| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2021 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Perpendicular distance from point to plane |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question requiring finding a plane equation from a point and line (using cross product), applying the point-to-plane distance formula with absolute values, and solving an equation involving two distance formulas. While the individual techniques are standard for FM students, the question requires careful algebraic manipulation across multiple parts and understanding of geometric relationships, placing it moderately above average difficulty. |
| Spec | 4.04b Plane equations: cartesian and vector forms4.04j Shortest distance: between a point and a plane |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(5\mathbf{i}+3\mathbf{j}-8\mathbf{k}\) and \(2\mathbf{i}-3\mathbf{j}-6\mathbf{k}\) lie in \(\Pi_1\) | B1 | Identifies 2 correct vectors lying in \(\Pi_1\) |
| \(\mathbf{n} = \begin{pmatrix}5\\3\\-8\end{pmatrix}\times\begin{pmatrix}2\\-3\\-6\end{pmatrix} = \begin{pmatrix}-18-24\\-(-30+16)\\-15-6\end{pmatrix}\) | M1 | Attempts the vector product between 2 correct vectors in \(\Pi_1\); if no working shown, look for at least 2 correct elements |
| \(= \begin{pmatrix}-42\\14\\-21\end{pmatrix}\) or e.g. \(\begin{pmatrix}6\\-2\\3\end{pmatrix}\) | A1 | Correct normal vector |
| \((6\mathbf{i}-2\mathbf{j}+3\mathbf{k})\cdot(\mathbf{i}+2\mathbf{j}+\mathbf{k}) = \ldots\) | dM1 | Attempts scalar product between their normal vector and position vector of a point in \(\Pi_1\); do not allow if "5" just appears without evidence of origin e.g. \(\mathbf{a}\cdot\mathbf{n}=\ldots\) where \(\mathbf{a}\) and \(\mathbf{n}\) defined earlier; depends on first method mark |
| \(6x-2y+3z=5\) * | A1* | Correct proof |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| 3 points on \(\Pi_1\) are \((1,2,1)\), \((3,-1,-5)\) and e.g. \((8,2,-13)\) | B1 | |
| \(a+2b+c=5\), \(3a-b-5=c\), \(8a+2b-13=c \Rightarrow a=\ldots, b=\ldots, c=\ldots\) | M1 | Solves simultaneously for \(a\), \(b\) and \(c\) using correct points |
| \(a=2, b=-\frac{2}{3}, c=\frac{5}{3}\) | A1 | Correct values |
| \(2x-\frac{2}{3}y+z=\frac{5}{3}\) | dM1 | Forms Cartesian equation |
| \(6x-2y+3z=5\) * | A1* | Correct proof |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((1,2,1)\to 6x-2y+3z=6-4+3=5\); shows \((1,2,1)\) lies on \(\Pi_1\) | B1 | |
| \(\frac{x-3}{5}=\frac{y+1}{3}=\frac{z+5}{-8}\to\mathbf{r}=\begin{pmatrix}3\\-1\\-5\end{pmatrix}+\lambda\begin{pmatrix}5\\3\\-8\end{pmatrix}\) | M1A1 | M1: Converts \(l\) to correct parametric form as part of attempt at this alternative; allow 1 slip with one of the elements; A1: Correct form |
| \(6(3+5\lambda)-2(-1+3\lambda)+3(-5-8\lambda)=5\); shows \(l\) lies in \(\Pi_1\) | dM1 | |
| \(P\) lies in \(\Pi_1\) and \(l\) lies in \(\Pi_1\) so \(6x-2y+3z=5\) * | A1* | All correct with conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(d = \frac{\ | 6(2)-2k+3(-7)-5\ | }{\sqrt{6^2+2^2+3^2}}\) |
| \(= \frac{1}{7}\ | -2k-14\ | = \frac{2}{7}\ |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Distance \(O\) to \(\Pi_1\) is \(\frac{5}{\sqrt{6^2+2^2+3^2}}\); Distance \(O\) to parallel plane containing \(Q\) is \(\frac{(6\mathbf{i}-2\mathbf{j}+3\mathbf{k})\cdot(2\mathbf{i}+k\mathbf{j}-7\mathbf{k})}{\sqrt{6^2+2^2+3^2}} = \frac{-9-2k}{7}\) | M1 | Correct method for the shortest distance |
| \(d = \left | \frac{5}{7}-\frac{-9-2k}{7}\right | = \frac{1}{7}\ |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(d = \left | \frac{\overrightarrow{PQ}\cdot\mathbf{n}}{ | \mathbf{n} |
| \(= \left | \frac{-42+14k-28+168}{49}\right | = \left |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{2}{7}\ | k+7\ | = \frac{\ |
| \(\frac{2}{7}(k+7) = "\frac{1}{9}(12-4k)" \Rightarrow k=\ldots\) or \(\frac{2}{7}(k+7) = "\frac{1}{9}(4k-12)" \Rightarrow k=\ldots\) | dM1 | Attempts to solve one of these equations where distance from \(Q\) to \(\Pi_2\) is of the form \(ak+b\) where \(a\) and \(b\) are non-zero; or squares both sides and attempts to solve resulting quadratic |
| \(k = -\frac{21}{23}\) or \(k=21\) | A1 | One correct value; must be 21 but allow equivalent exact fractions for \(-\frac{21}{23}\) |
| \(k = -\frac{21}{23}\) and \(k=21\) | A1 | Both correct values; must be 21, allow equivalent exact fractions for \(-\frac{21}{23}\), and no other values |
## Question 7(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $5\mathbf{i}+3\mathbf{j}-8\mathbf{k}$ and $2\mathbf{i}-3\mathbf{j}-6\mathbf{k}$ lie in $\Pi_1$ | B1 | Identifies 2 correct vectors lying in $\Pi_1$ |
| $\mathbf{n} = \begin{pmatrix}5\\3\\-8\end{pmatrix}\times\begin{pmatrix}2\\-3\\-6\end{pmatrix} = \begin{pmatrix}-18-24\\-(-30+16)\\-15-6\end{pmatrix}$ | M1 | Attempts the vector product between 2 correct vectors in $\Pi_1$; if no working shown, look for at least 2 correct elements |
| $= \begin{pmatrix}-42\\14\\-21\end{pmatrix}$ or e.g. $\begin{pmatrix}6\\-2\\3\end{pmatrix}$ | A1 | Correct normal vector |
| $(6\mathbf{i}-2\mathbf{j}+3\mathbf{k})\cdot(\mathbf{i}+2\mathbf{j}+\mathbf{k}) = \ldots$ | dM1 | Attempts scalar product between their normal vector and position vector of a point in $\Pi_1$; do not allow if "5" just appears without evidence of origin e.g. $\mathbf{a}\cdot\mathbf{n}=\ldots$ where $\mathbf{a}$ and $\mathbf{n}$ defined earlier; depends on first method mark |
| $6x-2y+3z=5$ * | A1* | Correct proof |
**Alternative 1 for (a):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| 3 points on $\Pi_1$ are $(1,2,1)$, $(3,-1,-5)$ and e.g. $(8,2,-13)$ | B1 | |
| $a+2b+c=5$, $3a-b-5=c$, $8a+2b-13=c \Rightarrow a=\ldots, b=\ldots, c=\ldots$ | M1 | Solves simultaneously for $a$, $b$ and $c$ using correct points |
| $a=2, b=-\frac{2}{3}, c=\frac{5}{3}$ | A1 | Correct values |
| $2x-\frac{2}{3}y+z=\frac{5}{3}$ | dM1 | Forms Cartesian equation |
| $6x-2y+3z=5$ * | A1* | Correct proof |
**Alternative 2 for (a):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(1,2,1)\to 6x-2y+3z=6-4+3=5$; shows $(1,2,1)$ lies on $\Pi_1$ | B1 | |
| $\frac{x-3}{5}=\frac{y+1}{3}=\frac{z+5}{-8}\to\mathbf{r}=\begin{pmatrix}3\\-1\\-5\end{pmatrix}+\lambda\begin{pmatrix}5\\3\\-8\end{pmatrix}$ | M1A1 | M1: Converts $l$ to correct parametric form as part of attempt at this alternative; allow 1 slip with one of the elements; A1: Correct form |
| $6(3+5\lambda)-2(-1+3\lambda)+3(-5-8\lambda)=5$; shows $l$ lies in $\Pi_1$ | dM1 | |
| $P$ lies in $\Pi_1$ and $l$ lies in $\Pi_1$ so $6x-2y+3z=5$ * | A1* | All correct with conclusion |
---
## Question 7(b):
**Way 1:**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $d = \frac{\|6(2)-2k+3(-7)-5\|}{\sqrt{6^2+2^2+3^2}}$ | M1 | Correct method for the shortest distance |
| $= \frac{1}{7}\|-2k-14\| = \frac{2}{7}\|k+7\|$ * | A1* | Correct completion |
**Way 2:**
| Answer/Working | Marks | Guidance |
|---|---|---|
| Distance $O$ to $\Pi_1$ is $\frac{5}{\sqrt{6^2+2^2+3^2}}$; Distance $O$ to parallel plane containing $Q$ is $\frac{(6\mathbf{i}-2\mathbf{j}+3\mathbf{k})\cdot(2\mathbf{i}+k\mathbf{j}-7\mathbf{k})}{\sqrt{6^2+2^2+3^2}} = \frac{-9-2k}{7}$ | M1 | Correct method for the shortest distance |
| $d = \left|\frac{5}{7}-\frac{-9-2k}{7}\right| = \frac{1}{7}\|2k+14\| = \frac{2}{7}\|k+7\|$ * | A1* | Correct completion |
**Way 3:**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $d = \left|\frac{\overrightarrow{PQ}\cdot\mathbf{n}}{|\mathbf{n}|}\right| = \left|\frac{(\mathbf{i}+(k-2)\mathbf{j}-8\mathbf{k})\cdot(-42\mathbf{i}+14\mathbf{j}-21\mathbf{k})}{\sqrt{42^2+14^2+21^2}}\right|$ | M1 | Correct method for the shortest distance |
| $= \left|\frac{-42+14k-28+168}{49}\right| = \left|\frac{14k+98}{49}\right| = \frac{2}{7}\|k+7\|$ * | A1* | Correct completion |
---
## Question 7(c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{2}{7}\|k+7\| = \frac{\|8(2)-4k-7+3\|}{\sqrt{8^2+4^2+1^2}}$; correctly attempts distance between $(2,k,-7)$ and $\Pi_2$ and sets equal to result from (b) | M1 | May see alternative methods e.g. finds a point on $\Pi_2$ and uses vector method |
| $\frac{2}{7}(k+7) = "\frac{1}{9}(12-4k)" \Rightarrow k=\ldots$ or $\frac{2}{7}(k+7) = "\frac{1}{9}(4k-12)" \Rightarrow k=\ldots$ | dM1 | Attempts to solve one of these equations where distance from $Q$ to $\Pi_2$ is of the form $ak+b$ where $a$ and $b$ are non-zero; or squares both sides and attempts to solve resulting quadratic |
| $k = -\frac{21}{23}$ or $k=21$ | A1 | One correct value; must be 21 but allow equivalent exact fractions for $-\frac{21}{23}$ |
| $k = -\frac{21}{23}$ **and** $k=21$ | A1 | Both correct values; must be 21, allow equivalent exact fractions for $-\frac{21}{23}$, and no other values |\begin{enumerate}
\item The point $P$ has coordinates $( 1,2,1 )$
\end{enumerate}
The line $l$ has Cartesian equation
$$\frac { x - 3 } { 5 } = \frac { y + 1 } { 3 } = \frac { z + 5 } { - 8 }$$
The plane $\Pi _ { 1 }$ contains the point $P$ and the line $l$.\\
(a) Show that a Cartesian equation for $\Pi _ { 1 }$ is
$$6 x - 2 y + 3 z = 5$$
The point $Q$ has coordinates $( 2 , k , - 7 )$, where $k$ is a constant.\\
(b) Show that the shortest distance between $\Pi _ { 1 }$ and $Q$ is
$$\frac { 2 } { 7 } | k + 7 |$$
The plane $\Pi _ { 2 }$ has Cartesian equation $8 x - 4 y + z = - 3$\\
Given that the shortest distance between $\Pi _ { 1 }$ and $Q$ is the same as the shortest distance between $\Pi _ { 2 }$ and $Q$,\\
(c) determine the possible values of $k$.\\
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\hfill \mbox{\textit{Edexcel F3 2021 Q7 [11]}}