RD Sharma Class 12 Exercise 23.2 Scalar and dot product Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Solutions Chapter23 Scaler and dot product - Other Exercise

• Chapter 23- Scaler and dot product - Ex-23.1

• Chapter 23- Scaler and dot product - Ex-MCQ

• Chapter 23- Scaler and dot product - Ex-FBQ

• Chapter 23- Scaler and dot product - Ex-VSA

Scalar or Dot Products Excercise: 23.2

Scaler and dot product exercise 23.2 question 1

Answer: $O P^{2}+O Q^{2}=\frac{5}{9} A B^{2}$
Hint: Use $\Delta O A B$
Given: $\text { In } \Delta O A B, \angle A O B=90^{\circ} \text { and if } P \; \& \; Q$are points of trisection of $AB-PT$
$O P^{2}+O Q^{2}=\frac{5}{9} A B^{2}$
Solution:

$\overrightarrow{O A}, \overrightarrow{O B}$
$\begin{aligned} &\overrightarrow{O A}, \overrightarrow{O B}=0 \\\\ &\overrightarrow{A B} \cdot \overrightarrow{A B}=\overrightarrow{O A} \cdot \overrightarrow{O B} \\\\ &\overrightarrow{O P}=\overrightarrow{O B}-\frac{2}{3} \overrightarrow{A B} \\\\ &\overrightarrow{O Q}=\overrightarrow{O B}-\frac{1}{3} \overrightarrow{A B} \end{aligned}$
$\begin{aligned} &\overrightarrow{O P}=\overrightarrow{O B}-\frac{2}{3}(\overrightarrow{O B}-\overrightarrow{O A}) \\\\ &\overrightarrow{O P}=\frac{\overrightarrow{O B}}{3}+\frac{2}{3} \overrightarrow{O A} \\\\ &\overrightarrow{O Q}=\overrightarrow{O B}-\frac{1}{3}(\overrightarrow{O B}-\overrightarrow{O A}) \\ \end{aligned}$
$\frac{2}{3} \overrightarrow{O B}+\frac{\overrightarrow{O A}}{3}$
$\begin{aligned} &O P^{2}+O Q^{2}=\frac{5}{9} A B^{2} \Rightarrow O P^{2}=|\overrightarrow{O P}|^{2}=\overrightarrow{O P} \cdot \overrightarrow{O P} \\\\ &O Q^{2}=|\overrightarrow{O Q}|^{2}=\overrightarrow{O Q} \cdot \overrightarrow{O Q} \\\\ &\overrightarrow{O P} \cdot \overrightarrow{O P}=\left(\frac{\overrightarrow{O B}}{3}+\frac{2}{3} \overrightarrow{O A}\right)\left(\frac{\overrightarrow{O B}}{3}, \frac{2}{3} \overrightarrow{O A}\right) \end{aligned}$
$\begin{aligned} &=\frac{|\overrightarrow{O B}|^{2}}{9}+\frac{4}{9}|\overrightarrow{O A}|^{2} \\\\ &O Q \cdot O Q=\frac{4}{9}|\overrightarrow{O B}|^{2}+\frac{|\overrightarrow{O A}|^{2}}{9} \end{aligned}$
$\begin{aligned} &O P^{2}+O Q^{2}=\frac{5}{9}\left(|O B|^{2}+|\overrightarrow{O A}|^{2}\right)=|A B|^{2} \\\\ &O P^{2}+O Q^{2}=\frac{5}{9}|A B|^{2} \end{aligned}$

Scaler and dot product exercise 23.2 question 2

Answer: Yes, quadrant bisects at $90^{0}$ each other
Hint: Use rhombus
Given: Prove that if diagonal of quadrant bisects each other at right angle
Then it is rhombus.
Solution: Take quadrilateral $ABCD$ , $AC$and $BD$ are diagonals which intersect at 0

$In \; \triangle \; A O B \; and \; \triangle A O D\\\\ \begin{array}{ll} D O=O B & \text { (O is midpoint) } \\ A O=A O & \text { (common side) } \end{array}$
$\angle A O B=\angle A O D \quad\left(90^{\circ}\right. Right \; angle ) \\\\So, \triangle A O B \cong \triangle A O D\\\\ So, A B=A D$
Similarly, $AB=BC=CD=AD$

can proved which mean that $ABCD$ is rhombus

Scaler and dot product exercise 23.2 question 5

Answer: $|P Q|=|P S|$
Hint: Use vector
Given: P.T by vector the quadrilateral obtained by joining
mid point of adjacent side of rectangle is rhombus.
Solution:

$ABCD$ is rectangle.
Let $P, Q, R, S$ be mid-point of side $AB, BC, CD, DA.$
Now
$\overrightarrow{P Q}=\overrightarrow{P B}+\overrightarrow{B Q}=\frac{1}{2} \overrightarrow{A B}+\frac{1}{2} \overrightarrow{B C}=\frac{1}{2}(\overrightarrow{A B}+\overrightarrow{B C})=\frac{1}{2} \overrightarrow{A C}$ $\rightarrow (1)$
$\overrightarrow{S R}=\overrightarrow{S D}+\overrightarrow{D R}=\frac{1}{2} \overrightarrow{A D}+\frac{1}{2} \overrightarrow{D C}=\frac{1}{2}(\overrightarrow{A D}+\overrightarrow{D C})=\frac{1}{2} \overrightarrow{A C} \quad \rightarrow(2)$
From (1) & (2)
$\overrightarrow{P Q}=\overrightarrow{S R}$
So side PQ and SR are equal and parallel
$\begin{aligned} &(|\overrightarrow{P Q}|)^{2}=\overrightarrow{P Q} \cdot \overrightarrow{P Q} \\\\ &(|\overrightarrow{P Q}|)^{2}=(\overrightarrow{P B}+\overrightarrow{B Q})(\overrightarrow{P B}+\overrightarrow{B Q}) \\\\ &(|\overrightarrow{P Q}|)^{2}=(|\overrightarrow{P B}|)^{2}+2 \cdot \overrightarrow{P B} \overrightarrow{B Q}+(|\overrightarrow{B Q}|)^{2} \end{aligned}$
$\begin{aligned} &(|\overrightarrow{P Q}|)^{2}=(|\overrightarrow{P B}|)^{2}+0+(|\overrightarrow{B Q}|)^{2} \quad(\overrightarrow{P B} \perp \overrightarrow{B Q}) \\\\ &(|\overrightarrow{P Q}|)^{2}=(|\overrightarrow{P B})^{2}+(|\overrightarrow{B Q}|)^{2} \rightarrow(3) \end{aligned}$
Also,
$\begin{aligned} &(|\overrightarrow{P S}|)^{2}=\overrightarrow{P S} \cdot \overrightarrow{P S} \\\\ &(|\overrightarrow{P S}|)^{2}=(\overrightarrow{P A}+\overrightarrow{A S})(\overrightarrow{P A}+\overrightarrow{A S}) \\\\ &(|\overrightarrow{P S}|)^{2}=(|\overrightarrow{P A}|)^{2}+2 \cdot \overrightarrow{P A} \cdot \overrightarrow{A S}+(|\overrightarrow{A S}|)^{2} \end{aligned}$
$\begin{aligned} &(|\overrightarrow{P S}|)^{2}=(|\overrightarrow{P A}|)^{2}+0+(|\overrightarrow{A S}|)^{2} \quad(\overrightarrow{P A} \perp \overrightarrow{A S}) \\\\ &(|\overrightarrow{P S}|)^{2}=(|\overrightarrow{P B}|)^{2}+(|\overrightarrow{B Q}|)^{2} \rightarrow(4) \end{aligned}$
From (3) & (4)
$\begin{aligned} &(|\overrightarrow{P Q}|)^{2}=(|\overrightarrow{P S}|)^{2} \\\\ &|\overrightarrow{P Q}|=|\overrightarrow{P S}| \end{aligned}$

Scaler and dot product exercise 23.2 question 6

Answer: $\angle A O B=\angle A O D=90^{\circ}$
Hint: Prove by 90
Given: P.T diagonal of rhombus are perpendicular bisector each other.
Solution:

$ABCD$ is rhombus in which $AB=BC=CD=DA$
$OA=OC$ and $OB=OD$
$Now \; in \; \angle B O C \; and \; \angle D O C\\\\ \mathrm{OB}=\mathrm{OD}, \mathrm{BC}=\mathrm{DC} \text { and } \mathrm{OC}=\mathrm{OC}$
$\begin{aligned} &\angle B O C=\angle D O C \\\\ &\angle B O C+\angle D O C=180^{\circ} \quad \text { (linear pair) } \\\\ &\angle B O C-\angle D O C=90^{\circ} \end{aligned}$
$Similarly \; \; \angle A O B=\angle A O D=90^{\circ}$
Hence the diagonal of rhombus bisect each other at right angle.

Scaler and dot product exercise 23.2 question 9

Answer: $\angle A D C=\angle A D B=90^{\circ}$
Hint: Firstly prove the two triangle form by median
Given: if the median to base of triangle is perpendicular to base of triangle is isosceles
Solution:

Let ABC an isosceles triangle with AB=AC and let AB be median
To base BC then D Is midpoint of BC
$\angle A B C=\angle A C B\\\\\ in\; \triangle \mathrm{ABD} \; \& \; \triangle \mathrm{ACD}\\\\ A B=A C \quad (equal \; side )$
$\begin{aligned} &\angle A B C=\angle A C B \quad \text { (equal angle }) \\\\ &B D=D C \quad(\mathrm{BC} \text { bisected by median } \mathrm{AD}) \end{aligned}$
Therefore by SAS criterion of congurence triangle
$\begin{aligned} &\triangle \mathrm{ABD} \cong \triangle \mathrm{ACD} \\\\ &\text { Hence } \angle A B C=\angle A C B \end{aligned}$ (by correcponding part of congurence)
$Let \; \angle A D C=\angle A D B=x\\\\ \angle A D B+\angle A D C=180^{\circ}$
Put the value we get
$\begin{aligned} &x+x=180^{\circ} \\\\ &2 x=180^{\circ} \\\\ &x=90^{\circ} \\\\ &\angle A D C=\angle A D B=90^{\circ} \end{aligned}$

Scaler and dot product exercise 23.2 question 10

Answer: $A B^{2}+B C^{2}+C D^{2}+D A^{2}=A C^{2}+B D^{2}+4 P Q^{2}$
Hint: Use mid point theorem
Given: In quadrilateral $ABCD$
Prove $A B^{2}+B C^{2}+C D^{2}+D A^{2}=A C^{2}+B D^{2}+4 P Q^{2}$ where $P \; \& \; Q$ are
Middle point of diagonal $AC$ and $BD$
Solution:


Let $ABCD$ be quadrilateral.
Take a as origin. Let position vector $B,C,D$ be $\vec{b}, \vec{c}, \vec{d}$ respectively
Then vector $P=\frac{\vec{c}}{2}$ (Mid point formula)
Position vector $\mathrm{Q}=\frac{\vec{b}+\vec{d}}{2}$ (Mid point)
Now,
$\begin{aligned} &A B^{2}+B C^{2}+C D^{2}+D A^{2} \\\\ &(|\vec{b}|)^{2}+(|\vec{c}-\vec{b}|)^{2}+(|\vec{d}-\vec{c}|)^{2}+(|\vec{d}|)^{2} \end{aligned}$
$\begin{aligned} &(|\vec{b}|)^{2}+(|\vec{c}|)^{2}-2 \vec{c} \cdot \vec{b}+(|\vec{b}|)^{2}+(|\vec{d}|)^{2}-2 \vec{d} \vec{c}+(|\vec{c}|)^{2}+(|\vec{d}|)^{2} \\\\ &2(|\vec{b}|)^{2}+2(|\vec{c}|)^{2}+2(|\vec{d}|)^{2}-2 \cdot \vec{b} \cdot \vec{c}-2 \cdot \vec{c} \cdot \vec{d} \quad \rightarrow(1) \end{aligned}$
Also,
$\begin{aligned} &A C^{2}+B D^{2}+4 P Q^{2} \\\\ &(|\overrightarrow{A C}|)^{2}+(|\overrightarrow{B D}|)^{2}+4(|\overrightarrow{P Q}|)^{2} \\\\ &(|\vec{c}|)^{2}+(|\bar{d}-\vec{b}|)^{2}+(|\vec{b}+\vec{d}|)^{2}-2(|\vec{b}+\vec{d}|)^{2} \cdot c+(|\vec{c}|)^{2} \end{aligned}$
$\begin{aligned} &2(|\vec{c}|)^{2}+2(|\bar{d}|)^{2}+2(|\vec{b}|)^{2}-2 \cdot \vec{b} \cdot \vec{c}-2 \vec{b} \cdot \vec{c}-2 \cdot \vec{c} \cdot \vec{d} \\\\ &A B^{2}+B C^{2}+C D^{2}+D A^{2}=A C^{2}+B D^{2}+4 P Q^{2} \end{aligned}$

The RD Sharma class 12 arrangement of Scalar and dab item exercise 23.2 is utilized by thousands of students and instructors for the practical information on maths. The RD Sharma class 12th exercise 23.2 comprises of 10 inquiries that concealment every one of the fundamental confirmations like-

Demonstrating of Pythagoras theorem by vector method

Demonstrating quadrilateral acquired by joining midpoints of nearby sides of a square shape is a rhombus.

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RD Sharma Chapter wise Solutions

• Chapter 1 - Relations
• Chapter 2 - Functions
• Chapter 3 - Inverse Trigonometric Functions
• Chapter 4 - Algebra of Matrices
• Chapter 5 - Determinants
• Chapter 6 - Adjoint and Inverse of a Matrix
• Chapter 7 - Solution of Simultaneous Linear Equations
• Chapter 8 - Continuity
• Chapter 9 - Differentiability
• Chapter 10 - Differentiation
• Chapter 11 - Higher Order Derivatives
• Chapter 12 - Derivative as a Rate Measurer
• Chapter 13 - Differentials, Errors and Approximations
• Chapter 14 - Mean Value Theorems
• Chapter 15 - Tangents and Normals
• Chapter 16 - Increasing and Decreasing Functions
• Chapter 17 - Maxima and Minima
• Chapter 18 - Indefinite Integrals
• Chapter 19 - Definite Integrals
• Chapter 20 - Areas of Bounded Regions
• Chapter 21 - Differential Equations
• Chapter 22 - Algebra of Vectors
• Chapter 23 - Scalar Or Dot Product
• Chapter 24 - Vector or Cross Product
• Chapter 25 - Scalar Triple Product
• Chapter 26 - Direction Cosines and Direction Ratios
• Chapter 27 - Straight Line in Space
• Chapter 28 - The Plane
• Chapter 29 - Linear programming
• Chapter 30- Probability
• Chapter 31 - Mean and Variance of a Random Variable