RD Sharma Solutions Class 12 Mathematics Chapter 22 MCQ

The RD Sharma solution chapters are said to be of the highest quality when it comes under the consideration of practicing maths with effective results from students all across the country. The Class 12 RD Sharma chapter 22 exercise MCQ solution are the multiple-choice questions, where each question has four options to choose from, thus making it easy for the students to confidently answer the questions. The RD Sharma class 12th exercise MCQ contains questions covering the basics and the tough level questions, all in the same solution paper.

RD Sharma Class 12 Solutions Chapter22 MCQ 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-FBQ

• Chapter 22 - Algebra of vectors - Ex-VSA

Algebra of Vectors Excercise: MCQ

Algebra of Vectors exercise Multiple choice question 1

Answer: $(1-\sqrt{3})\hat{i}+(1+\sqrt{3})\hat{j}$
Hint: You have to find circumcenter of $\mathrm{\Delta }ABC$
Given: In $\mathrm{\Delta }ABC,A=(0,0),B=(3,3\sqrt{3}),C=(-3\sqrt{3},3)$, magnitude $2\sqrt{2}$
Solution:

$\begin{array}{rl}& |\overrightarrow{AO}|=2\sqrt{2}\\ & |\overrightarrow{AO}|=|\overrightarrow{BO}|=|\overrightarrow{CO}|=2\sqrt{2}=R\end{array}$
Let Positive vector of be $x\hat{i}+y\hat{j}$
$\begin{array}{rl}& |\overrightarrow{AO}|=\sqrt{x^2+y^2}\\ \\ & x^2+y^2=8\end{array}$ ...................(1)
$\begin{array}{rl}& \text{ Also }|\overrightarrow{BO}|=|\overrightarrow{CO}|\\ \\ & \sqrt{(x-3{)}^2+(y-3\sqrt{3}{)}^2}=\sqrt{(x+3\sqrt{3}{)}^2+(y-3{)}^2}\\ \\ & x^2-6x+9+y^2-6\sqrt{3}y+27=x^2+6\sqrt{3}x+27+y^2-6y+9\\ \\ & y(6-6\sqrt{3})=x(6\sqrt{3}+6)\end{array}$
$y=\frac{x(\sqrt{3}+1)}{1-\sqrt{3}}$ ......................(2)
Substituting from (2) and (1) we get
$\begin{array}{rl}& (1-\sqrt{3}{)}^2{x}^2+(1+\sqrt{3}{)}^2{x}^2=8(1-\sqrt{3}{)}^2\\ \\ & x^2\times 8=8(1-\sqrt{3}{)}^2\\ \\ & x=1-\sqrt{3}\\ \\ & y=1+\sqrt{3}\end{array}$
The positive vector of O is $(1-\sqrt{3})\hat{i}+(1+\sqrt{3})\hat{j}$
$\overrightarrow{AO}=(1-\sqrt{3})\hat{i}+(1+\sqrt{3})\hat{j}$

Algebra of Vectors exercise Multiple choice question 3

Answer:$3AC=5CB$
Hint: $ON\perp AB$
Given: Forces $3\overrightarrow{OA},5\overrightarrow{OB}$ act along $OA$ and $OB$. If resultant pass through $C$ on $AB$ then
Solution: Draw $ON$ the perpendicular to line $AB$
Let i be unit actor along $ON$
The Resultant Force $\overrightarrow{R}=3\overrightarrow{OA}+5\overrightarrow{OB}$ ……….. (1)
The angle between $i$ and force
$\overrightarrow{R},3\overrightarrow{OA},5\overrightarrow{OB}\text{ are }\mathrm{\angle }CON,\mathrm{\angle }AON,\mathrm{\angle }BON$

$\overrightarrow{R}\overrightarrow{i}=3\overrightarrow{OA}\overrightarrow{i}+5\overrightarrow{OB}\overrightarrow{i}$
$\begin{array}{rl}& \text{ R.1.cos }\mathrm{\angle }CON=3\\ \\ & \overrightarrow{OA}\cdot 1.\cos \mathrm{\angle }BON\\ \\ & R\cdot \frac{ON}{OC}=3\overrightarrow{OA}\times \frac{ON}{OA}+5\overrightarrow{OB}\times \frac{ON}{OB}\\ \\ & \frac{R}{OC}=3+5,R=8\end{array}$
$\overrightarrow{OC}$ we know that
$\begin{array}{rl}& \overrightarrow{OA}=\overrightarrow{OC}+\overrightarrow{CA}\\ \\ & 3\overrightarrow{OA}=3\overrightarrow{OC}+3\overrightarrow{CA}\end{array}$ ..................(i)
$\begin{array}{rl}& \overrightarrow{OB}=\overrightarrow{OC}+\overrightarrow{CB}\\ \\ & 5\overrightarrow{OB}=5\overrightarrow{OC}+5\overrightarrow{CB}\end{array}$ ..................(ii)
On adding (i) and (ii)
$\begin{array}{rl}& 3\overrightarrow{OA}+5\overrightarrow{OB}=8\overrightarrow{OC}+3\overrightarrow{CA}+5\overrightarrow{CB}\end{array}$
$R=8\overrightarrow{OC}+3\overrightarrow{CA}+5\overrightarrow{CB}$
$8\overrightarrow{OC}=8\overrightarrow{OC}+3\overrightarrow{CA}+5\overrightarrow{CB}$
$|3\overrightarrow{AC}|=|5\overrightarrow{CB}|$
$3AC=5CB$

Algebra of Vectors exercise Multiple choice question 6

Answer: $4\overrightarrow{OG}$
Hint: Find $\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}=?$
Given: If $G$ is intersection of diagonal of parallelogram. $ABCD$ and $O$ is any point that $\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}=?$
Solution: Let consider the point $O$ as origin $G$ is the midpoint of $AC$

$\begin{array}{rl}& \overrightarrow{OG}=\frac{\overrightarrow{OA}+\overrightarrow{OC}}{2}\\ \\ & 2\overrightarrow{OG}=\overrightarrow{OA}+\overrightarrow{OC}\end{array}$ ................(1)
Also $G$ is midpoint of $BD$
$\begin{array}{rl}& \overrightarrow{OG}=\frac{\overrightarrow{OB}+\overrightarrow{OD}}{2}\\ \\ & 2\overrightarrow{OG}=\overrightarrow{OB}+\overrightarrow{OD}\end{array}$
On add (1) and (2)
$\begin{array}{rl}& 2\overrightarrow{OG}+2\overrightarrow{OG}=\overrightarrow{OA}+\overrightarrow{OC}+\overrightarrow{OB}+\overrightarrow{OD}\\ \\ & 4\overrightarrow{OG}=\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}\\ \\ & \overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}=4\overrightarrow{OG}\end{array}$

Algebra of Vectors exercise Multiple choice question 8

Answer: $\overrightarrow{b}+\overrightarrow{c}$
Hint: Find $\overrightarrow{AE}$
Given: In regular Hexagon $\text{ ABCDEF, }\overrightarrow{AB}=\overrightarrow{a},\overrightarrow{BC}=\overrightarrow{b},\overrightarrow{CD}=\overrightarrow{c}$ then $\overrightarrow{AE}=?$
Solution: We have
$\overrightarrow{AB}=\overrightarrow{a},\overrightarrow{BC}=\overrightarrow{b},\overrightarrow{CD}=\overrightarrow{c}$

$\overrightarrow{AC}=a+b$
In $\mathrm{\Delta }ACD$ we have
$\begin{array}{rl}& \overrightarrow{AC}+\overrightarrow{CD}=\overrightarrow{AD}\\ & \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{AD}\end{array}$
In $\mathrm{\Delta }AED$ we have
$\begin{array}{rl}& \overrightarrow{AD}+\overrightarrow{DE}=\overrightarrow{AE}\\ \\ & a+b+c-a=\overrightarrow{AE}\\ \\ & \overrightarrow{AE}=\overrightarrow{b}+\overrightarrow{c}\end{array}(DE=BA=-a)$

Algebra of Vectors exercise Multiple choice question 13

Answer: $4\overrightarrow{AB}$
Hint: Find $\overrightarrow{AD}+\overrightarrow{EB}+\overrightarrow{FC}$ equals - ?
Given: If $ABCDEF$ is regular Hexagon then $\overrightarrow{AD}+\overrightarrow{EB}+\overrightarrow{FC}$
Answer: We have

$\begin{array}{rl}& \overrightarrow{AD}+\overrightarrow{EB}+\overrightarrow{FC}\\ \\ & (\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD})+(\overrightarrow{ED}+\overrightarrow{DC}+\overrightarrow{CB})+\overrightarrow{FC}\end{array}$
$\begin{array}{c}\overrightarrow{AB}+(\overrightarrow{BC}+\overrightarrow{CB})+(\overrightarrow{CD}+\overrightarrow{DC})+\overrightarrow{ED}+\overrightarrow{FC}\\ \\ \overrightarrow{AB}+\overrightarrow{0}+\overrightarrow{0}+\overrightarrow{AB}+2\overrightarrow{AB}=4\overrightarrow{AB}\\ \\ \left(\begin{array}{l}\overrightarrow{ED}=\overrightarrow{AB},\\ \\ \overrightarrow{FC}=2\overrightarrow{AB}\end{array}\right)\end{array}$

Algebra of Vectors exercise Multiple choice question 19

Answer: $\overrightarrow{AB}+\overrightarrow{BC}-\overrightarrow{CA}=0$
Hint: According to diagram
Given: Which of the following is not true?
Solution:

$\text{ From the given diagram }$
$\begin{array}{rl}& \overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}\\ & \overrightarrow{AB}+\overrightarrow{BC}-\overrightarrow{CA}=\overrightarrow{AC}-\overrightarrow{CA}\\ & \Rightarrow \overrightarrow{AB}+\overrightarrow{BC}-\overrightarrow{CA}=2\overrightarrow{AC}\end{array}$
Hence, $\overrightarrow{AB}+\overrightarrow{BC}-\overrightarrow{CA}=0$ is not true.