quad \overrightarrow { O Q } = - 2 \mathbf { i } + 4 \mathbf { j } , \quad \overrightarrow { O R } = 4 \mathbf { i } + 2 \mathbf { j } + \mathbf { k } , \quad \overrightarrow { O S } = 3 \mathbf { i } + 5 \mathbf { j } - 6 \mathbf { k } .$$

(i) Find the equation of the plane containing $P , Q$ and $R$, giving your answer in the form $a x + b y + c z = d$.\\
(ii) The point $N$ is the foot of the perpendicular from $S$ to this plane. Find the position vector of $N$ and show that the length of $S N$ is 7 .

\hfill \mbox{\textit{CAIE P3 2004 Q11 [12]}}