| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | 3D geometry applications |
| Difficulty | Standard +0.3 This is a structured multi-part question on 3D coordinate geometry with clear scaffolding. Students find midpoint coordinates, verify perpendicularity using dot products, find a plane equation, and work with line equations. While it involves multiple steps and 3D visualization, each part follows standard techniques with significant guidance. The ratio division and final verification are routine calculations. Slightly easier than average due to the scaffolding and straightforward application of standard methods. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms4.04c Scalar product: calculate and use for angles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| P is \((0, 10, 30)\), Q is \((0, 20, 15)\), R is \((-15, 20, 30)\) | B2,1,0 | |
| \(\overrightarrow{PQ} = \begin{pmatrix}0-0\\20-10\\15-30\end{pmatrix} = \begin{pmatrix}0\\10\\-15\end{pmatrix}\) * | E1 | |
| \(\overrightarrow{PR} = \begin{pmatrix}-15-0\\20-10\\30-30\end{pmatrix} = \begin{pmatrix}-15\\10\\0\end{pmatrix}\) * | E1 [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{pmatrix}2\\3\\2\end{pmatrix}\cdot\begin{pmatrix}0\\10\\-15\end{pmatrix} = 0 + 30 - 30 = 0\) | M1 | Scalar product with 1 vector in the plane OR vector \(\times\) product oe |
| \(\begin{pmatrix}2\\3\\2\end{pmatrix}\cdot\begin{pmatrix}-15\\10\\0\end{pmatrix} = -30 + 30 + 0 = 0\) | E1 | |
| \(\Rightarrow \begin{pmatrix}2\\3\\2\end{pmatrix}\) is normal to the plane | ||
| \(\Rightarrow\) equation of plane is \(2x + 3y + 2z = c\) | M1 | \(2x + 3y + 2z = c\) or an appropriate vector form |
| At P (say), \(x=0\), \(y=10\), \(z=30\): \(c = 2\times0 + 3\times10 + 2\times30 = 90\) | M1dep | substituting to find \(c\) or completely eliminating parameters |
| \(\Rightarrow\) equation of plane is \(2x + 3y + 2z = 90\) | A1 cao [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| S is \(\left(-7\frac{1}{2}, 20, 22\frac{1}{2}\right)\) | B1 | |
| \(\overrightarrow{OT} = \overrightarrow{OP} + \frac{2}{3}\overrightarrow{PS}\) | M1 | Or \(\frac{1}{3}(\overrightarrow{OP} + \overrightarrow{OR} + \overrightarrow{OQ})\) oe ft their S |
| \(= \begin{pmatrix}0\\10\\30\end{pmatrix} + \frac{2}{3}\begin{pmatrix}-7\frac{1}{2}\\10\\-7\frac{1}{2}\end{pmatrix} = \begin{pmatrix}-5\\16\frac{2}{3}\\25\end{pmatrix}\) | A1ft | Or \(\frac{1}{3}\begin{pmatrix}0\\10\\30\end{pmatrix} + \frac{2}{3}\begin{pmatrix}-7\frac{1}{2}\\20\\22\frac{1}{2}\end{pmatrix}\) ft their S |
| So T is \(\left(-5, 16\frac{2}{3}, 25\right)\) * | E1 [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{r} = \begin{pmatrix}-5\\16\frac{2}{3}\\25\end{pmatrix} + \lambda\begin{pmatrix}2\\3\\2\end{pmatrix}\) | B1, B1 | \(\begin{pmatrix}-5\\16\frac{2}{3}\\25\end{pmatrix} + \ldots + \lambda\begin{pmatrix}2\\3\\2\end{pmatrix}\) |
| At C \((-30, 0, 0)\): \(-5 + 2\lambda = -30\), \(16\frac{2}{3} + 3\lambda = 0\), \(25 + 2\lambda = 0\) | M1 A1 | Substituting coordinates of C into vector equation. At least 2 relevant correct equations for \(\lambda\) |
| 1st and 3rd equations give \(\lambda = -12\frac{1}{2}\), not compatible with 2nd. So line does not pass through C. | E1 [5] | oe www |
# Question 4:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| P is $(0, 10, 30)$, Q is $(0, 20, 15)$, R is $(-15, 20, 30)$ | B2,1,0 | |
| $\overrightarrow{PQ} = \begin{pmatrix}0-0\\20-10\\15-30\end{pmatrix} = \begin{pmatrix}0\\10\\-15\end{pmatrix}$ * | E1 | |
| $\overrightarrow{PR} = \begin{pmatrix}-15-0\\20-10\\30-30\end{pmatrix} = \begin{pmatrix}-15\\10\\0\end{pmatrix}$ * | E1 [4] | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}2\\3\\2\end{pmatrix}\cdot\begin{pmatrix}0\\10\\-15\end{pmatrix} = 0 + 30 - 30 = 0$ | M1 | Scalar product with 1 vector in the plane OR vector $\times$ product oe |
| $\begin{pmatrix}2\\3\\2\end{pmatrix}\cdot\begin{pmatrix}-15\\10\\0\end{pmatrix} = -30 + 30 + 0 = 0$ | E1 | |
| $\Rightarrow \begin{pmatrix}2\\3\\2\end{pmatrix}$ is normal to the plane | | |
| $\Rightarrow$ equation of plane is $2x + 3y + 2z = c$ | M1 | $2x + 3y + 2z = c$ or an appropriate vector form |
| At P (say), $x=0$, $y=10$, $z=30$: $c = 2\times0 + 3\times10 + 2\times30 = 90$ | M1dep | substituting to find $c$ or completely eliminating parameters |
| $\Rightarrow$ equation of plane is $2x + 3y + 2z = 90$ | A1 cao [5] | |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| S is $\left(-7\frac{1}{2}, 20, 22\frac{1}{2}\right)$ | B1 | |
| $\overrightarrow{OT} = \overrightarrow{OP} + \frac{2}{3}\overrightarrow{PS}$ | M1 | Or $\frac{1}{3}(\overrightarrow{OP} + \overrightarrow{OR} + \overrightarrow{OQ})$ oe ft their S |
| $= \begin{pmatrix}0\\10\\30\end{pmatrix} + \frac{2}{3}\begin{pmatrix}-7\frac{1}{2}\\10\\-7\frac{1}{2}\end{pmatrix} = \begin{pmatrix}-5\\16\frac{2}{3}\\25\end{pmatrix}$ | A1ft | Or $\frac{1}{3}\begin{pmatrix}0\\10\\30\end{pmatrix} + \frac{2}{3}\begin{pmatrix}-7\frac{1}{2}\\20\\22\frac{1}{2}\end{pmatrix}$ ft their S |
| So T is $\left(-5, 16\frac{2}{3}, 25\right)$ * | E1 [4] | |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{r} = \begin{pmatrix}-5\\16\frac{2}{3}\\25\end{pmatrix} + \lambda\begin{pmatrix}2\\3\\2\end{pmatrix}$ | B1, B1 | $\begin{pmatrix}-5\\16\frac{2}{3}\\25\end{pmatrix} + \ldots + \lambda\begin{pmatrix}2\\3\\2\end{pmatrix}$ |
| At C $(-30, 0, 0)$: $-5 + 2\lambda = -30$, $16\frac{2}{3} + 3\lambda = 0$, $25 + 2\lambda = 0$ | M1 A1 | Substituting coordinates of C into vector equation. At least 2 relevant correct equations for $\lambda$ |
| 1st and 3rd equations give $\lambda = -12\frac{1}{2}$, not compatible with 2nd. So line does not pass through C. | E1 [5] | oe www |4 A computer-controlled machine can be programmed to make ats by entering the equation of the plane of the cut, and to drill holes by entering the equation of the line of the hole.
A $20 \mathrm {~cm} \times 30 \mathrm {~cm} \times 30 \mathrm {~cm}$ cuboid is to be at and drilled. The cuboid is positioned relative to $x$-, $y ^ { 2 }$ and z-axes as shown in Fig. 8.1.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{7a52b6ce-a0cc-421d-8eae-3b6cf098e381-4_416_702_463_322}
\captionsetup{labelformat=empty}
\caption{Fig.8.1}
\end{center}
\end{figure}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{7a52b6ce-a0cc-421d-8eae-3b6cf098e381-4_420_683_459_1044}
\captionsetup{labelformat=empty}
\caption{Fig. 8.2}
\end{center}
\end{figure}
Fkst, a plane out is made to remove the comer at $E$. The cut goes through the points $P . Q$ and $R$, which are the midpoints of the sides $\mathrm { ED } , \mathrm { EA }$ and EF respectively.\\
(i) Write down the coordinates of $\mathbf { P } , \mathbf { Q }$ and $\mathbf { R }$.
$$\text { Henceshowlhat } \mathbb { N } ^ { 1 } \left\{ \begin{array} { l }
1 \\
0 \\
0
\end{array} \right\} \text { and } \left\{ \begin{array} { l }
1 \\
0
\end{array} \right\}$$
(;) Show that the veeto, $\binom { \text { I } } { \text { ) } }$ is pc,pcndicula, to the plone through $\mathrm { P } , \mathrm { Q }$, nd R\\
Hence find the Cartesian equation of this plane.
A hole is then drilled perpendicular to lriangle $P Q R$, as shown in Fig. 82. The hole passes through the triangle at the point $T$ which divides the line $P S$ in the ratio 2 : $I$, where $S$ is the midpoint of $Q R$.\\
(iii) Write down the coordinates of S , and show that the point T has coordinates $( - 5,16,25 )$.\\
(iv) Write down a vector equation of the line of the drill hole.
Hence determine whether or not this line passes through C .
\hfill \mbox{\textit{OCR MEI C4 Q4 [18]}}