CAIE P3 2009 June — Question 9 11 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2009
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Cross Product & Distances
TypeLine lies in or parallel to plane
DifficultyStandard +0.3 This is a straightforward application of standard vector techniques: part (i) requires substituting a point and direction vector into the plane equation to find two unknowns (routine algebraic manipulation), while part (ii) is a standard perpendicular distance calculation using the cross product formula. Both parts follow well-established procedures with no novel insight required, making this slightly easier than average.
Spec4.04a Line equations: 2D and 3D, cartesian and vector forms4.04f Line-plane intersection: find point

9 The line \(l\) has equation \(\mathbf { r } = 4 \mathbf { i } + 2 \mathbf { j } - \mathbf { k } + t ( 2 \mathbf { i } - \mathbf { j } - 2 \mathbf { k } )\). It is given that \(l\) lies in the plane with equation \(2 x + b y + c z = 1\), where \(b\) and \(c\) are constants.
  1. Find the values of \(b\) and \(c\).
  2. The point \(P\) has position vector \(2 \mathbf { j } + 4 \mathbf { k }\). Show that the perpendicular distance from \(P\) to \(l\) is \(\sqrt { } 5\).

Question 9:
Part (i) EITHER method:
AnswerMarks Guidance
Answer/WorkingMark Guidance
Substitute coordinates of general point of \(l\) in equation of plane and equate constant terms, obtaining an equation in \(b\) and \(c\)M1*
Obtain a correct equation, e.g. \(8 + 2b - c = 1\)A1
Equate the coefficient of \(t\) to zero, obtaining an equation in \(b\) and \(c\)M1*
Obtain a correct equation, e.g. \(4 - b - 2c = 0\)A1
Part (i) OR method:
AnswerMarks Guidance
Answer/WorkingMark Guidance
Substitute \((4, 2, -1)\) in the plane equationM1*
Obtain a correct equation in \(b\) and \(c\), e.g. \(2b - c = -7\)A1
EITHER Find a second point on \(l\) and obtain an equation in \(b\) and \(c\)M1*
Obtain a correct equation in \(b\) and \(c\), e.g. \(b + 2c = 4\)A1
OR Calculate scalar product of a direction vector for \(l\) and a vector normal for the plane and equate to zeroM1*
Obtain a correct equation for \(b\) and \(c\)A1
Solve for \(b\) or for \(c\)M1(dep*)
Obtain \(b = -2\) and \(c = 3\)A1 [6 marks]
Part (ii) EITHER method:
AnswerMarks Guidance
Answer/WorkingMark Guidance
Find \(\overrightarrow{PQ}\) for a point \(Q\) on \(l\) with parameter \(t\), e.g. \(4\mathbf{i} - 5\mathbf{k} + t(2\mathbf{i} - \mathbf{j} - 2\mathbf{k})\)B1
Calculate scalar product of \(\overrightarrow{PQ}\) and a direction vector for \(l\) and equate to zeroM1
Solve and obtain \(t = -2\)A1
Carry out a complete method for finding the length of \(\overrightarrow{PQ}\)M1
Obtain the given answer \(\sqrt{5}\) correctlyA1
Part (ii) OR 1:
AnswerMarks Guidance
Answer/WorkingMark Guidance
Calling \((4, 2, -1)\) \(A\), state \(\overrightarrow{AP}\) (or \(\overrightarrow{PA}\)) in component form, e.g. \(4\mathbf{i} - 5\mathbf{k}\)B1
Calculate vector product of \(\overrightarrow{AP}\) and a direction vector for \(l\), e.g. \((4\mathbf{i} - 5\mathbf{k}) \times (2\mathbf{i} - \mathbf{j} - 2\mathbf{k})\)M1
Obtain correct answer, e.g. \(-5\mathbf{i} - 2\mathbf{j} - 4\mathbf{k}\)A1
Divide modulus of the product by that of the direction vectorM1
Obtain the given answer correctlyA1
Part (ii) OR 2:
AnswerMarks Guidance
Answer/WorkingMark Guidance
State \(\overrightarrow{AP}\) (or \(\overrightarrow{PA}\)) in component formB1
Use a scalar product to find the projection of \(\overrightarrow{AP}\) (or \(\overrightarrow{PA}\)) on \(l\)M1
Obtain correct answer in any form, e.g. \(\dfrac{18}{\sqrt{9}}\)A1
Use Pythagoras to find the perpendicularM1
Obtain the given answer correctlyA1
Part (ii) OR 3:
AnswerMarks Guidance
Answer/WorkingMark Guidance
State \(\overrightarrow{AP}\) (or \(\overrightarrow{PA}\)) in component formB1
Use a scalar product to find the cosine of \(PAQ\)M1
Obtain correct answer in any form, e.g. \(\dfrac{18}{\sqrt{41}\cdot\sqrt{9}}\)A1
Use trig to find the perpendicularM1
Obtain the given answer correctlyA1
Part (ii) OR 4:
AnswerMarks Guidance
Answer/WorkingMark Guidance
State \(\overrightarrow{AP}\) (or \(\overrightarrow{PA}\)) in component formB1
Find a second point \(B\) on \(l\) and use the cosine rule in triangle \(APB\) to find the cosine of \(A\), \(B\) or \(P\), or use a vector product to find the area of \(APB\)M1
Obtain correct answer in any formA1
Use trig or area formula to find the perpendicularM1
Obtain the given answer correctlyA1
Part (ii) OR 5:
AnswerMarks Guidance
Answer/WorkingMark Guidance
Find \(\overrightarrow{PQ}\) for a point \(Q\) on \(l\) with parameter \(t\), e.g. \(4\mathbf{i} - 5\mathbf{k} + t(2\mathbf{i} - \mathbf{j} - 2\mathbf{k})\)B1
Use correct method to express \(PQ^2\) (or \(PQ\)) in terms of \(t\)M1
Obtain a correct expression in any form, e.g. \((4+2t)^2 + (-t)^2 + (-5-2t)^2\)A1
Carry out a complete method for finding its minimumM1
Obtain the given answer correctlyA1 [5 marks]
# Question 9:

## Part (i) EITHER method:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitute coordinates of general point of $l$ in equation of plane and equate constant terms, obtaining an equation in $b$ and $c$ | M1* | |
| Obtain a correct equation, e.g. $8 + 2b - c = 1$ | A1 | |
| Equate the coefficient of $t$ to zero, obtaining an equation in $b$ and $c$ | M1* | |
| Obtain a correct equation, e.g. $4 - b - 2c = 0$ | A1 | |

## Part (i) OR method:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitute $(4, 2, -1)$ in the plane equation | M1* | |
| Obtain a correct equation in $b$ and $c$, e.g. $2b - c = -7$ | A1 | |
| **EITHER** Find a second point on $l$ and obtain an equation in $b$ and $c$ | M1* | |
| Obtain a correct equation in $b$ and $c$, e.g. $b + 2c = 4$ | A1 | |
| **OR** Calculate scalar product of a direction vector for $l$ and a vector normal for the plane and equate to zero | M1* | |
| Obtain a correct equation for $b$ and $c$ | A1 | |
| Solve for $b$ or for $c$ | M1(dep*) | |
| Obtain $b = -2$ and $c = 3$ | A1 | **[6 marks]** |

## Part (ii) EITHER method:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Find $\overrightarrow{PQ}$ for a point $Q$ on $l$ with parameter $t$, e.g. $4\mathbf{i} - 5\mathbf{k} + t(2\mathbf{i} - \mathbf{j} - 2\mathbf{k})$ | B1 | |
| Calculate scalar product of $\overrightarrow{PQ}$ and a direction vector for $l$ and equate to zero | M1 | |
| Solve and obtain $t = -2$ | A1 | |
| Carry out a complete method for finding the length of $\overrightarrow{PQ}$ | M1 | |
| Obtain the given answer $\sqrt{5}$ correctly | A1 | |

## Part (ii) OR 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Calling $(4, 2, -1)$ $A$, state $\overrightarrow{AP}$ (or $\overrightarrow{PA}$) in component form, e.g. $4\mathbf{i} - 5\mathbf{k}$ | B1 | |
| Calculate vector product of $\overrightarrow{AP}$ and a direction vector for $l$, e.g. $(4\mathbf{i} - 5\mathbf{k}) \times (2\mathbf{i} - \mathbf{j} - 2\mathbf{k})$ | M1 | |
| Obtain correct answer, e.g. $-5\mathbf{i} - 2\mathbf{j} - 4\mathbf{k}$ | A1 | |
| Divide modulus of the product by that of the direction vector | M1 | |
| Obtain the given answer correctly | A1 | |

## Part (ii) OR 2:

| Answer/Working | Mark | Guidance |
|---|---|---|
| State $\overrightarrow{AP}$ (or $\overrightarrow{PA}$) in component form | B1 | |
| Use a scalar product to find the projection of $\overrightarrow{AP}$ (or $\overrightarrow{PA}$) on $l$ | M1 | |
| Obtain correct answer in any form, e.g. $\dfrac{18}{\sqrt{9}}$ | A1 | |
| Use Pythagoras to find the perpendicular | M1 | |
| Obtain the given answer correctly | A1 | |

## Part (ii) OR 3:

| Answer/Working | Mark | Guidance |
|---|---|---|
| State $\overrightarrow{AP}$ (or $\overrightarrow{PA}$) in component form | B1 | |
| Use a scalar product to find the cosine of $PAQ$ | M1 | |
| Obtain correct answer in any form, e.g. $\dfrac{18}{\sqrt{41}\cdot\sqrt{9}}$ | A1 | |
| Use trig to find the perpendicular | M1 | |
| Obtain the given answer correctly | A1 | |

## Part (ii) OR 4:

| Answer/Working | Mark | Guidance |
|---|---|---|
| State $\overrightarrow{AP}$ (or $\overrightarrow{PA}$) in component form | B1 | |
| Find a second point $B$ on $l$ and use the cosine rule in triangle $APB$ to find the cosine of $A$, $B$ or $P$, or use a vector product to find the area of $APB$ | M1 | |
| Obtain correct answer in any form | A1 | |
| Use trig or area formula to find the perpendicular | M1 | |
| Obtain the given answer correctly | A1 | |

## Part (ii) OR 5:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Find $\overrightarrow{PQ}$ for a point $Q$ on $l$ with parameter $t$, e.g. $4\mathbf{i} - 5\mathbf{k} + t(2\mathbf{i} - \mathbf{j} - 2\mathbf{k})$ | B1 | |
| Use correct method to express $PQ^2$ (or $PQ$) in terms of $t$ | M1 | |
| Obtain a correct expression in any form, e.g. $(4+2t)^2 + (-t)^2 + (-5-2t)^2$ | A1 | |
| Carry out a complete method for finding its minimum | M1 | |
| Obtain the given answer correctly | A1 | **[5 marks]** |

---
9 The line $l$ has equation $\mathbf { r } = 4 \mathbf { i } + 2 \mathbf { j } - \mathbf { k } + t ( 2 \mathbf { i } - \mathbf { j } - 2 \mathbf { k } )$. It is given that $l$ lies in the plane with equation $2 x + b y + c z = 1$, where $b$ and $c$ are constants.\\
(i) Find the values of $b$ and $c$.\\
(ii) The point $P$ has position vector $2 \mathbf { j } + 4 \mathbf { k }$. Show that the perpendicular distance from $P$ to $l$ is $\sqrt { } 5$.

\hfill \mbox{\textit{CAIE P3 2009 Q9 [11]}}