RD Sharma Class 12 Exercise 22.3 Algebra of Vectors Solutions Maths

RD Sharma Class 12 Exercise 22.3 Algebra of Vectors Solutions Maths - Download PDF Free Online

Edited By Satyajeet Kumar | Updated on Jan 24, 2022 06:19 PM IST

RD Sharma solutions have exercise-wise solutions for every chapter. For example, RD Sharma Class 12th Exercise 22.3 has seven inquiries alongside examples to simplify the excursion for each understudy. Moreover, position Vectors of the place of convergence, coplanar are discussed in this specific exercise which would be helpful for students.

Latest: JEE Main: high scoring chapters | Past 10 year's papers

Don't Miss: Most scoring concepts for NEET | NEET papers with solutions

New: Aakash iACST Scholarship Test. Up to 90% Scholarship. Register Now

This Story also Contains

RD Sharma Class 12 Solutions Chapter22 Algebra of vectors - Other Exercise
Algebra of Vectors Excercise: 22.3
RD Sharma Chapter wise Solutions

Algebra-based math of vectors is one of the main chapters in Class 12 Mathematics. Here we concentrate on the variable-based math of vector amounts. Scalars and vectors are two kinds of actual quantities. RD Sharma solutions have come to help students who face trouble in this chapter.

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

Algebra of Vectors Excercise: 22.3

Algebra of Vectors Exercise 22.3 Question 1

Answer:
$\frac{5}{3}\vec{a},3\vec{a}+4\vec{b}$
Hint: if a point R divides a line PQ into m:n internally then point $P=\frac{m\vec{OR+n\vec{OP}}}{m-n}$ and externally then point $R=\frac{m\vec{OQ+n\vec{OP}}}{m-n}$
Given:
$\vec{OP}=2\vec{a}+\vec{b}$
$\vec{OQ}=\vec{a}-2\vec{b}$
Solution:

$(i) \overrightarrow{O R}( internally )$
$=\frac{1(\vec{a}-2 \vec{b})+2(2 \vec{a}+\vec{b})}{1+2}$
$=\frac{\vec{a}-2 \vec{b}+4 \vec{a}+2 \vec{b}}{3}$
$=\frac{5 \vec{a}}{3}$
$(ii) \overrightarrow{O R}( Externally )$
$=\frac{1(\vec{a}-2 \vec{b})-2(2 \vec{a}+\vec{b})}{1-2}$
$=\frac{\vec{a}-2 \vec{b}-4 \vec{b}-2 \vec{b}}{-1}$
$=\frac{-3 \vec{a}-4 \vec{b}}{-1}$
$=3 \vec{a}+4 \vec{b}$

Algebra of Vectors Exercise 22.3 Question 2

Answer:
$-\vec{a}-7\vec{b}$
Hint: $if \: a \: point \: Z \: divides\: a \: line\: XY \: into\: m:n\: ratio \: externally \: then \: point$ $Z=\frac{m\vec{OY-n\vec{OX}}}{m-n}$
Given:
$\vec{OX}=3\vec{a}+\vec{b}$
$\vec{OY}=\vec{a}-3\vec{b}$
Solution: $\vec{d}$

$\vec{OZ}\: (Externally)$
$\begin{aligned} &=\frac{(2(\vec{a}-3 \vec{b})-1(3 \vec{a}+\vec{b}))}{2-1} \\ &=(2 \vec{a}-6 \vec{b}-3 \vec{a}-\vec{b}) \\ &=-\vec{a}-7 \vec{b} \end{aligned}$

Algebra of Vectors Exercise 22.3 Question 4

Answer: $3\vec{b}-2\vec{a},2\vec{a}-\vec{b}$
Hint: Vector law of addition, (two vectors one represented as two sides of the triangle with the order of magnitude and direction, then the third side of the triangle represents the magnitude and direction of the resultant vector.)
Given: $\vec{AC}=3\vec{AB}$
Solution:

$In \triangle \mathrm{OAB},$
$=\overrightarrow{O A}+\overrightarrow{A B}=\overrightarrow{O B}\: (By \: triangle\: law \: of \: vector \: addition)$
$\mathrm{AB}=\vec{b}-\vec{a}$
$In \triangle \mathrm{OAC}$
$=\overrightarrow{O C}=\overrightarrow{O A}+\overrightarrow{A C}$ $(By\: triangle \: law \: of\: vector \: addition)$

$=\overrightarrow{O A}+3 \overrightarrow{A B} \$ $[ given \overrightarrow{A C}=3 \overrightarrow{A B}]$
$=\vec{a}+3(\vec{b}-\vec{a})$
$=3 \vec{b}-2 \vec{a}$
$\text{Again in }\triangle O A D$
$\overrightarrow{O D}=\overrightarrow{O A}+\overrightarrow{A D} \ \ \ (\text{ By triangle law of vector addition})$
$=\vec{a}+\vec{a}-\vec{b} \quad[\because \overrightarrow{A D}=-\overrightarrow{A B}]$
$=2 \vec{a}-\vec{b}$

Algebra of Vectors Exercise 22.3 Question 5

Answer: $\begin{aligned} &\frac{3 \vec{a}+5 \vec{c}}{8}=\frac{2 \vec{b}+6 \vec{d}}{8} \\ \end{aligned}$
Hint: Use vector algebra
Given: $\begin{aligned} &\qquad 3 \vec{a}-2 \vec{b}+5 \vec{c}-6 \vec{d}=0 \\ \end{aligned}$
Solution: $\qquad 3\vec{a} +5 \vec{c}=2 \vec{b}+6\vec{d}$
$\begin{aligned} &\frac{3 \vec{a}+5 \vec{c}}{8}=\frac{2 \vec{b}+6 \vec{d}}{8} \end{aligned}$

AC and BD are intersect at the point.

A,B,C,D are co- planner.
P.V of point of intersection of line segment AC and BD are $\begin{aligned} &\frac{3 \vec{a}+5 \vec{c}}{8}\: or\: \frac{2 \vec{b}+6 \vec{d}}{8} \\ \end{aligned}$

Algebra of Vectors Exercise 22.3 Question 6

Answer: $\frac{-5 p+6 \vec{r}}{11} o r \frac{2 \vec{q}+9 \vec{s}}{11}$

Hint: Use vector algebra.
Given: $5 \vec{p}-2 \vec{q}+6 \vec{r}-9 \vec{s}=0$
Solution: $5 \vec{p}-6 \vec{r}=2 \vec{q}-9 \vec{s}$
$\frac{-5 p+6 \vec{r}}{11} =\frac{2 \vec{q}+9 \vec{s}}{11}$ ....(1)

As we know that, if the point C divides the line segment AB in the ratio m:n the $\overrightarrow{O C}=\frac{n \overrightarrow{O A}+m \overrightarrow{O B}}{m+n}$
let AB the point which PR in the ratio 6:5 and B be the point which divides QS in the ratio 9:2 .hence the P.V of AB are $\frac{6 \vec{p}+6 \vec{r}}{11}, \frac{2 \vec{q}+9 \vec{s}}{11}$
From (1)
The P.V of AB is same.
= point A and B intersect at the same point
= PQ and QS intersect at A
of two line segment intersect then it must be Co-planner.
P,Q,R,S are co-planner.

Algebra of Vectors Exercise 22.3 Question 7

Answer: – we need to prove that $I=\frac{\alpha \vec{a}+(\beta+\gamma) \vec{d}}{\alpha+\beta+\gamma}$
Hint: Use vector algebra.
Given: The vertices A,B,C of triangle ABC have respectively. P.V $\vec{a},\vec{b},\vec{c}$ with respect to the given origin 0.
Solution :

Let the PV of A,B and C w.r.t some origin 0 be $\vec{a},\vec{b},\vec{c}$ respectively. let DB be a point on BC where bisector of angle D meets.
let $\vec{d}$ be P.V of D which divide CB internally in the ratio β and $\gamma$ where $\beta =\mid \vec{AC}\mid$ and $\gamma =\mid \vec{AB}\mid$
Thus $\beta =\mid \vec{c}-\vec{a}\mid$ and $\gamma =\mid \vec{b}-\vec{a}\mid$ by section formula, the P.V of D is given by
$\begin{aligned} &=\overrightarrow{O D}=\frac{\beta \vec{b}+\gamma \vec{c}}{\beta+\gamma} \\ &=\alpha=|\vec{b}-\vec{c}| \end{aligned}$

In centre is the concurrent point angle bisector and in centre divides the line AD in the ratio
$=a=\beta -\gamma$
So the P.V of in centre is given by $I=\frac{\alpha \vec{a}+(\beta+\gamma) \vec{d}}{\alpha+\beta+\gamma}$

A team of experts prepares Class 12th RD Sharma Chapter 22 Exercise 22.3 Solutions using the most recent NCERT textbook and marking schemes. When a student refers to RD Sharma Class 12 Solutions Algebra of Vectors Ex. 22.3, all concepts are clear, and the student understands the exams better. When a student refers to RD Sharma Class 12 Solutions Chapter 22 ex 22.3, the following benefits will accrue:

1. Expertly crafted:

Subject experts create RD Sharma Class 12th Chapter 22 Exercise 22.3 in accordance with the most recent marking pattern and CBSE guidelines. Every student must practice RD Sharma Class 12th Exercise 22.3 to achieve a high score. These solutions explain the topics mentioned above so that even a newcomer can learn and solve them.

2. Efficient

RD Sharma Class 12th Exercise 22.3 offers the best in class solutions for students. It is updated to the latest version of the book and contains all the new concepts. As it follows the CBSE syllabus, students can use it to prepare efficiently and save time. Moreover, referring to this material will help them gain more knowledge on the subject and understand different aspects of the chapter.

3. Easy to understand

The solutions provided in this material contain step-by-step answers which students can refer to increase their knowledge. Students can learn about various alternatives to solving a problem and choose the one which best suits them.

4. Exam-oriented

RD Sharma Class 12 Solutions Ex. 22.3 material contains solutions that follow the CBSE syllabus and are exam-oriented. As RD Sharma books are widely used in schools, this material's questions can also appear in exams. This is an efficient and paperless alternative for preparation.

5. Free of Cost

These solutions are available on career 360's website for free. Students can access them by searching the book name and exercise number on the website. So instead of buying expensive materials physically, students can take advantage of this website and score good marks in exams.

Students will find all of these solutions for free at Career360, the best website to visit for all of the benefits associated with board exam questions and competitive exams. With Career360, a student will understand, learn, and perform exceptionally well in exams. These solutions have benefited thousands of students, making Career360 the ideal choice for all students.