In the diagram O A → = a and O B → = b . Also A C → = 2 a and A D → = 3 b - a . Write A B → in terms of a and b .

Given, Consider the △ O A B , O B = b and O A = a Triangle law of vector addition states that when two vectors are represented as two sides of the triangle with the order of magnitude and direction, then the third side of the triangle represents the magnitude and direction of the resultant vector. Thus, we get A B → = O B → - O A → ∴ A B → = b - a Therefore, the answer is b - a

In the diagram O A → = a and O B → = b . Also A C → = 2 a and A D → = 3 b - a . Prove that △ O A B and △ O D C are similar triangles.

Given, Consider the △ A C D , A D → = 3 b - a and A C → = 2 a Triangle law of vector addition states that when two vectors are represented as two sides of the triangle with the order of magnitude and direction, then the third side of the triangle represents the magnitude and direction of the resultant vector. Thus, we get C D → = C A → + A D → ⇒ C D → = - 2 a + 3 b - a ⇒ C D → = 3 b - 3 a So, C D → = 3 A B → Thus, C D is parallel to A B , so the triangles △ O A B and △ O D C are similar.

In the diagram O P → = p , O Q → = q , Q R = 1 2 O P and S Q = 1 3 P Q . R Q is parallel to O P . Find in terms of p and q : P Q → .

Given, Consider the △ O P Q , O P → = p and O Q → = q Triangle law of vector addition states that when two vectors are represented as two sides of the triangle with the order of magnitude and direction, then the third side of the triangle represents the magnitude and direction of the resultant vector. Thus, we get P Q → = P O → + O Q → ⇒ P Q → = - O P → + O Q → ∴ P Q → = - p + q Thus, the answer is q - p .

In the diagram O P → = p , O Q → = q , Q R = 1 2 O P and S Q = 1 3 P Q . R Q is parallel to O P . Find in terms of p and q : P S → .

Given, Consider the △ O P Q , O P → = p and O Q → = q Triangle law of vector addition states that when two vectors are represented as two sides of the triangle with the order of magnitude and direction, then the third side of the triangle represents the magnitude and direction of the resultant vector. Thus, we get P Q → = P O → + O Q → ⇒ P Q → = - O P → + O Q → ∴ P Q → = - p + q It is given that S Q = 1 3 P Q Clearly, we can see that P S → = P Q → - S Q → P S → = - p + q - 1 3 - p + q ⇒ P S → = - p + q + 1 3 p - 1 3 q ⇒ P S → = 2 3 - p + q Hence, the answer is 2 3 - p + q .

In the diagram O P → = p , O Q → = q , Q R = 1 2 O P and S Q = 1 3 P Q . R Q is parallel to O P . Find in terms of p and q : O S → .

Given, Consider the △ O P Q , O P → = p and O Q → = q Triangle law of vector addition states that when two vectors are represented as two sides of the triangle with the order of magnitude and direction, then the third side of the triangle represents the magnitude and direction of the resultant vector. Thus, we get P Q → = P O → + O Q → ⇒ P Q → = - O P → + O Q → ∴ P Q → = - p + q Now, consider △ S O Q , we get O Q → = q , also given that S Q = 1 3 P Q = 1 3 - p + q . By applying triangle law of vector addition, we get O S → = O Q → + Q S → ⇒ O S → = q - 1 3 - p + q [ Q S → = - S Q → ] ⇒ O S → = q - q 3 + 1 3 p ⇒ O S → = 2 3 q + 1 3 p Hence, the answer is 2 3 q + 1 3 p

In the diagram O P → = p , O Q → = q , Q R = 1 2 O P and S Q = 1 3 P Q . R Q is parallel to O P . Find in terms of p and q : O R → .

Given, Consider the △ O P Q , O P → = p and O Q → = q Triangle law of vector addition states that when two vectors are represented as two sides of the triangle with the order of magnitude and direction, then the third side of the triangle represents the magnitude and direction of the resultant vector. Thus, we get P Q → = P O → + O Q → ⇒ P Q → = - O P → + O Q → ∴ P Q → = - p + q Now, consider △ S O Q , we get O Q → = q , also given that S Q = 1 3 P Q = 1 3 - p + q . By applying triangle law of vector addition, we get O S → = O Q → + Q S → ⇒ O S → = q - 1 3 - p + q [ Q S → = - S Q → ] ⇒ O S → = q - q 3 + 1 3 p ⇒ O S → = 2 3 q + 1 3 p Now, consider △ S Q R , S Q = 1 3 q - 1 3 p and Q R = 1 2 O P = 1 2 p By applying triangle law of vector addition, we get S R → = S Q → + Q R → S R → = 1 3 q - 1 3 p + 1 2 p S R → = 1 3 q + 1 6 p Now, O R → = O S → + S R → ⇒ O R → = 2 3 q + 1 3 p + 1 3 q + 1 6 p ∴ O R → = q + 1 2 p Hence, the answer is q + 1 2 p .

E M B I B E

Cambridge IGCSE® Mathematics Core and Extended Coursebook Second Edition>Shape, Space and Measures>Chapter 23 - Vectors and Transformations>Exercise 23.8

Karen Morrison and Nick Hamshaw Solutions for Exercise 8: Exercise 23.8

Author:Karen Morrison & Nick Hamshaw

Karen Morrison Mathematics Solutions for Exercise - Karen Morrison and Nick Hamshaw Solutions for Exercise 8: Exercise 23.8

Attempt the free practice questions from Exercise 8: Exercise 23.8 with hints and solutions to strengthen your understanding. Cambridge IGCSE® Mathematics Core and Extended Coursebook Second Edition solutions are prepared by Experienced Embibe Experts.

Questions from Karen Morrison and Nick Hamshaw Solutions for Exercise 8: Exercise 23.8 with Hints & Solutions

MEDIUM

Upper Secondary: IGCSE

IMPORTANT

Cambridge IGCSE® Mathematics Core and Extended Coursebook Second Edition>Chapter 23 - Vectors and Transformations>Exercise 23.8>Q 8

Vector $v$ represents the velocity (in $km / h$ ) of a person jogging. The sides of each block on the grid represents a speed of $1 km / h$ . Calculate the speed at which the person was jogging.

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EASY

Upper Secondary: IGCSE

IMPORTANT

Cambridge IGCSE® Mathematics Core and Extended Coursebook Second Edition>Chapter 23 - Vectors and Transformations>Exercise 23.8>Q 9

In the diagram $\vec{O A} = a$ and $\vec{O B} = b$ . Also $\vec{A C} = 2 a$ and $\vec{A D} = 3 b - a$ . Write $\vec{A B}$ in terms of $a$ and $b$ .

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EASY

Upper Secondary: IGCSE

IMPORTANT

Cambridge IGCSE® Mathematics Core and Extended Coursebook Second Edition>Chapter 23 - Vectors and Transformations>Exercise 23.8>Q 9

In the diagram $\vec{O A} = a$ and $\vec{O B} = b$ . Also $\vec{A C} = 2 a$ and $\vec{A D} = 3 b - a$ . $\vec{O D} = n b$ where $n$ is a whole number. Find $n$ .

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EASY

Upper Secondary: IGCSE

IMPORTANT

Cambridge IGCSE® Mathematics Core and Extended Coursebook Second Edition>Chapter 23 - Vectors and Transformations>Exercise 23.8>Q 9

In the diagram $\vec{O A} = a$ and $\vec{O B} = b$ . Also $\vec{A C} = 2 a$ and $\vec{A D} = 3 b - a$ . Prove that $△ O A B$ and $△ O D C$ are similar triangles.

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EASY

Upper Secondary: IGCSE

IMPORTANT

Cambridge IGCSE® Mathematics Core and Extended Coursebook Second Edition>Chapter 23 - Vectors and Transformations>Exercise 23.8>Q 10

In the diagram $\vec{O P} = p, \vec{O Q} = q, Q R = \frac{1}{2} O P$ and $S Q = \frac{1}{3} P Q$ . $R Q$ is parallel to $O P$ . Find in terms of $p$ and $q$ : $\vec{P Q}$ .

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EASY

Upper Secondary: IGCSE

IMPORTANT

Cambridge IGCSE® Mathematics Core and Extended Coursebook Second Edition>Chapter 23 - Vectors and Transformations>Exercise 23.8>Q 10

In the diagram $\vec{O P} = p, \vec{O Q} = q, Q R = \frac{1}{2} O P$ and $S Q = \frac{1}{3} P Q$ . $R Q$ is parallel to $O P$ . Find in terms of $p$ and $q$ : $\vec{P S}$ .

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EASY

Upper Secondary: IGCSE

IMPORTANT

Cambridge IGCSE® Mathematics Core and Extended Coursebook Second Edition>Chapter 23 - Vectors and Transformations>Exercise 23.8>Q 10

In the diagram $\vec{O P} = p, \vec{O Q} = q, Q R = \frac{1}{2} O P$ and $S Q = \frac{1}{3} P Q$ . $R Q$ is parallel to $O P$ . Find in terms of $p$ and $q$ : $\vec{O S}$ .

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MEDIUM

Upper Secondary: IGCSE

IMPORTANT

Cambridge IGCSE® Mathematics Core and Extended Coursebook Second Edition>Chapter 23 - Vectors and Transformations>Exercise 23.8>Q 10

In the diagram $\vec{O P} = p, \vec{O Q} = q, Q R = \frac{1}{2} O P$ and $S Q = \frac{1}{3} P Q$ . $R Q$ is parallel to $O P$ . Find in terms of $p$ and $q$ : $\vec{O R}$ .

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