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  • Explain Solution R.D. Sharma Class 12 Chapter 22 Algebra of Vectors Exercise 22.2 Question 10 Maths Textbook Solution.

Explain Solution R.D. Sharma Class 12 Chapter 22 Algebra of Vectors  Exercise 22.2 Question 10 Maths Textbook Solution.

Answers (1)

Answer:

We need to prove that

Hint:

Use triangle law of addition.

Given :

ABCDEF is a hexagon

Solution:

Given ,

ABCDEF is a regular hexagon.

From the triangle law of addition,

\begin{aligned} &\overrightarrow{A C}+\overrightarrow{C D}=\overrightarrow{A D}\\ \end{aligned}

\begin{aligned} &\overrightarrow{A C}+\overrightarrow{A F}=\overrightarrow{A D}\\ \end{aligned}

             \begin{aligned} &\because|\overrightarrow{C D}|=|\overrightarrow{A F}|\\ \end{aligned}

\begin{aligned} &\overrightarrow{C D} \text { parallel to } \overrightarrow{A F}: C D=A F\\ \end{aligned}

\begin{aligned} &\overrightarrow{A B}+\overrightarrow{A C}+\overrightarrow{A F}+\overrightarrow{A D}+\overrightarrow{A C}+\overrightarrow{A E}\\ &=\overrightarrow{A B}+\overrightarrow{A D}+\overrightarrow{A D}+\overrightarrow{A C} \quad \because|\overrightarrow{A E}|=|\overrightarrow{E D}|\\ &=2 \overrightarrow{A D}+\overrightarrow{E D}+\overrightarrow{A E A E} \text { parallel to } \overrightarrow{E D} \therefore \overrightarrow{A B}=\overrightarrow{E D}\\ &=2 \overrightarrow{A D}+\overrightarrow{A D}\\ &=3 \overrightarrow{A D} \quad[\overrightarrow{A B}+\overrightarrow{A E}=\overrightarrow{E D}+\overrightarrow{A E}=\overrightarrow{A D}\\ &=3(2 \cdot \overrightarrow{A O}) \quad[O \text { is the mid point and also the mid point of } \mathrm{AD}, \mathrm{OA}=\mathrm{OD}, \mathrm{AD}=2 \cdot \overrightarrow{\mathrm{AO}}]\\ &=6 \overrightarrow{A O} \end{aligned}

(proved)

Posted by

infoexpert21

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