| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2009 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Line lies in or parallel to plane |
| Difficulty | Standard +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. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04f Line-plane intersection: find point |
| Answer | Marks | Guidance |
|---|---|---|
| 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 |
| Answer | Marks | Guidance |
|---|---|---|
| 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] |
| Answer | Marks | Guidance |
|---|---|---|
| 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 |
| Answer | Marks | Guidance |
|---|---|---|
| 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 |
| Answer | Marks | Guidance |
|---|---|---|
| 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 |
| Answer | Marks | Guidance |
|---|---|---|
| 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 |
| Answer | Marks | Guidance |
|---|---|---|
| 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 |
| Answer | Marks | Guidance |
|---|---|---|
| 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] |
# 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]** |
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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]}}