Question 1

OCR MEI C4 — Question 1 6 marks

Exam Board	OCR MEI
Module	C4 (Core Mathematics 4)
Marks	6
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Vectors 3D & Lines
Type	Linear combinations of vectors
Difficulty	Moderate -0.5 This is a straightforward linear combination problem requiring students to set up and solve simultaneous equations from vector components, followed by a standard geometric interpretation. The algebraic manipulation is routine and the deduction (coplanarity) is a direct textbook result, making it slightly easier than average.
Spec	1.10d Vector operations: addition and scalar multiplication 1.10e Position vectors: and displacement

1 The points \(\mathrm { A } , \mathrm { B }\) and C have coordinates \(\mathrm { A } ( 3,2 , - 1 ) , \mathrm { B } ( - 1,1,2 )\) and \(\mathrm { C } ( 10,5 , - 5 )\), relative to the origin O . Show that \(\overrightarrow { \mathrm { OC } }\) can be written in the form \(\lambda \overrightarrow { \mathrm { OA } } + \mu \overrightarrow { \mathrm { OB } }\), where \(\lambda\) and \(\mu\) are to be determined. What can you deduce about the points \(\mathrm { O } , \mathrm { A } , \mathrm { B }\) and C from the fact that \(\overrightarrow { \mathrm { OC } }\) can be expressed as a combination of \(\overrightarrow { \mathrm { OA } }\) and \(\overrightarrow { \mathrm { OB } }\) ?

Show mark scheme Show mark scheme source

Part 1:

Answer	Marks
M1	Required form, can be seen from two or more correct equations
M1	Forming at least two equations and attempting to solve or equivalent
A1	Verifying third equation, do not give benefit of doubt
A1	Accept a statement such as verification
A1	Must clearly show that the solutions satisfy all the equations
[6]	Independent of all above marks

Part 2:

\(4\mathbf{j} - 3\mathbf{k} = \lambda \mathbf{a} + \mu \mathbf{b}\)

\(= \lambda(2\mathbf{i} + \mathbf{j} - \mathbf{k}) + \mu(4\mathbf{i} - 2\mathbf{j} + \mathbf{k})\)

Answer	Marks
M1	Equating components

\(\Rightarrow 0 = 2\lambda + 4\mu\)

\(4 = \lambda - 2\mu\)

\(-3 = -\lambda + \mu\)

Answer	Marks
M1	At least two correct equations

\(\Rightarrow \lambda = -2\mu, 2\lambda = 4 \Rightarrow \lambda = 2, \mu = -1\)

Answer	Marks
A1	(blank)
A1, A1	(blank)
[5]	(blank)

Part 3:

Answer	Marks
B1	Seen or indicated, condone wrong sense

\(\overrightarrow{BA} = \begin{pmatrix} -4 \\ 1 \\ -3 \end{pmatrix}\), \(\overrightarrow{BC} = \begin{pmatrix} 2 \\ 5 \\ -1 \end{pmatrix}\)

Answer	Marks
M1	Scalar product

\(\overrightarrow{BA} \cdot \overrightarrow{BC} = (-4) \times 2 + 1 \times 5 + (-3) \times (-1) = -8 + 5 + 3 = 0\)

Answer	Marks
A1	Equal to 0

\(\Rightarrow \angle ABC = 90°\)

Answer	Marks
M1	Area of triangle formula or equivalent
M1	Length formula

Area of triangle \(= \frac{1}{2} \times BA \times BC = \frac{1}{2} \times \sqrt{(-4)^2 + 1^2 + (-3)^2} \times \sqrt{2^2 + 5^2 + (-1)^2}\)

\(= \frac{1}{2} \times \sqrt{26} \times \sqrt{30} = 13.96\) square units

Answer	Marks
A1	Accept 14.0 and \(\sqrt{195}\)
[6]	(blank)

# Question 1

**Part 1:**

M1 | Required form, can be seen from two or more correct equations

M1 | Forming at least two equations and attempting to solve or equivalent

A1 | Verifying third equation, do not give benefit of doubt

A1 | Accept a statement such as verification

A1 | Must clearly show that the solutions satisfy all the equations

[6] | Independent of all above marks

**Part 2:**

$4\mathbf{j} - 3\mathbf{k} = \lambda \mathbf{a} + \mu \mathbf{b}$

$= \lambda(2\mathbf{i} + \mathbf{j} - \mathbf{k}) + \mu(4\mathbf{i} - 2\mathbf{j} + \mathbf{k})$

M1 | Equating components

$\Rightarrow 0 = 2\lambda + 4\mu$

$4 = \lambda - 2\mu$

$-3 = -\lambda + \mu$

M1 | At least two correct equations

$\Rightarrow \lambda = -2\mu, 2\lambda = 4 \Rightarrow \lambda = 2, \mu = -1$

A1 | (blank)

A1, A1 | (blank)

[5] | (blank)

**Part 3:**

B1 | Seen or indicated, condone wrong sense

$\overrightarrow{BA} = \begin{pmatrix} -4 \\ 1 \\ -3 \end{pmatrix}$, $\overrightarrow{BC} = \begin{pmatrix} 2 \\ 5 \\ -1 \end{pmatrix}$

M1 | Scalar product

$\overrightarrow{BA} \cdot \overrightarrow{BC} = (-4) \times 2 + 1 \times 5 + (-3) \times (-1) = -8 + 5 + 3 = 0$

A1 | Equal to 0

$\Rightarrow \angle ABC = 90°$

M1 | Area of triangle formula or equivalent

M1 | Length formula

Area of triangle $= \frac{1}{2} \times BA \times BC = \frac{1}{2} \times \sqrt{(-4)^2 + 1^2 + (-3)^2} \times \sqrt{2^2 + 5^2 + (-1)^2}$

$= \frac{1}{2} \times \sqrt{26} \times \sqrt{30} = 13.96$ square units

A1 | Accept 14.0 and $\sqrt{195}$

[6] | (blank)

Show LaTeX source

1 The points $\mathrm { A } , \mathrm { B }$ and C have coordinates $\mathrm { A } ( 3,2 , - 1 ) , \mathrm { B } ( - 1,1,2 )$ and $\mathrm { C } ( 10,5 , - 5 )$, relative to the origin O . Show that $\overrightarrow { \mathrm { OC } }$ can be written in the form $\lambda \overrightarrow { \mathrm { OA } } + \mu \overrightarrow { \mathrm { OB } }$, where $\lambda$ and $\mu$ are to be determined.

What can you deduce about the points $\mathrm { O } , \mathrm { A } , \mathrm { B }$ and C from the fact that $\overrightarrow { \mathrm { OC } }$ can be expressed as a combination of $\overrightarrow { \mathrm { OA } }$ and $\overrightarrow { \mathrm { OB } }$ ?

\hfill \mbox{\textit{OCR MEI C4  Q1 [6]}}

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