What is the algebra of vectors?

Algebra of vectors is a type of algebra in which the components are in vector form, and their algebraic operations comply with vector laws.

How many questions are present in this exercise of RD Sharma solutions?

RD Sharma Class 12th Exercise 22.2 has 13 questions that a student needs to practice.

What are the concepts that are discussed in RD Sharma Algebra of vectors Class 12th solutions?

The central concepts that are discussed are as follows: Introduction Basic Concepts Types of Vectors Addition of Vectors Multiplication of a vector by a scalar

Where can I find RD Sharma Solutions for Class 12 Maths Chapter 22 exercise-by-exercise answers?

Students in Class 12 should select appropriate study materials to help them solve textbook problems efficiently. In Career360, the solutions to both chapter and exercise problems are available. It can be referred to by students while solving problems to quickly clear their doubts

What are the Vector Addition Laws?

The two main vector addition laws are as follows: A + B = B + A is a commutative law. A + (B + C) = (A + B) + C is a Associative Law.

RD Sharma Class 12 Exercise 22.2 Algebra of Vectors Solutions Maths - Download PDF Free Online

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RD Sharma Class 12 Exercise 22.2 Algebra of Vectors Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Exercise 22.2 Algebra of Vectors Solutions Maths - Download PDF Free Online

Updated on 24 Jan 2022, 06:19 PM IST

The study of vectors and scalars is the focus of Vector Algebra for Class 12. Several quantities involve both magnitude and direction. Vectors are quantities that have magnitude as well as direction. Vector Quantities are the names given to such quantities—displacement, velocity, force, weight, acceleration, momentum, electric intensity, and so on. RD Sharma solutions When a student is facing difficulty, RD Sharma Class 12th Exercise 22.2 comes to help them.

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RD Sharma Class 12 Solutions Chapter22 Algebra of vectors - Other Exercise
Algebra of Vectors Excercise: 22.2
Algebra of Vectors Exercise 22.2 Question 9
RD Sharma Chapter wise Solutions

RD Sharma Class 12th Chapter 22 Exercise 22.2 has exercise-wise questions in this book. This particular exercise has 10 questions from the chapter 'Algebra of Vectors.' The concepts discussed in this exercise are Collinear and Non-Collinear Points, truth values, how to solve vectors, Proving like ABCD is a IIgm, and many more. Practicing these questions is vital for every student.

RD Sharma Class 12 Solutions Chapter22 Algebra of vectors - Other Exercise

Algebra of Vectors Excercise: 22.2

Algebra of Vectors Exercise 22.2 Question 1

Answer:
The value of $\vec{PR}=\vec{a}+\vec{b}$
Hint:
Use,$\vec{PQ}+\vec{QR}=\vec{PR}$
Given:
P,Q,R are collinear.
Solution:
95789

P,Q,R are collinear, they lie on the same line
$\begin{aligned} &\overrightarrow{P Q}=\vec{a} \text { and } \overrightarrow{P R}=\vec{b} \\ &\overrightarrow{P Q}+\overrightarrow{Q R}=\overrightarrow{P R} \\ &\vec{a}+\vec{b}=\overrightarrow{P R} \\ &\overrightarrow{P R}=\vec{a}+\vec{b} \end{aligned}$

Algebra of Vectors Exercise 22.2 Question 2

Answer:
The possiblilities are
(i) $\vec{a}+\vec{b}+\vec{c}=0$
(ii) $\vec{a}=\vec{b}+\vec{c}$
(iii) $\vec{c}=\vec{a}+\vec{b}$
(iv) $\vec{b}=\vec{a}+\vec{c}$
Hint:
Use triangle law of addition.
Given:
$\vec{a},\vec{b},\vec{c}$ are three sides of the triangle
Solution: tri

$\begin{aligned} &\overrightarrow{A B}=\vec{a} \\ &\overrightarrow{B C}=\vec{b} \\ &\therefore \overrightarrow{A B}+\overrightarrow{B C}=\overrightarrow{A C} \\ &\vec{a}+\vec{b}=-\vec{c} \\ &\vec{a}+\vec{b}+\vec{c}=0 \\ \end{aligned}$

The other possibilities are
$\begin{aligned} &\vec{a}=\vec{b}+\vec{c} \\ &\vec{c}=\vec{a}+\vec{b} \\ &\vec{b}=\vec{a}+\vec{c} \end{aligned}$

Algebra of Vectors Exercise 22.2 Question 3

Answer:

$\vec{a}+\vec{b}$ and $\vec{d}-\vec{b}$ are the diagonals of parallelogram. whose adjacent sides are $\vec{a}$ and $\vec{b}$
Hint:
Use triangle law of vector.
Given :
$\vec{a},\vec{b},\vec{c}$are three sides of triangle
Solution:
$\vec{a}$ and $\vec{b}$ are two collinear vectors having the same initial point.
Let $AB= \vec{a}$ and $AD= \vec{b}$
let us draw a parallelogram with AB and AD as any of two sides of the parallelogram.
We know in parallelogram opposite sides are equal hence,
$DC= \vec{a}$ and $BC= \vec{b}$
Now consider $\Delta ABC$ , applying
Triangle law of vector we got,
95792
$\begin{aligned} &\overrightarrow{A B}+\overrightarrow{B C}=\overrightarrow{A C}\\ &\vec{a}+\vec{b}=\overrightarrow{A C}\\ \end{aligned}$
Similarly in $\Delta ABD$
$\begin{aligned} &\overrightarrow{A D}+\overrightarrow{D B}=\overrightarrow{A B}\\ &\vec{b}+D B=\vec{a}\\ &\therefore \vec{a}-\vec{b}=\overrightarrow{D B} \end{aligned}$

Algebra of Vectors Exercise 22.2 Question 4

Answer:

Either $m$ is $0$ or $\vec{a}=0$

Hint:

Use $ab=0$

Given :$m\vec{a}=0$

Solution:

Given, $m\vec{a}=0$

Since if $ab=0$

Either $a=0$ or $b=0$

Therefore, if $m\vec{a}=0$

Either $m=0\: or \: \vec{a}=0$

Algebra of Vectors Exercise 22.2 Question 5

Answer:

True
False
False

Hint:
use vector algebra.
Given :
$\vec{a},\vec{b}$ are two vectors.
Solution
$(i) \; \; \; \; \; \; \: \: \: \vec{a}=-\vec{b}$
$|\vec{a}|=|\vec{b}|$
$\vec{a}=-\vec{b}$
$|\vec{a}|=|-\vec{b}|$
$|\vec{a}|=|-||\vec{b}|$
$|\vec{a}|=1 \cdot|\vec{b}|$
$|\vec{a}|=|\vec{b}|$
$(i)\: \: \: \: \: Is \; \; \; true$
$(ii) \quad|\vec{a}|=|\vec{b}|=\vec{a}=\pm \vec{b}$

$(ii)\: \: is \: \: false$
$(iii) \: \quad|\vec{a}|=|\vec{b}|=\vec{a}=\pm \vec{b}$
$|\vec{a}|=|\vec{b}|$
$a=b$
$(iii)\: false$.

Algebra of Vectors Exercise 22.2 Question 6

Answer:
$\vec{BA}+\vec{BC}+\vec{CD}+\vec{DA}=2\vec{BA}$
Hint:
Form the triangle in quadrilateral. Use triangle law of inequality.
Given :
ABCD are quadrilateral.
Solution:

In $\Delta ABC,$
By the triangle law of inequality
$\vec{BC}+\vec{CA}=\vec{BA}$
$\vec{CA}=\vec{BA}-\vec{BC}(1)$

Again in $\Delta ADC,$

By triangle law of inequality

$\vec{CA}=\vec{CD}+\vec{DA}.......(2)$

From (1) and (2)

$\begin{aligned} &\overrightarrow{C D}+\overrightarrow{D A}=\overrightarrow{B A}-\overrightarrow{B C} \\ &\overrightarrow{B A}+\overrightarrow{B C}+\overrightarrow{C D}+\overrightarrow{D A}=\overrightarrow{B A}+\overrightarrow{B A} \\ &\overrightarrow{B A}+\overrightarrow{B C}+\overrightarrow{C D}+\overrightarrow{D A}=2 \overrightarrow{B A} \end{aligned}$

Algebra of Vectors Exercise 22.2 Question 7

Answer:
We need to prove that
Hint:
Use triangle law of vector.
Given :
ABCDE is a pentagon.
Solution:
Given , ABCDE is a pentagon.
95798

In $\Delta ABC$,by tringle law
$\vec{AB}+\vec{BC}=\vec{AC}.......(1)$
In $\Delta ACD$ by tringle law
$\vec{AC}+\vec{CD}=\vec{AD}.......(2)$
In $\Delta AED$,by tringle law
$\vec{AD}+\vec{DE}=\vec{AE}......(3)$
or $\vec{AE}+\vec{ED}=\vec{AD}.........(4)$
In $\Delta ADC$,by tringle law
$\overrightarrow{A D}+\overrightarrow{D C}=\overrightarrow{A C} \ldots \ldots \ldots (5)$
$(i) \overrightarrow{A D}+\overrightarrow{D C}=\overrightarrow{A C}$
$=\overrightarrow{A C}+\overrightarrow{E C}+E A$
$=\overrightarrow{A E}+\overrightarrow{E A}$
$=\overrightarrow{A E}-\overrightarrow{A E}$
$=0$
$(ii) L.H.S, \overrightarrow{A B}+\overrightarrow{A E}+\overrightarrow{B C}+\overrightarrow{D C}+\overrightarrow{E D}+\overrightarrow{A C}$
$=(\overrightarrow{A B}+\overrightarrow{B C})+(\overrightarrow{A E}+\overrightarrow{E D})+\overrightarrow{A C}+\overrightarrow{D C}$
$=\overrightarrow{A C}+\overrightarrow{A D}+\overrightarrow{D C}+\overrightarrow{A C} \quad [using (1)and(4) ]$
$=\overrightarrow{A C}+\overrightarrow{A C}+\overrightarrow{A C}=3 \overrightarrow{A C} (proved)$

Algebra of Vectors Exercise 22.2 Question 8

Answer:
We need to prove that
Hint:
Used equal and adjacent side law.
Given :
Let 0 be the centre of the octagon with vertices ABCDEFGH.
Solution:
since 0 is the mid-point.
95799

$\begin{aligned} &\overrightarrow{O A}=-\overrightarrow{O E}\\ &\overrightarrow{O A}+\overrightarrow{O E}=0 \ldots \ldots \ldots \text { (1) }\\ &\overrightarrow{O B}=-\overrightarrow{O F}\\ &\overrightarrow{O B}+\overrightarrow{O F}=0 \ldots \ldots \ldots . .\\ &\overrightarrow{O C}=-\overrightarrow{O G}\\ &\overrightarrow{O C}+\overrightarrow{O G}=0 \ldots \ldots \ldots .\\ &\overrightarrow{O P}=-\overrightarrow{O H}\\ &\overrightarrow{O P}+\overrightarrow{O H}=0 \ldots \ldots \ldots \ldots . \text { (4) }\\ \end{aligned}$

Adding (1) (2) (3) and (4)
$\begin{aligned} &\overrightarrow{O A}+\overrightarrow{O E}+\overrightarrow{O B}+\overrightarrow{O F}+\overrightarrow{O C}+\overrightarrow{O G}+\overrightarrow{O P}+\overrightarrow{O H}=\overrightarrow{0} \end{aligned}$

Hence proved.

Algebra of Vectors Exercise 22.2 Question 9

Answer:
We need to prove that
Hint:
Use triangle law.
Given :
$\vec{AD}+\vec{PB}+\vec{PD}=\vec{PC}$
$\vec{AD}+\vec{PB}+\vec{PD}-\vec{PC}$
Solution:
95800

In $\Delta PDC$,by triangle law
$\vec{PD}+\vec{DC}=\vec{PC}$
In $\Delta PAB$,by triangle law
$\vec{PA}+\vec{PB}=\vec{AB}$
$\therefore$ABCD is a parallelogram.

Algebra of Vectors Exercise 22.2 Question 10

Answer:
We need to prove that
Hint:
Use triangle law of addition.
Given :
ABCDEF is a hexagon
Solution:
Given ,
ABCDEF is a regular hexagon.
95801

From the triangle law of addition,

$\begin{aligned} &\overrightarrow{A C}+\overrightarrow{C D}=\overrightarrow{A D}\\ \end{aligned}$

$\begin{aligned} &\overrightarrow{A C}+\overrightarrow{A F}=\overrightarrow{A D}\\ \end{aligned}$

$\begin{aligned} &\because|\overrightarrow{C D}|=|\overrightarrow{A F}|\\ \end{aligned}$

$\begin{aligned} &\overrightarrow{C D} \text { parallel to } \overrightarrow{A F}: C D=A F\\ \end{aligned}$

$\begin{aligned} &\overrightarrow{A B}+\overrightarrow{A C}+\overrightarrow{A F}+\overrightarrow{A D}+\overrightarrow{A C}+\overrightarrow{A E}\\ &=\overrightarrow{A B}+\overrightarrow{A D}+\overrightarrow{A D}+\overrightarrow{A C} \quad \because|\overrightarrow{A E}|=|\overrightarrow{E D}|\\ &=2 \overrightarrow{A D}+\overrightarrow{E D}+\overrightarrow{A E A E} \text { parallel to } \overrightarrow{E D} \therefore \overrightarrow{A B}=\overrightarrow{E D}\\ &=2 \overrightarrow{A D}+\overrightarrow{A D}\\ &=3 \overrightarrow{A D} \quad[\overrightarrow{A B}+\overrightarrow{A E}=\overrightarrow{E D}+\overrightarrow{A E}=\overrightarrow{A D}\\ &=3(2 \cdot \overrightarrow{A O}) \quad[O \text { is the mid point and also the mid point of } \mathrm{AD}, \mathrm{OA}=\mathrm{OD}, \mathrm{AD}=2 \cdot \overrightarrow{\mathrm{AO}}]\\ &=6 \overrightarrow{A O} \end{aligned}$

(proved)

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