What are the norms of the vector?

The length of the vector is referred to as the vector or the vector's magnitude. The length of a vector is a non-negative number that describes the extent of the vector in space, and is sometimes referred to as the vectors of magnitude or the norm

What are the directions of a vector?

The direction of a vector is the orientation of the vector, that is, the angle it makes with the x-axis.

What is the sum of vectors called?

The sum of two or more vectors is called the resultant. The resultant of two vectors can be found using either the parallelogram method or the triangle method.

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RD Sharma Solutions Class 12 Mathematics Chapter 22 FBQ

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RD Sharma Solutions Class 12 Mathematics Chapter 22 FBQ

RD Sharma Solutions Class 12 Mathematics Chapter 22 FBQ

Updated on 24 Jan 2022, 06:20 PM IST

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.

This Story also Contains

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

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

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 $\vec{a}, \vec{b}, \vec{c}$ represent the side of a triangle taken in order that $\vec{a}+\vec{b}+\vec{c}------$
Solution:-

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

Algebra of Vectors exercise Fill in the blanks question 2

Answer:- $\overrightarrow{A C}=\vec{a}+\vec{b}, \overrightarrow{B D}=\vec{b}+\vec{a}$
Hint:- To solve this equation ,we use right angle triangle formula.
Given:- In a parallelogram $\mathrm{ABCD} \text { If } \overrightarrow{A B}=a \text { and } \overrightarrow{B C}=b \text {, then } \overrightarrow{A C}=\ldots \text { and } \overrightarrow{B D}=\ldots$
Solution:- In $\Delta ABC$

According to right triangle
$\begin{aligned} &\overrightarrow{A B}+\overrightarrow{B C}=\overrightarrow{A C}\\\\ &\vec{a}+\vec{b}=\overrightarrow{A C}\\\\ &\overrightarrow{A C}=\vec{a}+\vec{b}\\\\ &\text { In } \Delta B D A \end{aligned}$
According to right triangle
$\begin{aligned} &\overrightarrow{A B}+\overrightarrow{B D}=\overrightarrow{A D} \\\\ &\vec{a}+\overrightarrow{B D}=\vec{b} \\\\ &\overrightarrow{B D}=\vec{b}-\vec{a} \end{aligned}$

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 $\vec{a}, \vec{b}, \vec{c}$ are the position vectors of the vectors of a triangle having its centroid at the origin then $\vec{a}+\vec{b}+\vec{c}=\ldots .$
Solution: Here the coordinates of $D = (a+b)/2$

$\begin{gathered} \frac{m \vec{b}+n \vec{a}}{m+n} \\\\ \frac{m}{n}=2 \end{gathered}$
$\begin{aligned} &\frac{2\left(\frac{\vec{a}+\vec{b}}{2}\right)+1(\vec{c})}{3} \\\\ &=G\left(\frac{\vec{a}+\vec{b}+\vec{c}}{3}\right) \end{aligned}$
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 $\vec{a}+\vec{b}+\vec{c}=\cdots \cdots \cdots$
Solution:-$\overrightarrow{O C}=\vec{c}$

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

Algebra of Vectors exercise Fill in the blanks question 5

Answer:- 9
Hint:- To solve this question we use $\vec{b}=\lambda \vec{a}$ where $\lambda$ is constant.
Given:- If the vector $\vec{a}=\hat{i}-2 \hat{j}+3 \hat{k} \text { and } \vec{b}=3 \hat{i}-6 \hat{j}+m \hat{k}$ are collinear then $m= ----$
$\begin{aligned} &\vec{a}=\hat{i}-2 \hat{j}+3 \hat{k} \\\\ &\vec{b}=3 \hat{i}-6 \hat{j}+m \hat{k} \end{aligned}$
Using $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

Then, $\frac{1}{3}=\frac{-2}{-6}=\frac{3}{m}$

$\frac{1}{3}=\frac{3}{m}$
Therefore, $m = 9.$

Algebra of Vectors exercise Fill in the blanks question 6

Answer:-$\vec{B}=9 \hat{i}-18 \hat{j}+6 \hat{k}$
Hint:-To solve this equation we use $\hat{A}=\frac{\vec{A}}{|\vec{A}|}$
Given:-Vector of magnitude $21$ units in the direction of the vector $3 \hat{i}-6 \hat{j}+2 \hat{k}$
Solution:- we have Vector of magnitude $21$ units in the direction of the vector $3 \hat{i}-6 \hat{j}+2 \hat{k}$
$\begin{aligned} &=\frac{3 \hat{i}-6 \hat{j}+2 \hat{k}}{\sqrt{9+36+4}} \\\\ &\hat{A}=\frac{3 \hat{i}-6 \hat{j}+2 \hat{k}}{\sqrt{49}} \end{aligned}$
$\begin{aligned} &\hat{A}=\frac{3 \hat{i}}{7}-\frac{6 \hat{j}}{7}+\frac{2 \hat{k}}{7} \\\\ &\vec{B}=|B| \hat{A} \end{aligned}$
$\begin{aligned} &\vec{B}=21\left[\frac{3 \hat{i}}{7}-\frac{6 \hat{j}}{7}+\frac{2 \hat{k}}{7}\right] \\\\ &\vec{B}=9 \hat{i}-18 \hat{j}+6 \hat{k} \end{aligned}$

Algebra of Vectors exercise Fill in the blanks question 7

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

$\begin{aligned} &\vec{b}-\vec{a}+\vec{c}-\vec{a}=k\left[\frac{\vec{b}+\vec{c}}{2}-\vec{a}\right] \\\\ &\vec{b}+\vec{c}-2 \vec{a}=k \frac{(\vec{b}+\vec{c}-2 \vec{a})}{2} \\\\ &\mathrm{k}=2 \end{aligned}$

Algebra of Vectors exercise Fill in the blanks question 8

Answer:- $\frac{1}{3}, \frac{-2}{3}, \frac{2}{3}$
Hint:- To solve this question we use,$\vec{a}=|\vec{a}|$
Given:- The cosines of the angles made by the vector $\hat{i}-2 \hat{j}+2 \hat{k}$ with the coordinates axes are
Solution:- Let $\vec{a}$ makes $\alpha,\beta \; \&\; \gamma$ with $x,y$ and $z$ axis respectively
$\begin{aligned} &\vec{a}=\hat{i}-2 \hat{j}+2 \hat{k} \\\\ &|\vec{a}|=\sqrt{1+4+4}=\sqrt{9}=3 \\\\ &\text { Dc of } \vec{a}=\left(\frac{1}{3}, \frac{-2}{3}, \frac{2}{3}\right)=(l, m, n) \end{aligned}$

Algebra of Vectors exercise Fill in the blanks question 9

Answer:-$3 \sqrt{2} \hat{i}+3 \hat{j}+3 \hat{k}$
Hint:-To solve this equation, we use standard formula $l^{2}+n^{2}+m^{2}=1 .$ .
Given:-A vector of magnitude 6; making angle $\frac{\pi }{4}$ with x-axis; $\frac{\pi }{3}$ with y axis and an acute angle with z axis
Solution:-we have $|\vec{a}|=6$
$\begin{aligned} &\text { So } l=\cos \alpha=\frac{\sqrt{2}}{2} \\\\ &m=\cos \beta=\frac{1}{2} \\\\ &n=\cos \gamma=\cos \theta \end{aligned}$

We know that $l^{2}+n^{2}+m^{2}=1$

$\begin{aligned} &\left(\frac{\sqrt{2}}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}+n^{2}=1 \\\\ &\frac{2}{4}-\frac{1}{4}+n^{2}=1 \\\\ &\frac{3}{4}+n^{2}=1 \end{aligned}$
$\begin{aligned} &n^{2}=1-\frac{3}{4}=\frac{4-3}{4}=\frac{1}{4} \\\\ &n=\pm \frac{1}{2} \\\\ &\hat{a}=l \hat{i}+m \hat{j}+n \hat{k}=\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{2} \hat{j}+\frac{1}{2} \hat{k} \end{aligned}$
$\begin{aligned} &\hat{a}=|\hat{a}|(\hat{i}+m \hat{j}+n \hat{k}) \\\\ &=6\left(\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{2} \hat{j}+\frac{1}{2} \hat{k}\right) \\\\ &=3 \sqrt{2} \hat{i}+3 \hat{j}+3 \hat{k} \end{aligned}$

Algebra of Vectors exercise Fill in the blanks question 10

Answer:- $x+y+z=2$
Hint:-To solve this equation, we use substitute method
Given:- If the vector $\vec{a}=2 \hat{i}-(y+z) \hat{j}+5 \hat{k} \text { and } \vec{b}=(x+y) \hat{i}+3 \hat{j}+(z+x) \hat{k}$ are then $x+y+z=$____
Solution:- $\vec{a}=\vec{b}$
$2 \hat{i}-(y+z) \hat{j}+5 \hat{k}=(x+y) \hat{i}+3 \hat{j}+(z+x) \hat{k}$
Comparing we get
$\begin{array}{ll} x+y=2 & \text {...( } i) \\\\ -(y+z)=3 & \text {...(ii) } \\\\ , z+x=5 & \text {...(iii) } \end{array}$
From eq (ii) we make y the subject to the equation
Substitute for $y=-3-z$ in the equation(i)
$\begin{aligned} &x-3-z=2 \quad ......(iv)\\ \\ &x-z=5 \end{aligned}$
Solving eq(iii) & (iv)
$\begin{aligned} &z+x+x-z=5+5 \\\\ &2 x=10 \\\\ &x=5 \end{aligned}$
Put the values of x in eq(iii)
$\begin{aligned} &z+5=5 \\\\ &z=0 \end{aligned}$
Put the values of z in eq(ii)
$\begin{aligned} &\mathrm{y}=-3 \\\\ &\text { so, } \mathrm{x}+\mathrm{y}+\mathrm{z}=5+(-3)+0 \\\\ &\quad=5-3 \\\\ &\quad=2 \end{aligned}$

Algebra of Vectors exercise Fill in the blanks question 11

Answer:- $\vec{c}=3 \vec{b}-2 \vec{a}$
Hint:-To solve this equation, we use magnitude length.
Given:- If $\vec{a}$ and $\vec{b}$ are the position vector of A and B respectively, then the position vector of a point $C$and $AB$ produced such that $\overrightarrow{A C}=3 \overrightarrow{A B}$
Solution:-here
$\begin{aligned} &\overrightarrow{A B}=\vec{B}-\vec{A}=\vec{b}-\vec{a} \\\\ &\overrightarrow{A C}=\vec{C}-\vec{A}=\vec{c}-\vec{a} \end{aligned}$
We have $\overrightarrow{A C}=3 \overrightarrow{A B}$
$|\overrightarrow{A B}|=x$
Then $\overrightarrow{A C}=3 x$
$\frac{A C}{B C}=\frac{m}{n} \rightarrow \bar{c}=\frac{m \vec{b}-n \vec{a}}{m-n}$
External dimension
$\bar{c}=\frac{3 \vec{b}-2 \vec{a}}{3-2}=\frac{3 \vec{b}-2 \vec{a}}{1}=3 \vec{b}-2 \vec{a}$

Algebra of Vectors exercise Fill in the blanks question 12

Answer:- $\overrightarrow{A C}+\overrightarrow{A F}-\overrightarrow{A B}=\overrightarrow{A E}$
Hint:-To solve this equation, we use reverse process of vectors
Given:-In a rectangular hexagon $\text { ABCDEF, } \overrightarrow{A C}+\overrightarrow{A F}-\overrightarrow{A B}$

Solution:-
we have $\overrightarrow{A E}=\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{C D}+\overrightarrow{D E}$ …(i)
From figure we can see
$\begin{aligned} &\overrightarrow{A B}+\overrightarrow{B C}=\overrightarrow{A C}\quad \quad\quad\quad\quad\quad\quad...(ii)\\ \\&\overrightarrow{C D}=\overrightarrow{A F} \quad[\because A F \| C D,-A F=C D]\quad ...(iii)\\ \\&\overrightarrow{D E}=-\overrightarrow{A B}\quad\quad\quad\quad\quad\quad ...(iv) \end{aligned}$
Using (ii) , (iii) and (iv) in (i), we get
$\begin{aligned} &\overrightarrow{A E}=\overrightarrow{A C}+\overrightarrow{A F}-\overrightarrow{A B} \\\\ &\text { Hence } \overrightarrow{A C}+\overrightarrow{A F}-\overrightarrow{A B}=\overline{A E} \end{aligned}$

Algebra of Vectors exercise Fill in the blanks question 13

Answer: $1$
Hint:-To solve this equation, we split $\overline{O A} \text { in } a, b, c$ forms
Given:-If $O,A,B,C and \; D$are five points $3 \overrightarrow{O D}+\overrightarrow{D A}+\overrightarrow{D B}+\overrightarrow{D C}=k(\overline{O A}+\overrightarrow{O B}+\overrightarrow{O C}) \text { then } \mathrm{k}=$____
Solution: $O=\vec{o}, A=\vec{a}, B=\vec{b}, C=\vec{c}, D=\vec{d}$
$\begin{aligned} &3 \overrightarrow{O D}+\overrightarrow{D A}+\overrightarrow{D B}+\overrightarrow{D C}=k(\overline{O A}+\overrightarrow{O B}+\overrightarrow{O C}) \\\\ &3(d-o)+(a-d)+(b-d)+(c-d)=k[(a-o)+(b-o)+(c-o)] \end{aligned}$
$\begin{aligned} &3 d-3 o+a-d+b-d+c-d=k(a-o+b-o+c-o) \\\\ &(-30+a+b+c)=k(-3 o+a+b+c) \\\\ &k=1 \end{aligned}$

Algebra of Vectors exercise Fill in the blanks question 14

Answer:- $3$
Hint:-To solve this equation we do $\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}$ method
Given:-$A, B, C, D, E$ are five coplanar, such that $\overrightarrow{D A}+\overrightarrow{D B}+\overrightarrow{D C}+\overrightarrow{A E}+\overrightarrow{B E}+\overrightarrow{C E}=K \overrightarrow{D E} \text { then } \mathrm{k} \text { is.... }$
Solution:- $\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}$
$\overrightarrow{D A}+\overrightarrow{D B}+\overrightarrow{D C}+\overrightarrow{A E}+\overrightarrow{B E}+\overrightarrow{C E}=K \overrightarrow{D E}$
$\begin{aligned} &\Rightarrow(\overrightarrow{O A}-\overrightarrow{O D})+(\overrightarrow{O B}-\overrightarrow{O D})+(\overrightarrow{O C}-\overrightarrow{O D})+(\overrightarrow{O E}-\overrightarrow{O A})+(\overrightarrow{O E}-\overrightarrow{O B})+(\overrightarrow{O E}-\overrightarrow{O C})=k \overrightarrow{D E} \\\\ &\Rightarrow 3(\overrightarrow{O E}-\overrightarrow{O D})=k \overrightarrow{D E} \\ \end{aligned}$
$\Rightarrow 3 \overrightarrow{D E}=k \overrightarrow{D E} \\$
$\Rightarrow 3=k \\$
$\Rightarrow k=3$

Algebra of Vectors exercise Fill in the blanks question 15

Answer:- $\frac{1}{3}$
Hint:-To solve this we know that two lines are collinear
Given:-The vector $\vec{a}$ and $\vec{b}$ are non collinear if $(x-2) \vec{a}+\vec{b} \text { and }(2 x+1) \vec{a}-\vec{b}$ are collinear then x……
Solution:- Let $a_{1} x+b_{1} y+c_{1}=0, a_{2} x+b_{2} y+c_{2}=0$
For collinear
$\begin{aligned} &\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}} \\\\ &\vec{c}=(x-2) \vec{a}+\vec{b} \\\\ &\vec{b}=(2 x+1) \vec{a}-\vec{b} \end{aligned}$
Hence
$\begin{aligned} \\ &\frac{x-2}{2 x+1}=\frac{1}{-1}\\\\ &-x+2=2 x+1\\\\ &2-1=2 x+x \end{aligned}$
$\begin{aligned} &1=3 x \\\\ &x=\frac{1}{3} \end{aligned}$

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 $\overrightarrow{A B}+\overrightarrow{A E}+\overrightarrow{B C}+\overrightarrow{D C}+\overrightarrow{E D}=K \overrightarrow{A C}$ then K IS….

Solution:-
$\begin{aligned} \overrightarrow{A B}+\overrightarrow{A E}+\overrightarrow{B C}+\overrightarrow{D C}+\overrightarrow{E D} \\ \end{aligned}$
$\begin{aligned} &=(\overrightarrow{A B}+\overrightarrow{B C})+(\overrightarrow{E D}+\overrightarrow{D C})+\overrightarrow{A E} \\\\ &=\overrightarrow{A C}+(\overrightarrow{E C}+\overrightarrow{A E}) \end{aligned}$ [∴Triangle Law of Vector Addition]
$\begin{aligned} &=\overrightarrow{A C}+\overrightarrow{A C} \\\\ &=2 \overrightarrow{A C} \\\\ &\therefore K=2 \end{aligned}$

Algebra of Vectors exercise Fill in the blanks question 17

Answer:-$\lambda =4$
Hint:-To solve this question we have to connect or add $AP,BP ,CP\; and\; DP$ vectors
Given:-Let P be the point of intersection of the diagonal of a parallelogram $ABCD$ and $O$ is any point If $\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}+\overrightarrow{O D}=\lambda \overrightarrow{O P}$
Solution:-

$\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}+\overrightarrow{O D}=\lambda \overline{O P}$
$\begin{aligned} &=(\overrightarrow{O A}+\overrightarrow{A P})+(\overrightarrow{O B}+\overrightarrow{B P})+(\overrightarrow{O C}-\overrightarrow{C P})+(\overrightarrow{O D}+\overrightarrow{D P}) \\\\ &=\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}+(\overrightarrow{A P}+\overline{C P})+(\overline{B P}+\overrightarrow{D P}) \\\\ &=\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}+\overrightarrow{O D}=4 \overline{O P} \\\\ &\lambda=4 \end{aligned}$

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 $BC,CA \; and \; AB$ respectively of $\Delta ABC$ then $\overrightarrow{A D}+\overrightarrow{B E}+\overrightarrow{C \vec{F}}=---$
Solution:-
In $\triangle A B C, D, E, F$ are mid pts.

Let $\vec{a}, \vec{b}, \vec{c}$ be the position vector of $\vec{A}, \vec{B}, \vec{C}$
Using mid point formula.
As $D$ is a mid point of $BC$
$D=\frac{\vec{b}+\vec{c}}{2}$
Similarly E and F are the mid points of AC and AB respectively
$\begin{aligned} &\therefore E=\frac{\vec{a}+\vec{c}}{2} \\\\ &F=\frac{\vec{a}+\vec{b}}{2} \end{aligned}$
Now,
$\overrightarrow{A D}+\overrightarrow{B E}+\overrightarrow{C F}=\overrightarrow{O D}-\overrightarrow{O A}+\overrightarrow{O E}-\overrightarrow{O B}+\overrightarrow{O F}-\overrightarrow{O C}$
$\begin{aligned} &=\overrightarrow{O D}+\vec{O} E+\overrightarrow{O F}-[\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}] \\\\ &=\frac{\vec{b}+\vec{c}}{2}+\frac{\vec{a}+\vec{c}}{2}+\frac{\vec{a}+\vec{b}}{2}-(\vec{a}+\vec{b}+\vec{c}) \end{aligned}$
$\begin{aligned} &=\frac{2(\vec{a}+\vec{b}+\vec{c})}{2}-(\vec{a}+\vec{b}+\vec{c}) \\\\ &=(\vec{a}+\vec{b}+\vec{c})-(\vec{a}+\vec{b}+\vec{c}) \\\\ &=0 \end{aligned}$

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:-
$\begin{aligned} &\text { } \overrightarrow{A D}+\overrightarrow{B \bar{E}}+\overrightarrow{C \vec{F}}\\\\ &\text { In } \Delta A B D\; \; \overrightarrow{A B}+\overrightarrow{B D}+\overrightarrow{D A}=0 \end{aligned}$

$\begin{aligned} &\overrightarrow{E C}=\frac{1}{2} \overrightarrow{A C} \\\\ &\overrightarrow{B D}=\frac{1}{2} \overrightarrow{B C} \\\\ &\overrightarrow{F B}=\frac{1}{2} \overrightarrow{A B} \end{aligned}$
$\begin{aligned} &\therefore \overrightarrow{A D}=\overrightarrow{A B}+\frac{1}{2} \overrightarrow{B C} \\\\ &\text { In } \Delta B E C \; \; \overrightarrow{B C}+\overrightarrow{C E}+\overrightarrow{E B}=0 \end{aligned}$
$\begin{aligned} &\therefore \overrightarrow{B E}=\overrightarrow{B C}+\frac{1}{2} \overrightarrow{C A} \\\\ &\text { In } \Delta A C F \overrightarrow{A F}+\overrightarrow{F C}+\overrightarrow{C A}=0 \\\\ &\therefore \overrightarrow{C F}=\overrightarrow{C A}+\frac{1}{2} \overrightarrow{A B} \end{aligned}$
$\begin{aligned} &\overrightarrow{A B}+\frac{1}{2} \overrightarrow{B C}+\overrightarrow{B C}+\frac{1}{2} \overrightarrow{C A}+\overrightarrow{C A}+\frac{1}{2} \overrightarrow{A B} \\\\ &\Delta A B C \\\\ &\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{C A}=0 \end{aligned}$
$\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{C A}+\frac{1}{2}(\overrightarrow{B C}+\overrightarrow{C A}+\overrightarrow{A B})=0$

Algebra of Vectors exercise Fill in the blanks question 21

Answer:- $k \in\left(-1, \frac{1}{2}\right) \cup\left(-\frac{1}{2}, 1\right)$
Hint:-To solve this question, we use magnitude formula.
Given:-The value of k for which $|k \vec{a}|<|\vec{a}| \text { and } k \vec{a}+\frac{1}{2} \vec{a}$is parallel to $\vec{a}$ holds true are ---
Solution:- $|k \vec{a}|<|\vec{a}|$
$\begin{aligned} &\Rightarrow k|\vec{a}|<|\vec{a}| \\\\ &\Rightarrow k<1 \\ \\&\Rightarrow-1<k<1 k \in(-1,1) \end{aligned}$$\begin{aligned} &\Rightarrow k|\vec{a}|<|\vec{a}| \\\\ &\Rightarrow k<1 \\ \\&\Rightarrow-1<k<1 k \in(-1,1) \end{aligned}$
$\begin{aligned} &k \vec{a}+\frac{1}{2} \vec{a} \text { is parallel to } a\\\\ &\text { If we put } k=\frac{-1}{2} \text { then } k \vec{a}+\frac{1}{2} \vec{a} \end{aligned}$
$\begin{aligned} &=-\frac{1}{2} \vec{a}+\frac{1}{2} \vec{a}=0 \\\\ &k \neq \frac{-1}{2} \end{aligned}$
$k \in\left(-1, \frac{1}{2}\right) \cup\left(-\frac{1}{2}, 1\right)$

Algebra of Vectors exercise Fill in the blanks question 22

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

$\begin{aligned} &\Rightarrow \angle D A C=\angle C A B=\theta \\\\ &\text { Rhombus } \Rightarrow A B=B C \\\\ &|\vec{a}|=|\vec{b}| \end{aligned}$

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 $\overrightarrow{O A}=2 \hat{i}-\hat{j}-\hat{k} \text { and } \overrightarrow{O B}=2 \hat{i}-\hat{j}+2 \hat{k}$ respectively. The position vectors of a point $P$which divides the line segment joining $A$ and $B$ in the ratio $2:1$ is ----
Solution:- $\vec{a}=2 \hat{i}-\hat{j}-\hat{k}$

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

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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