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.

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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 $I=\frac{\alpha \vec{a}+(\beta+\gamma) \vec{d}}{\alpha+\beta+\gamma}$

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