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Name: Explain Solution R.D.Sharma Class 12 Chapter 22 Algebra of Vectors Exercise 22.2 Question 9 Maths Textbook Solution.
Uploaded: Sun, 2021-08-29 19:53
Description: Explain Solution R.D.Sharma Class 12 Chapter 22 Algebra of Vectors Exercise 22.2 Question 9 Maths Textbook Solution.

#Algebra of Vectors
#R.D. Sharma
#School
#Maths
#CBSE 12 Class
#RD Sharma Vol. 2 Class 12

#1.3
#10.1
#Fill in the blanks
#Very Short Answer type
#Multiple Choice Questions (MCQs)
#20.4
#22.1
#22.2
#22.3
#22.4
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#22.8
#22.9

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Explain Solution R.D.Sharma Class 12 Chapter 22 Algebra of Vectors Exercise 22.2 Question 9 Maths Textbook Solution.

Answers (1)

Answer:

We need to prove that

Hint:

Use triangle law.

Given :

$\vec{AD}+\vec{PB}+\vec{PD}=\vec{PC}$

$\vec{AD}+\vec{PB}+\vec{PD}-\vec{PC}$

Solution:

In $\Delta PDC$ ,by triangle law

$\vec{PD}+\vec{DC}=\vec{PC}$

In $\Delta PAB$ ,by triangle law

$\vec{PA}+\vec{PB}=\vec{AB}$

$\therefore$ ABCD is a parallelogram.

Posted by

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Explain Solution R.D.Sharma Class 12 Chapter 22 Algebra of Vectors Exercise 22.2 Question 9 Maths Textbook Solution.

Answers (1)

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