RD Sharma Solutions Class 12 Mathematics Chapter 22 FBQ

The RD Sharma class 12 solution of Algebra of vectors exercise FBQ is one of the highly recommended solutions for the Class 12 CBSE students, it is preferred over NCERT when it comes for the preparation of maths. The RD Sharma class 12th exercise FBQ comes under consideration when a student faces difficulty in solving any question of the Algebra of vectors. The Class 12 RD Sharma chapter 22 exercise FBQ solution gives a brief explanation of each and every concept that has been mentioned in the book.

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

• Chapter 22 - Algebra of vectors - Ex-22.1

• Chapter 22- Algebra of vectors - Ex-22.2

• Chapter 22- Algebra of vectors - Ex-22.3

• Chapter 22 - Algebra of vectors - Ex-22.4

• Chapter 22 - Algebra of vectors - Ex-22.5

• Chapter 22 - Algebra of vectors - Ex-22.6

• Chapter 22 - Algebra of vectors - Ex-22.7

• Chapter 22 - Algebra of vectors - Ex-22.8

• Chapter 22 - Algebra of vectors - Ex-22.9

• Chapter 22 - Algebra of vectors - Ex-MCQ

• Chapter 22 - Algebra of vectors - Ex-VSA

Algebra of Vectors Excercise: FBQ

Algebra of Vectors exercise Fill in the blanks question 1

Answer:-$a+b+c=0$
Hint:- To solve this equation, we solve this by
Answer:- If $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ represent the side of a triangle taken in order that $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}------$
Solution:-

$\begin{array}{rl}& \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\\ \\ & =\overrightarrow{AC}+\overrightarrow{c}\\ \\ & =\overrightarrow{AC}+\overrightarrow{CA}\end{array}$
$\begin{array}{rl}& =\overrightarrow{AC}-\overrightarrow{AC}\\ \\ & =0\\ \\ & \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0\end{array}$

Algebra of Vectors exercise Fill in the blanks question 2

Answer:- $\overrightarrow{AC}=\overrightarrow{a}+\overrightarrow{b},\overrightarrow{BD}=\overrightarrow{b}+\overrightarrow{a}$
Hint:- To solve this equation ,we use right angle triangle formula.
Given:- In a parallelogram $\mathrm{ABCD}\text{ If }\overrightarrow{AB}=a\text{ and }\overrightarrow{BC}=b\text{, then }\overrightarrow{AC}=\dots \text{ and }\overrightarrow{BD}=\dots$
Solution:- In $\mathrm{\Delta }ABC$

According to right triangle
$\begin{array}{rl}& \overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}\\ \\ & \overrightarrow{a}+\overrightarrow{b}=\overrightarrow{AC}\\ \\ & \overrightarrow{AC}=\overrightarrow{a}+\overrightarrow{b}\\ \\ & \text{ In }\mathrm{\Delta }BDA\end{array}$
According to right triangle
$\begin{array}{rl}& \overrightarrow{AB}+\overrightarrow{BD}=\overrightarrow{AD}\\ \\ & \overrightarrow{a}+\overrightarrow{BD}=\overrightarrow{b}\\ \\ & \overrightarrow{BD}=\overrightarrow{b}-\overrightarrow{a}\end{array}$

Algebra of Vectors exercise Fill in the blanks question 3

Answer:-$0$
Hint:- We use the section formula for finding coordinates of centroid.
Given:- If $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are the position vectors of the vectors of a triangle having its centroid at the origin then $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\dots .$
Solution: Here the coordinates of $D=(a+b)/2$


$\begin{array}{c}\frac{m\overrightarrow{b}+n\overrightarrow{a}}{m+n}\\ \\ \frac{m}{n}=2\end{array}$
$\begin{array}{rl}& \frac{2\left(\frac{\overrightarrow{a}+\overrightarrow{b}}{2}\right)+1(\overrightarrow{c})}{3}\\ \\ & =G\left(\frac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}}{3}\right)\end{array}$
Here, given that centroid is at the origin $(a+b+c)/3=0$
Therefore, $a+b+c=0$

Algebra of Vectors exercise Fill in the blanks question 4

Answer:-$0$
Hint:- To solve this equation we will make triangle, then we will find modulus.
Given:-If $a,b,c$ are the positive vectors of vertices of an equilateral triangle having its circumcentre at the origin, then $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\cdots \cdots \cdots$
Solution:-$\overrightarrow{OC}=\overrightarrow{c}$

Centroid of triangle $=(a+b+c)/3$
For an equilateral triangle the circumcentre coincide with the centroid. Therefore,
$\begin{array}{rl}& \frac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}}{3}=0\\ \\ & \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0\end{array}$

Algebra of Vectors exercise Fill in the blanks question 7

Answer:- $2$
Hint:- To solve this equation we take $\overrightarrow{a}+\overrightarrow{b}=\lambda$
Given:- If $D$ is the mid point of side $BC$ of $\mathrm{\Delta }ABC$ and $\overrightarrow{AB}+\overrightarrow{AC}=k\overrightarrow{AD}$ , Then $K$
Solution:- Let $a,b,c$ are the position vectors of $AB,BC,andCA$ respectively.
$\overrightarrow{AB}+\overrightarrow{AC}=k\overrightarrow{AD}$

$\begin{array}{rl}& \overrightarrow{b}-\overrightarrow{a}+\overrightarrow{c}-\overrightarrow{a}=k[\frac{\overrightarrow{b}+\overrightarrow{c}}{2}-\overrightarrow{a}]\\ \\ & \overrightarrow{b}+\overrightarrow{c}-2\overrightarrow{a}=k\frac{(\overrightarrow{b}+\overrightarrow{c}-2\overrightarrow{a})}{2}\\ \\ & \mathrm{k}=2\end{array}$

Algebra of Vectors exercise Fill in the blanks question 12

Answer:- $\overrightarrow{AC}+\overrightarrow{AF}-\overrightarrow{AB}=\overrightarrow{AE}$
Hint:-To solve this equation, we use reverse process of vectors
Given:-In a rectangular hexagon $\text{ ABCDEF, }\overrightarrow{AC}+\overrightarrow{AF}-\overrightarrow{AB}$

Solution:-
we have $\overrightarrow{AE}=\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}+\overrightarrow{DE}$ …(i)
From figure we can see
$\begin{array}{rl}& \overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}...(ii)\\ \\ & \overrightarrow{CD}=\overrightarrow{AF}[\because AF\Vert CD,-AF=CD]...(iii)\\ \\ & \overrightarrow{DE}=-\overrightarrow{AB}...(iv)\end{array}$
Using (ii) , (iii) and (iv) in (i), we get
$\begin{array}{rl}& \overrightarrow{AE}=\overrightarrow{AC}+\overrightarrow{AF}-\overrightarrow{AB}\\ \\ & \text{ Hence }\overrightarrow{AC}+\overrightarrow{AF}-\overrightarrow{AB}=\overline{AE}\end{array}$

Algebra of Vectors exercise Fill in the blanks question 16

Answer:-$2$
Hint:- To solve this we use triangle law of vector addition
Given:-If $A,B,C,D,E$ are five point in a plane such that $\overrightarrow{AB}+\overrightarrow{AE}+\overrightarrow{BC}+\overrightarrow{DC}+\overrightarrow{ED}=K\overrightarrow{AC}$ then K IS….

Solution:-
$\begin{array}{r}\overrightarrow{AB}+\overrightarrow{AE}+\overrightarrow{BC}+\overrightarrow{DC}+\overrightarrow{ED}\end{array}$
$\begin{array}{rl}& =(\overrightarrow{AB}+\overrightarrow{BC})+(\overrightarrow{ED}+\overrightarrow{DC})+\overrightarrow{AE}\\ \\ & =\overrightarrow{AC}+(\overrightarrow{EC}+\overrightarrow{AE})\end{array}$ [∴Triangle Law of Vector Addition]
$\begin{array}{rl}& =\overrightarrow{AC}+\overrightarrow{AC}\\ \\ & =2\overrightarrow{AC}\\ \\ & \therefore K=2\end{array}$

Algebra of Vectors exercise Fill in the blanks question 17

Answer:-$\lambda =4$
Hint:-To solve this question we have to connect or add $AP,BP,CPandDP$ vectors
Given:-Let P be the point of intersection of the diagonal of a parallelogram $ABCD$ and $O$ is any point If $\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}=\lambda \overrightarrow{OP}$
Solution:-

$\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}=\lambda \overline{OP}$
$\begin{array}{rl}& =(\overrightarrow{OA}+\overrightarrow{AP})+(\overrightarrow{OB}+\overrightarrow{BP})+(\overrightarrow{OC}-\overrightarrow{CP})+(\overrightarrow{OD}+\overrightarrow{DP})\\ \\ & =\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+(\overrightarrow{AP}+\overline{CP})+(\overline{BP}+\overrightarrow{DP})\\ \\ & =\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}=4\overline{OP}\\ \\ & \lambda =4\end{array}$

Algebra of Vectors exercise Fill in the blanks question 19

Answer:-$0$
Hint:-To solve this equation we use mid-point formula
Given:-If $D,E,F$ are mid points of the side $BC,CAandAB$ respectively of $\mathrm{\Delta }ABC$ then $\overrightarrow{AD}+\overrightarrow{BE}+\overrightarrow{C\overrightarrow{F}}=---$
Solution:-
In $\mathrm{△}ABC,D,E,F$ are mid pts.
Let $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ be the position vector of $\overrightarrow{A},\overrightarrow{B},\overrightarrow{C}$
Using mid point formula.
As $D$ is a mid point of $BC$
$D=\frac{\overrightarrow{b}+\overrightarrow{c}}{2}$
Similarly E and F are the mid points of AC and AB respectively
$\begin{array}{rl}& \therefore E=\frac{\overrightarrow{a}+\overrightarrow{c}}{2}\\ \\ & F=\frac{\overrightarrow{a}+\overrightarrow{b}}{2}\end{array}$
Now,
$\overrightarrow{AD}+\overrightarrow{BE}+\overrightarrow{CF}=\overrightarrow{OD}-\overrightarrow{OA}+\overrightarrow{OE}-\overrightarrow{OB}+\overrightarrow{OF}-\overrightarrow{OC}$
$\begin{array}{rl}& =\overrightarrow{OD}+\overrightarrow{O}E+\overrightarrow{OF}-[\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}]\\ \\ & =\frac{\overrightarrow{b}+\overrightarrow{c}}{2}+\frac{\overrightarrow{a}+\overrightarrow{c}}{2}+\frac{\overrightarrow{a}+\overrightarrow{b}}{2}-(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})\end{array}$
$\begin{array}{rl}& =\frac{2(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})}{2}-(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})\\ \\ & =(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})-(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})\\ \\ & =0\end{array}$

Algebra of Vectors exercise Fill in the blanks question 20

Answer:- $0$
Hint:- To solve this equation we use mid-point formula
Given:-The algebraic sum of the vector directed from the vectors to the mid points of the opposite side is equal to
Solution:-
$\begin{array}{rl}& \overrightarrow{AD}+\overrightarrow{B\overline{E}}+\overrightarrow{C\overrightarrow{F}}\\ \\ & \text{ In }\mathrm{\Delta }ABD\overrightarrow{AB}+\overrightarrow{BD}+\overrightarrow{DA}=0\end{array}$

$\begin{array}{rl}& \overrightarrow{EC}=\frac{1}{2}\overrightarrow{AC}\\ \\ & \overrightarrow{BD}=\frac{1}{2}\overrightarrow{BC}\\ \\ & \overrightarrow{FB}=\frac{1}{2}\overrightarrow{AB}\end{array}$
$\begin{array}{rl}& \therefore \overrightarrow{AD}=\overrightarrow{AB}+\frac{1}{2}\overrightarrow{BC}\\ \\ & \text{ In }\mathrm{\Delta }BEC\overrightarrow{BC}+\overrightarrow{CE}+\overrightarrow{EB}=0\end{array}$
$\begin{array}{rl}& \therefore \overrightarrow{BE}=\overrightarrow{BC}+\frac{1}{2}\overrightarrow{CA}\\ \\ & \text{ In }\mathrm{\Delta }ACF\overrightarrow{AF}+\overrightarrow{FC}+\overrightarrow{CA}=0\\ \\ & \therefore \overrightarrow{CF}=\overrightarrow{CA}+\frac{1}{2}\overrightarrow{AB}\end{array}$
$\begin{array}{rl}& \overrightarrow{AB}+\frac{1}{2}\overrightarrow{BC}+\overrightarrow{BC}+\frac{1}{2}\overrightarrow{CA}+\overrightarrow{CA}+\frac{1}{2}\overrightarrow{AB}\\ \\ & \mathrm{\Delta }ABC\\ \\ & \overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}=0\end{array}$
$\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}+\frac{1}{2}(\overrightarrow{BC}+\overrightarrow{CA}+\overrightarrow{AB})=0$

Algebra of Vectors exercise Fill in the blanks question 22

Answer:-$|\overrightarrow{a}|=|\overrightarrow{b}|$
Hint:-To solve this equation we use rhombus in parallelogram.
Given:-The vector $\overrightarrow{a}+\overrightarrow{b}$ bisects the angle between the non-collinear $\overrightarrow{a}$ and $\overrightarrow{b}$ if ---
Solution:-Non-collinear vector $\overrightarrow{a}$ & $\overrightarrow{b}$ have found parallelogram $ABCD$
$AC=\overrightarrow{a}+\overrightarrow{b}$

$\begin{array}{rl}& \Rightarrow \mathrm{\angle }DAC=\mathrm{\angle }CAB=\theta \\ \\ & \text{ Rhombus }\Rightarrow AB=BC\\ \\ & |\overrightarrow{a}|=|\overrightarrow{b}|\end{array}$

Algebra of Vectors exercise Fill in the blanks question 23

Answer:-$2\hat{i}-\hat{j}+\hat{k}$
Hint:-To solve this equation we use standard formula.
Given:-The position vectors of two points $A$ and $B$ are $\overrightarrow{OA}=2\hat{i}-\hat{j}-\hat{k}\text{ and }\overrightarrow{OB}=2\hat{i}-\hat{j}+2\hat{k}$ respectively. The position vectors of a point $P$which divides the line segment joining $A$ and $B$ in the ratio $2:1$ is ----
Solution:- $\overrightarrow{a}=2\hat{i}-\hat{j}-\hat{k}$

$\begin{array}{rl}& \overrightarrow{b}=2\hat{i}-\hat{j}+2\hat{k}\\ \\ & m=2;n=1\end{array}$
Section formula:
Internal $=\frac{m\overrightarrow{b}+n\overrightarrow{a}}{m+n}$
$\begin{array}{rl}& \Rightarrow \frac{m\overrightarrow{b}+n\overrightarrow{a}}{m+n}\\ \\ & =\frac{2(2\hat{i}-\hat{j}+2\hat{k})+1(2\hat{i}-\hat{j}-\hat{k})}{2+1}\end{array}$
$\begin{array}{rl}& =\frac{4\hat{i}-2\hat{j}+4\hat{k}+2\hat{i}-\hat{j}-\hat{k}}{3}\\ \\ & =\frac{6\hat{i}-3\hat{j}+3\hat{k}}{3}\\ \\ & =2\hat{i}-\hat{j}+\hat{k}\end{array}$

The RD Sharma class 12 solution chapter 22 exercise FBQ is the fill-in-the-blank type question that covers the essential concepts of this chapter. The RD Sharma class 12th exercise FBQ consists of a total of 23 questions giving knowledge about the concepts mentioned below-

• Scalar and Vector product

• Addition and subtraction of vectors

• Multiplication and division by a scalar quantity

• Collinear and non-collinear

• Unit vectors and position vector

• Coplanar points

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RD Sharma Chapter wise Solutions

• Chapter 1 - Relations
• Chapter 2 - Functions
• Chapter 3 - Inverse Trigonometric Functions
• Chapter 4 - Algebra of Matrices
• Chapter 5 - Determinants
• Chapter 6 - Adjoint and Inverse of a Matrix
• Chapter 7 - Solution of Simultaneous Linear Equations
• Chapter 8 - Continuity
• Chapter 9 - Differentiability
• Chapter 10 - Differentiation
• Chapter 11 - Higher Order Derivatives
• Chapter 12 - Derivative as a Rate Measurer
• Chapter 13 - Differentials, Errors and Approximations
• Chapter 14 - Mean Value Theorems
• Chapter 15 - Tangents and Normals
• Chapter 16 - Increasing and Decreasing Functions
• Chapter 17 - Maxima and Minima
• Chapter 18 - Indefinite Integrals
• Chapter 19 - Definite Integrals
• Chapter 20 - Areas of Bounded Regions
• Chapter 21 - Differential Equations
• Chapter 22 - Algebra of Vectors
• Chapter 23 - Scalar Or Dot Product
• Chapter 24 - Vector or Cross Product
• Chapter 25 - Scalar Triple Product
• Chapter 26 - Direction Cosines and Direction Ratios
• Chapter 27 - Straight Line in Space
• Chapter 28 - The Plane
• Chapter 29 - Linear programming
• Chapter 30- Probability
• Chapter 31 - Mean and Variance of a Random Variable