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Need Solution for R.D.Sharma Maths Class 12 Chapter 22 Algebra of Vectors Exercise 22.2 Question 7 Maths Textbook Solution.

Answers (1)

Answer:

We need to prove that

Hint:

Use triangle law of vector.

Given :

ABCDE is a pentagon.

Solution:

Given , ABCDE is a pentagon.

In $\Delta ABC$ ,by tringle law

$\vec{AB}+\vec{BC}=\vec{AC}.......(1)$

In $\Delta ACD$ by tringle law

$\vec{AC}+\vec{CD}=\vec{AD}.......(2)$

In $\Delta AED$ ,by tringle law

$\vec{AD}+\vec{DE}=\vec{AE}......(3)$

or $\vec{AE}+\vec{ED}=\vec{AD}.........(4)$

In $\Delta ADC$ ,by tringle law

$\overrightarrow{A D}+\overrightarrow{D C}=\overrightarrow{A C} \ldots \ldots \ldots (5)$

$(i) \overrightarrow{A D}+\overrightarrow{D C}=\overrightarrow{A C}$

$=\overrightarrow{A C}+\overrightarrow{E C}+E A$

$=\overrightarrow{A E}+\overrightarrow{E A}$

$=\overrightarrow{A E}-\overrightarrow{A E}$

$=0$

$(ii) L.H.S, \overrightarrow{A B}+\overrightarrow{A E}+\overrightarrow{B C}+\overrightarrow{D C}+\overrightarrow{E D}+\overrightarrow{A C}$

$=(\overrightarrow{A B}+\overrightarrow{B C})+(\overrightarrow{A E}+\overrightarrow{E D})+\overrightarrow{A C}+\overrightarrow{D C}$

$=\overrightarrow{A C}+\overrightarrow{A D}+\overrightarrow{D C}+\overrightarrow{A C} \quad [using (1)and(4) ]$

$=\overrightarrow{A C}+\overrightarrow{A C}+\overrightarrow{A C}=3 \overrightarrow{A C} (proved)$

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