RD Sharma Solutions Class 12 Mathematics Chapter 22 MCQ

Edited By Satyajeet Kumar | Updated on Jan 24, 2022 06:20 PM IST

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.

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

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

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 $\Delta ABC$
Given: In $\Delta A B C, A=(0,0), B=(3,3 \sqrt{3}), C=(-3 \sqrt{3}, 3)$ , magnitude $2\sqrt{2}$
Solution:

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

Algebra of Vectors exercise Multiple choice question 2

Answer: $\vec{b}-\vec{a}$
Hint: $AD=2BC$
Given: If $\vec{a}, \vec{b}$ are vector forming consecutive side of rectangular Hexagon $ABCDEF$ the vector representing side $CD$ is
Solution: Let $ABCDEF$ be rectangular Hexagon such that
$\overrightarrow{A B}=\vec{a} \text { and } \overrightarrow{B C}=\vec{b}$
We know
$AD$ is parallel $BC$ such that $AD = 2BC$
$\overrightarrow{A D}=2 \overrightarrow{B C}=2 \vec{b}$
In $\Delta ABC$ , we know
$\begin{aligned} &\overrightarrow{A B}+\overrightarrow{B C}=\overrightarrow{A C} \\\\ &\vec{a}+\vec{b}=\overrightarrow{A C} \end{aligned}$
In $\Delta ACD$ , we have
$\begin{aligned} &\overrightarrow{A C}+\overrightarrow{C D}=\overrightarrow{A D} \\\\ &\overrightarrow{C D}=\overrightarrow{A D}-\overrightarrow{A C} \\\\ &\overrightarrow{C D}=2 \vec{b}-(\vec{a}+\vec{b}) \\\\ &\overrightarrow{C D}=\vec{b}-\vec{a} \end{aligned}$

Algebra of Vectors exercise Multiple choice question 3

Answer: $3AC = 5CB$
Hint: $O N \perp A B$
Given: Forces $3 \overrightarrow{O A}, 5 \overrightarrow{O B}$ 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 $\vec{R}=3 \overrightarrow{O A}+5 \overrightarrow{O B}$ ……….. (1)
The angle between $i$ and force
$\vec{R}, 3 \overrightarrow{O A}, 5 \overrightarrow{O B} \text { are } \angle C O N, \angle A O N, \angle B O N$

$\vec{R} \vec{i}=3 \overrightarrow{O A} \vec{i}+5 \overrightarrow{O B} \vec{i}$
$\begin{aligned} &\text { R.1.cos } \angle C O N=3 \\\\ &\overrightarrow{O A} \cdot 1 . \cos \angle B O N \\\\ &R \cdot \frac{O N}{O C}=3 \overrightarrow{O A} \times \frac{O N}{O A}+5 \overrightarrow{O B} \times \frac{O N}{O B} \\\\ &\frac{R}{O C}=3+5, R=8 \end{aligned}$
$\overrightarrow{O C}$ we know that
$\begin{aligned} &\overrightarrow{O A}=\overrightarrow{O C}+\overrightarrow{C A} \\\\ &3 \overrightarrow{O A}=3 \overrightarrow{O C}+3 \overrightarrow{C A} \end{aligned}$ ..................(i)
$\begin{aligned} &\overrightarrow{O B}=\overrightarrow{O C}+\overrightarrow{C B} \\\\ &5 \overrightarrow{O B}=5 \overrightarrow{O C}+5 \overrightarrow{C B} \end{aligned}$ ..................(ii)
On adding (i) and (ii)
$\begin{aligned} &3 \overrightarrow{O A}+5 \overrightarrow{O B}=8 \overrightarrow{O C}+3 \overrightarrow{C A}+5 \overrightarrow{C B} \\ \end{aligned}$
$R=8 \overrightarrow{O C}+3 \overrightarrow{C A}+5 \overrightarrow{C B} \\$
$8 \overrightarrow{O C}=8 \overrightarrow{O C}+3 \overrightarrow{C A}+5 \overrightarrow{C B} \\$
$|3 \overrightarrow{A C}|=|5 \overrightarrow{C B}| \\$
$3 A C=5 C B$

Algebra of Vectors exercise Multiple choice question 4

Answer: None of these
Hint: Find $\vec{a}+\vec{b}+\vec{c}$
Given: If $\vec{a}, \vec{b}, \vec{c}$ are three non-vectors no two of which are collinear and vector $\vec{a}+\vec{b}$ is collinear with $\vec{c}, \vec{b}+\vec{c}$ is collinear with $\vec{a}$ then $\vec{a}+\vec{b}+\vec{c}=?$
Solution: Let $a+b=x c$ …………. (1)
$b+c=y a$ …………. (2)
Then $a+b+c=?$
Multiply equation (2) by 2
$2(b+c)=(y a) 2$ ...............(3)
$2 b+2 c=2 a y$
Subtract equation (3) by eq (1)
$\begin{aligned} &2 b+2 c-2 a y-a-b+x c \\\\ &b+(2+x) c+(-2 y-1) a=0 \end{aligned}$
Comparing both sides
$\begin{aligned} &b=0 \\ &(2+x) c=0 \\ &2 c=-x c \\ &x=-2 \end{aligned}$
$\begin{aligned} &(-2 y-1) a=0 \\ &-2 y a=1 \\ &y=\frac{1}{-2 a} \end{aligned}$

Algebra of Vectors exercise Multiple choice question 5

Answer: $-40$
Hint: Find a
Given: If $A(60 \hat{i}+3 \hat{j}), B(40 \hat{i}-8 \hat{j}) \text { and } C(a \hat{i}-52 \hat{j})$ are collinear then equal to
Solution:
$\begin{aligned} A &=60 \hat{i}+3 \hat{j} \\ B &=40 \hat{i}-8 \hat{j} \\ C &=a \hat{i}-52 \hat{j} \end{aligned}$
Now find $AB$ and $BC$
$\begin{aligned} &A B=-20 \hat{i}-11 \hat{j} \\ &B C=(a-40) \hat{i}-44 \hat{j} \end{aligned}$
Take cross product
$\begin{aligned} &A B \times B C=(-20 \hat{i}-11 \hat{j}) \times(a-40) \hat{i}-44 \hat{j} \\ &0 \hat{i}+0 \hat{j}+(880+11(a-40))=0 \\ &a-40=-80 \\ &a=-80+40 \\ &a=-40 \end{aligned}$

Algebra of Vectors exercise Multiple choice question 6

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

$\begin{aligned} &\overrightarrow{O G}=\frac{\overrightarrow{O A}+\overrightarrow{O C}}{2} \\\\ &2 \overrightarrow{O G}=\overrightarrow{O A}+\overrightarrow{O C} \end{aligned}$ ................(1)
Also $G$ is midpoint of $BD$
$\begin{aligned} &\overrightarrow{O G}=\frac{\overrightarrow{O B}+\overrightarrow{O D}}{2} \\\\ &2 \overrightarrow{O G}=\overrightarrow{O B}+\overrightarrow{O D} \end{aligned}$
On add (1) and (2)
$\begin{aligned} &2 \overrightarrow{O G}+2 \overrightarrow{O G}=\overrightarrow{O A}+\overrightarrow{O C}+\overrightarrow{O B}+\overrightarrow{O D} \\\\ &4 \overrightarrow{O G}=\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}+\overrightarrow{O D} \\\\ &\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}+\overrightarrow{O D}=4 \overrightarrow{O G} \end{aligned}$

Algebra of Vectors exercise Multiple choice question 7

Answer: Unit vector
Hint: $\hat{k}$ is which vector
Given: The vector $\cos \alpha \cos \beta \hat{i}+\cos \alpha \sin \beta \hat{i}+\sin \alpha \hat{k}$
Solution:
$\begin{aligned} &|\vec{A}|=\sqrt{(\cos \alpha \cos \beta)^{2}+(\cos \alpha \sin \beta)^{2}+\sin ^{2} \alpha} \\\\ &=\sqrt{\cos ^{2} \alpha\left(\cos ^{2} \beta+\sin ^{2} \beta\right)+\sin ^{2} \alpha}\\\\ &=\sqrt{\cos ^{2} \alpha+\sin ^{2} \alpha}=1 \text { (Unit vector) } \end{aligned}$ $\quad\left(\cos ^{2} \beta+\sin ^{2} \beta=1\right)$

Algebra of Vectors exercise Multiple choice question 8

Answer: $\vec{b}+\vec{c}$
Hint: Find $\overrightarrow{A E}$
Given: In regular Hexagon $\text { ABCDEF, } \overrightarrow{A B}=\vec{a}, \overrightarrow{B C}=\vec{b}, \overrightarrow{C D}=\vec{c}$ then $\overrightarrow{A E}=?$
Solution: We have
$\overrightarrow{A B}=\vec{a}, \overrightarrow{B C}=\vec{b}, \overrightarrow{C D}=\vec{c}$

$\overrightarrow{A C}=a+b$
In $\Delta ACD$ we have
$\begin{aligned} &\overrightarrow{A C}+\overrightarrow{C D}=\overrightarrow{A D} \\ &\vec{a}+\vec{b}+\vec{c}=\overrightarrow{A D} \end{aligned}$
In $\Delta AED$ we have
$\begin{aligned} &\overrightarrow{A D}+\overrightarrow{D E}=\overrightarrow{A E} \\\\ &a+b+c-a=\overrightarrow{A E} \\\\ &\overrightarrow{A E}=\vec{b}+\vec{c} \end{aligned} \quad(D E=B A=-a)$

Algebra of Vectors exercise Multiple choice question 9

Answer: $\alpha+\beta+\gamma=1$
Hint: Find plane of vector equation
Given: The vector equation of plane passing through $\vec{a}, \vec{b}, \vec{c}, \vec{r}=\alpha \vec{a}+\beta \vec{b}+\gamma \vec{c}$ provide that
Solution: $\vec{r}=\alpha \vec{a}+\beta \vec{b}+\gamma \vec{c}$ ................(1)
$\begin{aligned} &\vec{r}=\vec{a}+\lambda_{1} \vec{a}+\lambda_{1} \vec{c}-\lambda_{1} \vec{b}-\lambda_{2} \vec{a}\\\\ &\vec{r}=\vec{a}\left(\lambda_{1}-\lambda_{2}\right)+\vec{b}\left(-\lambda_{1}\right)+\lambda_{2} \vec{c} \end{aligned}$ ..............(2)
By comparing (1) and (2)
$\begin{aligned} &\alpha+\beta+\gamma=1+\lambda_{1}-\lambda_{2}-\lambda_{1}+\lambda_{2} \\ \\ &\alpha+\beta+\gamma=1 \end{aligned}$

Algebra of Vectors exercise Multiple choice question 10

Answer: $\overrightarrow{O O^{1}}$
Hint: $\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}$
Given : If $O$ and $O^{1}$ are circumcenter and orthocenter of $\Delta ABC$ , then $\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}$
Solution: According to properties of triangle circumcenter $O$ in $\Delta ABC$ the sum of distance from circumcentre to orthocentre
$\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}=\overrightarrow{O O^{1}}$

Algebra of Vectors exercise Multiple choice question 11

Answer: Parallelogram
Hint: Find $ABCD$
Given: If $\vec{a}, \vec{b}, \vec{c} \; \&\; \vec{d}$ are position vectors of points $A,B,C$ and $D$ such that no three of them are collinear $\vec{a}+\vec{c}=\vec{b}+\vec{d}$ then $ABCD$ is
Solution: Given $\vec{a}+\vec{c}=\vec{b}+\vec{d}$
$\begin{aligned} &\vec{c}-\vec{d}=\vec{b}-\vec{a} \\ \\ &\overrightarrow{A B}=\overrightarrow{D C} \text { and } \vec{a}+\vec{c}=\vec{b}+\vec{d} \\ \\ &\vec{c}-\vec{d}=\vec{b}-\vec{a} \\ \\ &\overrightarrow{A D}=\overrightarrow{B C} \end{aligned}$
Since $\vec{a}+\vec{c}=\vec{b}+\vec{d}$
$\frac{1}{2}(\vec{a}+\vec{c})=\frac{1}{2}(\vec{b}+\vec{d})$
So position vector midpoint of $BD =$ position vector of midpoint of $AC$
Hence bisect each other
The given point $ABCD$ is parallelogram.

Algebra of Vectors exercise Multiple choice question 12

Answer: $\frac{1}{3}(\vec{a}+\vec{b})$
Hint: Find the centroid $G$
Given: Let G is centroid of $\Delta A B C \; \; \overrightarrow{A B}=\vec{a}, \overrightarrow{A C}=\vec{b}$ then bisector $\overrightarrow{A G}$ in terms of $\vec{a} \; \&\; \vec{b}$
Solution: Take $A$ as origin them position vector $A, B, C$ are given $0, a$ and $b$ respectively
The centroid $=\frac{0+a+b}{3}=\frac{a+b}{3}=\frac{1}{3}(a+b)$

Algebra of Vectors exercise Multiple choice question 13

Answer: $4 \overrightarrow{A B}$
Hint: Find $\overrightarrow{A D}+\overrightarrow{E B}+\overrightarrow{F C}$ equals - ?
Given: If $ABCDEF$ is regular Hexagon then $\overrightarrow{A D}+\overrightarrow{E B}+\overrightarrow{F C}$
Answer: We have

$\begin{aligned} &\overrightarrow{A D}+\overrightarrow{E B}+\overrightarrow{F C} \\ \\ &(\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{C D})+(\overrightarrow{E D}+\overrightarrow{D C}+\overrightarrow{C B})+\overrightarrow{F C} \end{aligned}$
$\begin{gathered} \overrightarrow{A B}+(\overrightarrow{B C}+\overrightarrow{C B})+(\overrightarrow{C D}+\overrightarrow{D C})+\overrightarrow{E D}+\overrightarrow{F C} \\ \\ \overrightarrow{A B}+\overrightarrow{0}+\overrightarrow{0}+\overrightarrow{A B}+2 \overrightarrow{A B}=4 \overrightarrow{A B} \\ \\ \left(\begin{array}{l} \overrightarrow{E D}=\overrightarrow{A B}, \\ \\ \overrightarrow{F C}=2 \overrightarrow{A B} \end{array}\right) \end{gathered}$

Algebra of Vectors exercise Multiple choice question 14

Answer: Form as isosceles triangle.
Hint: $ABC$ for which triangle
Given: The position vector of points $A, B$ and $C$ are $2 \vec{i}+\vec{j}-\vec{k}, 3 \vec{i}-2 \vec{j}+\vec{k}, i+4 j-3 k$
Solution:
$\begin{aligned} &\overrightarrow{O A}=2 \hat{i}+\hat{j}-\hat{k}, \\\\ &\overrightarrow{O B}=3 \hat{i}-2 \hat{j}+\hat{k}, \\\\ &\overrightarrow{O C}=\hat{i}+4 \hat{j}-3 \hat{k} \end{aligned}$
$\begin{aligned} &\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A} \\\\ &=(3-2) \hat{i}+(-2-1) \hat{j}+(1+1) \hat{k} \\\\ &=\hat{i}-3 \hat{j}+2 \hat{k} \end{aligned}$
$\begin{aligned} &|\overrightarrow{A B}|=\sqrt{1^{2}+3^{2}+2^{2}}=\sqrt{1+9+4}=\sqrt{14} \\\\ &\overrightarrow{B C}=\overrightarrow{O C}-\overrightarrow{O B} \\\\ &=(1-3) \hat{i}+(4+2) \hat{j}+(-3-1) \hat{k} \\\\ &=-2 \hat{i}+6 \hat{j}-4 \hat{k} \end{aligned}$
$\begin{aligned} &|\overrightarrow{B C}|=\sqrt{(2)^{2}+6^{2}+(-4)^{2}}=\sqrt{4+36+16}=\sqrt{56}=2 \sqrt{14} \\\\ &\overrightarrow{C A}=\overrightarrow{O A}-\overrightarrow{O C} \\\\ &=(2-1) \hat{i}+(1-4) \hat{j}+(-1+3) \hat{k} \end{aligned}$
$\begin{aligned} &=\hat{i}-3 \hat{j}+2 \hat{k} \\\\ &|\overrightarrow{C A}|=\sqrt{1^{2}+3^{2}+2^{2}}=\sqrt{1+9+4}=\sqrt{14} \end{aligned}$

Hence $\overrightarrow{A B}\; \& \; \overrightarrow{C A}=\sqrt{14}$
It forms isosceles triangle.

Algebra of Vectors exercise Multiple choice question 15

Answer: $(2,-3)$
Hint: $(x,y)$ value find
Given: If three points $A, B , C$ have position vectors $\begin{aligned} &\hat{i}+x \hat{j}+3 \hat{k} \\\\ &3 \hat{i}+4 \hat{j}+7 \hat{k} \\ \\ &y \hat{i}-2 \hat{j}-5 \hat{k} \end{aligned}$ respective are collinear , then $(x,y)=$
Solution: If
$\begin{aligned} &\hat{i}+x \hat{j}+3 \hat{k} \\\\ &3 \hat{i}+4 \hat{j}+7 \hat{k} \\\\ &y \hat{i}-2 \hat{j}-5 \hat{k} \end{aligned}$
$\begin{aligned} &\overrightarrow{A B}=3 \hat{i}+4 \hat{j}+7 \hat{k}-\hat{i}-x \hat{j}-3 \hat{k}=2 \hat{i}+(4-x) \hat{j}+4 \hat{k} \\\\ &\overrightarrow{B C}=y \hat{i}-2 \hat{j}-5 \hat{k}-3 \hat{i}-4 \hat{j}-7 \hat{k}=(y-3) \hat{i}-6 \hat{j}-12 \hat{k} \end{aligned}$
Since vectors are collinear.
$\begin{aligned} &\overrightarrow{A B}=\lambda \overrightarrow{B C} \\\\ &2 \hat{i}+(4-x) \hat{j}+4 \hat{k}=\lambda((y-3) \hat{i}-6 \hat{j}-12 \hat{k}) \\\\ &2=\lambda(y-3) \end{aligned}$ ...................(1)
$(4-x)=-6 \lambda$ ....................(2)
$4=-12 \lambda \Rightarrow \lambda=\frac{-1}{3}$

Substitute $\lambda$ in (1) and (2)

$\begin{aligned} &2=\frac{-1}{3}(y-3) \Rightarrow-6=y-3 \Rightarrow y=-3 \\\\ &4-x=-6 \times \frac{-1}{3} \Rightarrow 4-x=2 \Rightarrow x=2 \\\\ &x=2, y=-3 \end{aligned}$

Algebra of Vectors exercise Multiple choice question 16

Answer: $2 \overrightarrow{A B}$
Hint: $AC-BD=?$
Given: $ABCD$ is parallelogram with $AC$ and $BS$ as diagonal
Solution: $ABCD$ is parallelogram So $A B\|C D\; \&\; A D\| B C$
$\overrightarrow{A C}=\overrightarrow{A B}-\overrightarrow{B C}$ (Subtraction theorem)
$\overrightarrow{B D}=\overrightarrow{B A}-\overrightarrow{A D}$ (Subtraction theorem)
$\begin{aligned} &\overrightarrow{A C}-\overrightarrow{B D}=\overrightarrow{A B}-\overrightarrow{B C}-\overrightarrow{B A}+\overrightarrow{A D}\\\\ &=\overrightarrow{A B}-\overrightarrow{B C}+\overrightarrow{A B}+\overrightarrow{B C} \quad\quad\quad\quad\quad(A B\|C D\; \& \; A D\| B C) \end{aligned}$
= $2 \overrightarrow{A B}$

Algebra of Vectors exercise Multiple choice question 17

Answer: $\frac{1}{2}(\vec{a}-\vec{b})$
Hint: $OA=?$
Given: If $OACB$ is parallelogram $\overrightarrow{O C}=\vec{a}, \overrightarrow{A B}=\vec{b}$
Solution: $\overrightarrow{O C}=\vec{a}, \overrightarrow{A B}=\vec{b}$
$\overrightarrow{O B}+\overrightarrow{B C}=\overrightarrow{O C} \$ $(\overrightarrow{B C}=\overrightarrow{O A})$
$\begin{aligned} &\overrightarrow{O B}=\overrightarrow{O C}-\overrightarrow{O A} \\ \\ &\overrightarrow{O B}=\vec{a}-\overrightarrow{O A} \\ \\ &\overrightarrow{O A}+\overrightarrow{A B}=\overrightarrow{O B} \end{aligned}$
$\begin{aligned} &2 \overrightarrow{O A}=\vec{a}-\vec{b} \\\\ &\overrightarrow{O A}=\frac{\vec{a}-\vec{b}}{2} \end{aligned}$

Algebra of Vectors exercise Multiple choice question 18

Answer: Both $\vec{a} \; \& \; \vec{b}$ have same direction but different magnitude
Hint: Talk about $\vec{a} \; \& \; \vec{b}$
Given: If $\vec{a} \; \& \; \vec{b}$ are collinear vector, then which of the following are incorrect?
Solution: If $\vec{a} \; \& \; \vec{b}$ are two collinear vectors then they are parallel
Therefore, $\vec{b}=\lambda \vec{a}$
$\begin{aligned} &\text { If } \lambda=\pm 1 \text { then } a=\pm b\\\\ &\text { If } \vec{a}=a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k} \text { and } \vec{b}=b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k} \text { then } \vec{b}=\lambda \vec{a} \end{aligned}$
$\begin{aligned} &b_{1}=\lambda a_{1}, b_{2}=\lambda a_{2} \text { and } b_{3}=\lambda a_{3} \\\\ &\frac{b_{1}}{a_{1}}=\frac{b_{2}}{a_{2}}=\frac{b_{3}}{a_{3}}=\lambda \end{aligned}$
Thus, respective component of $\vec{a} \; \& \; \vec{b}$ proportional. However, vector $\vec{a} \; \& \; \vec{b}$ have different direction.
So Statement $D$ is incorrect
That both $\vec{a} \; \& \; \vec{b}$ have same direction but different magnitude.

Algebra of Vectors exercise Multiple choice question 19

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

$\text { From the given diagram }$
$\begin{aligned} &\overrightarrow{A B}+\overrightarrow{B C}=\overrightarrow{A C} \\ &\overrightarrow{A B}+\overrightarrow{B C}-\overrightarrow{C A}=\overrightarrow{A C}-\overrightarrow{C A} \\ &\Rightarrow \overrightarrow{A B}+\overrightarrow{B C}-\overrightarrow{C A}=2 \overrightarrow{A C} \end{aligned}$
Hence, $\overrightarrow{A B}+\overrightarrow{B C}-\overrightarrow{C A}=0$ is not true.

Algebra of Vectors exercise Multiple choice question 20

Answer: $\frac{1}{3}(4 \vec{a}+\vec{b})$
Hint: Measure by $1:2$
Given: The position vector of point which divide the join of point with position $\vec{a}+\vec{b} \text { and } 2 \vec{a}-\vec{b}$ in ratio $1:2$ is:
Solution: Apply section formula the position vector require point
$\frac{2(\vec{a}+\vec{b})+1(2 \vec{a}-\vec{b})}{2+1}=\frac{4 \vec{a}+\vec{b}}{3}$
Since position vector of point $R$ divide line segment joining the point $P$ and $Q$ where position vector are $\vec{p}\; \& \; \vec{q}$ in ration $m : n$
$=\frac{m \vec{q}+n \vec{p}}{m+n}$

Algebra of Vectors exercise Multiple choice question 21

Answer: $\hat{i}-\hat{j}+2 \hat{k}$
Hint: Initial vector find
Given: The vector with initial point $P(2,-3,5)$ terminal point $Q(3,-4,7)$ is
Solution: Require vector $\begin{aligned} \overrightarrow{A B}=&(3-2) \hat{i}+(-4+3) \hat{j}+(7-5) \hat{k} \\\\ \end{aligned}$
$=\hat{i}-\hat{j}+2 \hat{k}$

Algebra of Vectors exercise Multiple choice question 22

Answer: $\frac{\sqrt{35}}{2}$
Hint: The length of median by median formula
Given: The vector $2 \hat{j}+\hat{k} \& 3 \hat{i}-\hat{j}+4 \hat{k}$ represent two side $\mathrm{AB}\; \&\; \mathrm{AC} \text { of } \Delta A B C$ . Find length of median,
Solution:
$\begin{aligned} &\overrightarrow{B C}=\overrightarrow{A C}-\overrightarrow{A B} \\\\ &=(3 \hat{i}-\hat{j}+4 \hat{k})-(2 \hat{j}+\hat{k}) \\\\ &=3 \hat{i}-3 \hat{j}+3 \hat{k} \end{aligned}$
$\begin{aligned} &\overrightarrow{B D}=\frac{1}{2} B C \Rightarrow \frac{3}{2} \hat{i}-\frac{3}{2} \hat{j}+\frac{3}{2} \hat{k} \\\\ &\overrightarrow{A D}=A B+B D \\\\ &=(2 \hat{j}+\hat{k})+\left(\frac{3}{2} \hat{i}-\frac{3}{2} \hat{j}+\frac{3}{2} \hat{k}\right) \end{aligned}$
$\begin{gathered} \overrightarrow{A D}=\frac{3}{2} \hat{\imath}+\frac{1}{2} \hat{\jmath}+\frac{5}{2} \hat{k} \\\\ |A D|=\sqrt{\left(\frac{3}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}+\left(\frac{5}{2}\right)^{2}} \end{gathered}$
$=\sqrt{\frac{9}{4}+\frac{1}{4}+\frac{25}{4}}=\sqrt{\frac{35}{4}}=\frac{\sqrt{35}}{2}$

Algebra of Vectors exercise Multiple choice question 23

Answer: $[0,6]$
Hint: Find interval of $\left | \lambda \vec{a} \right |$
Given: If $|\vec{a}|=3 \text { and }-1 \leq \lambda \leq 2$ then $\left | \lambda \vec{a} \right |$ lies in
Solution: $\left | \lambda \vec{a} \right |=?$
$|\vec{a}|=3 \quad-1 \leq \lambda \leq 2$
$\begin{aligned} &0 \leq \lambda \vec{a} \leq 2 \\\\ &0 \leq 3 \lambda \leq 2 \\\\ &0 \leq \lambda \leq 6 \\\\ &\lambda \vec{a} \in[0,6] \end{aligned}$

Algebra of Vectors exercise Multiple choice question 25

Answer: $\frac{5}{4} \vec{a}$
Hint: According to ratio $3:1$
Given: The position vector of point which divide the join of points $2 \vec{a}-3 \vec{b} \; \& \; \vec{a}+\vec{b}$ in ratio $3:1$
Solution: Position vector $R=\frac{3(\vec{a}+\vec{b})+1(2 \vec{a}-3 \vec{b})}{3+1}$
Since point vector are joint by $R$ divide line segment join $P$ and $Q$ whose positive vector $\vec{p}\; \&\; \vec{q}$ in ratio $m:n$ given $\frac{m \vec{q}+n \vec{p}}{m+n}$

$R=\frac{5}{4} \vec{a}$

Algebra of Vectors exercise Multiple choice question 26

Answer: $\overrightarrow{0}$
Hint: $\overrightarrow{E A}+\overrightarrow{E B}+\overrightarrow{E C}+\overrightarrow{E D}=?$
Given: $ABCD$ is rhombus whose diagonal intersect E then $\overrightarrow{E A}+\overrightarrow{E B}+\overrightarrow{E C}+\overrightarrow{E D}$ equal
Solution: Since in rhombus diagonal bisects each other, $|\overrightarrow{E A}|=|\overrightarrow{E C}| \text { and }|\overrightarrow{E B}|=|\overrightarrow{E D}|$
But since they are opposite to each other they are opposite signs
$\begin{aligned} &\overrightarrow{E A}=-\overrightarrow{E C} \\\\ &\overrightarrow{E B}=-\overrightarrow{E D} \\\\ &\overrightarrow{E A}+\overrightarrow{E C}=0 \end{aligned}$ ...................(1)
$\begin{aligned} &\overrightarrow{E B}+\overrightarrow{E D}=0 \\ \end{aligned}$ ...................(2)
$\overrightarrow{E A}+\overrightarrow{E B}+\overrightarrow{E C}+\overrightarrow{E D}=\overrightarrow{0}$

Algebra of Vectors exercise Multiple choice question 27

Answer: $\sqrt{209}$
Hint: $|\vec{a}+\vec{b}+\vec{c}|$
Given: If $\vec{a}, \vec{b}, \vec{c}$ are position vectors of point $A(2,3,-4), B(3,-4,-5) \; \& \; C(3,2,-3)$
Then $|\vec{a}+\vec{b}+\vec{c}|$ is equal to
Solution:
$\begin{aligned} &\vec{a}=2 \hat{i}+3 \hat{j}-4 \hat{k} \\\\ &\vec{b}=3 \hat{i}-4 \hat{j}-5 \hat{k} \\\\ &\vec{c}=3 \hat{i}+2 \hat{j}-3 \hat{k} \end{aligned}$
$\begin{gathered} |\vec{a}+\vec{b}+\vec{c}|=(2+3+3) \hat{i}+(3-4+2) \hat{j}+(-4-5-3) \hat{k} \\\\ \end{gathered}$
$=8 \hat{i}+\hat{j}-12 \hat{k}$
Magnitude:
$\begin{aligned} &=\sqrt{8^{2}+1^{2}+(-12)^{2}} \\\\ &=\sqrt{64+1+144} \\\\ &=\sqrt{209} \end{aligned}$

The RD Sharma class 12 solution of Algebra of vectors exercise MCQ consists of a total of 27 questions, all covering the essential concepts of the chapter and you have to choose the correct option among all the options given in the question. RD Sharma solutions The RD Sharma class 12th exercise MCQ covers the below-given concepts of the Algebra of vectors chapter-

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

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Frequently Asked Question (FAQs)

1. What is the magnitude of a vector?

The magnitude of a Vector is the distance between the initial point P and the endpoint Q.

2. What is scalar in vector algebra?

A scalar is an element of a field that is used to define a vector space. A quantity described by multiple scalars, such as having both directions and magnitude, is called a vector.

3. What is the vector B formula?

Vectors A and B are said to be equal if lAl = lBl as well as their directions are said to be the same.

4. What are the types of vectors?

There are a few vectors mainly, zero vector, unit vector, position vector, co-initial vector, like and unlike vectors, coplanar vector, collinear vector, equal vector..

5. Can I take the help of the RD Sharma solution to solve homework?

Yes, as teachers refer to these solutions for assigning homework, it is helpful and less time-consuming.