OCR MEI C4 — Question 3 5 marks

Exam BoardOCR MEI
ModuleC4 (Core Mathematics 4)
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Lines & Planes
TypeCartesian equation of a plane
DifficultyStandard +0.3 This is a straightforward multi-part question requiring standard vector techniques: computing two vectors in the plane, verifying perpendicularity via dot product, then using the normal vector to write the Cartesian equation. All steps are routine applications of learned methods with no novel insight required, making it slightly easier than average.
Spec1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication4.04b Plane equations: cartesian and vector forms4.04c Scalar product: calculate and use for angles
3 Verify that the vector \(\mathbf { d } - \mathbf { j } + 4 \mathbf { k }\) is perpendicular to the plane through the points \(\mathrm { A } ( 2,0,1 ) , \mathrm { B } ( 1,2,2 )\) and \(\mathrm { C } ( 0 , - 4,1 )\). Hence find the cartesian equation of the plane. [5]
Question 3:
AnswerMarks Guidance
\(\vec{AB} = \begin{pmatrix}-1\\2\\1\end{pmatrix},\ \vec{AC} = \begin{pmatrix}-2\\-4\\0\end{pmatrix}\)M1 scalar product with any two directions in the plane
\(\mathbf{n}\cdot\vec{AB} = \begin{pmatrix}2\\-1\\4\end{pmatrix}\cdot\begin{pmatrix}-1\\2\\1\end{pmatrix} = 2\times(-1)+(-1)\times2+4\times1=0\)B1 evaluation needed
\(\mathbf{n}\cdot\vec{AC} = \begin{pmatrix}2\\-1\\4\end{pmatrix}\cdot\begin{pmatrix}-2\\-4\\0\end{pmatrix} = 2\times(-2)+(-1)\times(-4)+4\times0=0\)B1 evaluation needed
\(\Rightarrow \mathbf{n}\) is perpendicular to plane
Equation: \(\mathbf{r}\cdot\mathbf{n} = \mathbf{a}\cdot\mathbf{n}\), i.e. \(\begin{pmatrix}x\\y\\z\end{pmatrix}\cdot\begin{pmatrix}2\\-1\\4\end{pmatrix} = \begin{pmatrix}2\\0\\1\end{pmatrix}\cdot\begin{pmatrix}2\\-1\\4\end{pmatrix}\)M1 any complete method leading to Cartesian equation
\(\Rightarrow 2x - y + 4z = 8\)A1
[5] 
## Question 3:

$\vec{AB} = \begin{pmatrix}-1\\2\\1\end{pmatrix},\ \vec{AC} = \begin{pmatrix}-2\\-4\\0\end{pmatrix}$ | M1 | scalar product with any two directions in the plane

$\mathbf{n}\cdot\vec{AB} = \begin{pmatrix}2\\-1\\4\end{pmatrix}\cdot\begin{pmatrix}-1\\2\\1\end{pmatrix} = 2\times(-1)+(-1)\times2+4\times1=0$ | B1 | evaluation needed

$\mathbf{n}\cdot\vec{AC} = \begin{pmatrix}2\\-1\\4\end{pmatrix}\cdot\begin{pmatrix}-2\\-4\\0\end{pmatrix} = 2\times(-2)+(-1)\times(-4)+4\times0=0$ | B1 | evaluation needed

$\Rightarrow \mathbf{n}$ is perpendicular to plane | |

Equation: $\mathbf{r}\cdot\mathbf{n} = \mathbf{a}\cdot\mathbf{n}$, i.e. $\begin{pmatrix}x\\y\\z\end{pmatrix}\cdot\begin{pmatrix}2\\-1\\4\end{pmatrix} = \begin{pmatrix}2\\0\\1\end{pmatrix}\cdot\begin{pmatrix}2\\-1\\4\end{pmatrix}$ | M1 | any complete method leading to Cartesian equation

$\Rightarrow 2x - y + 4z = 8$ | A1 |
| [5] |

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3 Verify that the vector $\mathbf { d } - \mathbf { j } + 4 \mathbf { k }$ is perpendicular to the plane through the points $\mathrm { A } ( 2,0,1 ) , \mathrm { B } ( 1,2,2 )$ and $\mathrm { C } ( 0 , - 4,1 )$. Hence find the cartesian equation of the plane. [5]

\hfill \mbox{\textit{OCR MEI C4  Q3 [5]}}
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