Ratio division of line segment - Vectors Introduction & 2D

OCR MEI C4 Q4

8 marks Moderate -0.8

4 The points $\mathrm { A } , \mathrm { B }$ and C are given by the position vectors $\mathbf { a } = \binom { - 2 } { 1 } , \mathbf { b } = \binom { 0 } { 5 }$ and $\mathbf { c } = \binom { 4 } { 3 }$. M is the midpoint of AC .

Find the position vector of M .
Find the vector $\overrightarrow { B C }$.
Find the position vector of the point D such that $\overrightarrow { \mathrm { BC } } = \overrightarrow { \mathrm { AD } }$.
Show that D lies on BM .

OCR H240/01 2020 November Q5

8 marks Moderate -0.3

5 \includegraphics[max width=\textwidth, alt={}, center]{febe231d-200a-4957-b41b-de5b9be98b0a-5_424_583_255_244} The diagram shows points $A$ and $B$, which have position vectors $\mathbf { a }$ and $\mathbf { b }$ with respect to an origin $O$. $P$ is the point on $O B$ such that $O P : P B = 3 : 1$ and $Q$ is the midpoint of $A B$.

Find $\overrightarrow { P Q }$ in terms of $\mathbf { a }$ and $\mathbf { b }$. The line $O A$ is extended to a point $R$, so that $P Q R$ is a straight line.
Explain why $\overrightarrow { P R } = k ( 2 \mathbf { a } - \mathbf { b } )$, where $k$ is a constant.
Hence determine the ratio $O A : A R$.

Edexcel AS Paper 1 2022 June Q3

6 marks Moderate -0.8

The triangle $P Q R$ is such that $\overrightarrow { P Q } = 3 \mathbf { i } + 5 \mathbf { j }$ and $\overrightarrow { P R } = 13 \mathbf { i } - 15 \mathbf { j }$
1. Find $\overrightarrow { Q R }$
2. Hence find $| \overrightarrow { Q R } |$ giving your answer as a simplified surd.
The point $S$ lies on the line segment $Q R$ so that $Q S : S R = 3 : 2$
Find $\overrightarrow { P S }$

Edexcel PMT Mocks Q2

3 marks Easy -1.2

2. \includegraphics[max width=\textwidth, alt={}, center]{cb92f7b6-2ba5-4703-9595-9ba8570fc52b-04_656_725_283_635} \section*{Figure 1} Figure 1 shows a triangle $O A C$ where $O B$ divides $A C$ in the ratio $2 : 3$.
Show that $\mathbf { b } = \frac { 1 } { 5 } ( 3 \mathbf { a } + 2 \mathbf { c } )$

Edexcel Paper 2 2019 June Q10

6 marks Standard +0.3

10. Figure 7 Figure 7 shows a sketch of triangle $O A B$.
The point $C$ is such that $\overrightarrow { O C } = 2 \overrightarrow { O A }$.
The point $M$ is the midpoint of $A B$.
The straight line through $C$ and $M$ cuts $O B$ at the point $N$.
Given $\overrightarrow { O A } = \mathbf { a }$ and $\overrightarrow { O B } = \mathbf { b }$

Find $\overrightarrow { C M }$ in terms of $\mathbf { a }$ and $\mathbf { b }$
Show that $\overrightarrow { O N } = \left( 2 - \frac { 3 } { 2 } \lambda \right) \mathbf { a } + \frac { 1 } { 2 } \lambda \mathbf { b }$, where $\lambda$ is a scalar constant.
Hence prove that $O N : N B = 2 : 1$

Edexcel Paper 2 2020 October Q2

3 marks Easy -1.2

Relative to a fixed origin, points $P , Q$ and $R$ have position vectors $\mathbf { p } , \mathbf { q }$ and $\mathbf { r }$ respectively.

Given that

$\quad P , Q$ and $R$ lie on a straight line
$Q$ lies one third of the way from $P$ to $R$ show that

$$\mathbf { q } = \frac { 1 } { 3 } ( \mathbf { r } + 2 \mathbf { p } )$$

WJEC Unit 1 Specimen Q18

7 marks Moderate -0.8

The vectors $\mathbf{u}$ and $\mathbf{v}$ are defined by $\mathbf{u} = 2\mathbf{i} - 3\mathbf{j}$, $\mathbf{v} = -4\mathbf{i} + 5\mathbf{j}$.
1. Find the vector $4\mathbf{u} - 3\mathbf{v}$.
2. The vectors $\mathbf{u}$ and $\mathbf{v}$ are the position vectors of the points $U$ and $V$, respectively. Find the length of the line $UV$. [4]
Two villages $A$ and $B$ are 40 km apart on a long straight road passing through a desert. The position vectors of $A$ and $B$ are denoted by $\mathbf{a}$ and $\mathbf{b}$, respectively.
1. Village $C$ lies on the road between $A$ and $B$ at a distance 4 km from $B$. Find the position vector of $C$ in terms of $\mathbf{a}$ and $\mathbf{b}$.
2. Village $D$ has position vector $\frac{2}{9}\mathbf{a} + \frac{5}{9}\mathbf{b}$. Explain why village $D$ cannot possibly be on the straight road passing through $A$ and $B$. [3]

SPS SPS SM Mechanics 2022 February Q4

6 marks Moderate -0.8

Relative to a fixed origin $O$, • the point $A$ has position vector $\mathbf{5i + 3j - 2k}$ • the point $B$ has position vector $\mathbf{7i + j + 2k}$ • the point $C$ has position vector $\mathbf{4i + 8j - 3k}$

Find $|\overrightarrow{AB}|$ giving your answer as a simplified surd. [2]

Given that $ABCD$ is a parallelogram,

find the position vector of the point $D$. [2]

The point $E$ is positioned such that • $ACE$ is a straight line • $AC:CE = 2:1$

Find the coordinates of the point $E$. [2]

SPS SPS SM Pure 2023 June Q4

6 marks Moderate -0.8

Relative to a fixed origin $O$, • the point $A$ has position vector $5\mathbf{i} + 3\mathbf{j} - 2\mathbf{k}$ • the point $B$ has position vector $7\mathbf{i} + \mathbf{j} + 2\mathbf{k}$ • the point $C$ has position vector $4\mathbf{i} + 8\mathbf{j} - 3\mathbf{k}$

Find $|\vec{AB}|$ giving your answer as a simplified surd. [2] Given that $ABCD$ is a parallelogram,
find the position vector of the point $D$. [2] The point $E$ is positioned such that • $ACE$ is a straight line • $AC : CE = 2 : 1$
Find the coordinates of the point $E$. [2]

OCR H240/02 2018 December Q5

8 marks Moderate -0.3

Points $A$ and $B$ have position vectors $\mathbf{a}$ and $\mathbf{b}$. Point $C$ lies on $AB$ such that $AC : CB = p : 1$.

Show that the position vector of $C$ is $\frac{1}{p+1}(\mathbf{a} + p\mathbf{b})$. [3]

It is now given that $\mathbf{a} = 2\mathbf{i} + 3\mathbf{j} - 4\mathbf{k}$ and $\mathbf{b} = -6\mathbf{i} + 4\mathbf{j} + 12\mathbf{k}$, and that $C$ lies on the $y$-axis.

Find the value of $p$. [4]
Write down the position vector of $C$. [1]