RD Sharma Class 12 Exercise 23.2 Scalar and dot product Solutions Maths

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

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The Class 12 RD Sharma chapter 23 exercise 23.2 solution is helpful inside and out and is additionally exam-arranged, basic, and fundamental. The RD Sharma class 12th exercise 23.2 can be utilized for self-practice and assessment of imprints.

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

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RD Sharma Class 12 Solutions Chapter23 Scaler and dot product - Other Exercise
Scalar or Dot Products Excercise: 23.2
RD Sharma Chapter wise Solutions

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 3

Answer: $(A C)^{2}=(A B)^{2}+(B C)^{2}$
Hint: Pythagoras theorem
Given: P.T by vector method in right angle triangle the square of hypotenuse
is equal to sum of other two sides.
Solution: Let ABC is right angled triangle at B.
The side opposite to right angle is hypotenuse
Hence, $(A C)^{2}=(A B)^{2}+(B C)^{2}$
Hence, $\vec{a} \cdot \vec{b}=0 \quad \rightarrow(1)$
By triangle law of vector addition
$A C=\vec{b}-\vec{a}$
Squaring both sides,
$\begin{aligned} &(A C)^{2}=(\vec{b}-\vec{a})^{2} \\\\ &(A C)^{2}=(\vec{a})^{2}+(\vec{b})^{2}-2(\vec{a} \cdot \vec{b}) \end{aligned}$
Use eqn(1) we get Now since and be position vector of AB & BC
$\begin{aligned} &(A C)^{2}=(\vec{a})^{2}+(\vec{b})^{2}-0 \\\\ &(A C)^{2}=(\vec{a})^{2}+(\vec{b})^{2} \\\\ &(A C)^{2}=(A B)^{2}+(B C)^{2} \end{aligned}$

Scaler and dot product exercise 23.2 question 4

Answer: $2\left(|a|^{2}+|b|^{2}\right)$
Hint: Let diagonal vector $(\vec{a}+\vec{b}) \text { and }(\vec{a}-\vec{b})$
Given: P.T by vector method that sum of square of diagonal of parallelogram
Equal to sum of square of sides
Solution: Let adjacent side of parallelogram $\vec{a} \text { and } \vec{b}$
Then diagonal vector will be $(\vec{a}+\vec{b}) \text { and }(\vec{a}-\vec{b})$
Sum of squares of diagonal
$(\vec{a}+\vec{b})^{2}+(\vec{a}-\vec{b})^{2}=2\left(|a|^{2}+|b|^{2}\right)$
Sum of squares of side
$2\left(|a|^{2}+|b|^{2}\right)$
Hence proved

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 7

Answer: $|\overrightarrow{A B}|=|\overrightarrow{A D}|$
Hint: Diagonals of a rectangle are perpendicular
Given: If the rectangle is a square
Solution: Let ABCD be a rectangle.
Suppose the position vectors of points B and D be $\vec{a} \text { and } \vec{b}$
Now
$\overrightarrow{A C}=\overrightarrow{A D}+\overrightarrow{B C}=\vec{a}+\vec{b}$
Also $\overrightarrow{ BD}=\vec{a} - \vec{b}$
Since ABCD is a rectangle , so $\overrightarrow{ AB}\perp \overrightarrow{ AD}$
$\vec{a} \cdot \vec{b}$
Now , diagonals AC and BD are perpendicular if $\overrightarrow{ AC}\cdot \overrightarrow{ BD}=0$
$\begin{aligned} &\text { If }(\vec{a}+\vec{b}) \cdot(\vec{a} \cdot \vec{b})=0 \\ &|\vec{a}|^{2}-|\vec{b}|^{2}=0 \\ &|\vec{a}|^{2}=|\vec{b}|^{2} \\ &|\overrightarrow{A B}|=|\overrightarrow{A D}| \end{aligned}$

ABCD is a square

Thus the diagonal of a rectangle are perpendicular if and only if the rectangle is a square.

Scaler and dot product exercise 23.2 question 8

Answer: $A B^{2}+A C^{2}=2\left(A D^{2}+C D^{2}\right)$
Hint: Use Pythagoras theorem
Given: $\text { if AD is median of } \triangle A B C \text { using vector }$
$\text { P.T } A B^{2}+A C^{2}=2\left(A D^{2}+C D^{2}\right)$
Solution: $A D \text { is median of } \triangle \mathrm{ABC}$
Draw $\mathrm{AE} \perp \mathrm{BC}$
In right angled $\Delta ACE$
$(A B)^{2}=(A E)^{2}+(B E)^{2}$ (By Pythagoras Theorem)
In right angled triangle $\Delta ACE$
$(A C)^{2}=(A E)^{2}+(E C)^{2}$
Add (1) & (2)
$\begin{aligned} &(A B)^{2}+(A C)^{2}=(A E)^{2}+(B E)^{2}+(A E)^{2}+(E C)^{2} \\\\ &(A B)^{2}+(A C)^{2}=2(A E)^{2}+(B D-E D)^{2}+(E D+D C)^{2} \end{aligned}$
$\begin{aligned} &(A B)^{2}+(A C)^{2}=2(A E)^{2}+(B D)^{2}-2 B D \cdot E D+(E D)^{2}+(E D)^{2}+2 E D \cdot D C+(D C)^{2} \\\\ &(A B)^{2}+(A C)^{2}=2(A E)^{2}+2(E D)^{2}+(D C)^{2} \end{aligned}$
$\begin{aligned} &(A B)^{2}+(A C)^{2}=2(A E)^{2}+2(E D)^{2}+2(B D)^{2} \quad(B D=D C) \\\\ &(A B)^{2}+(A C)^{2}=2\left((A E)^{2}+(E D)^{2}+(B D)^{2}\right) \\\\ &(A B)^{2}+(A C)^{2}=2\left(A D^{2}+C D^{2}\right) \end{aligned}$

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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