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Provide Solution For  R.D. Sharma Maths Class 12 Chapter 22 Algebra of Vectors  Exercise 22.2 Question 6 Maths Textbook Solution.

Answers (1)

Answer:

\vec{BA}+\vec{BC}+\vec{CD}+\vec{DA}=2\vec{BA}

Hint:

Form the triangle in quadrilateral. Use triangle law of inequality.

Given :

ABCD are quadrilateral.

Solution:

In \Delta ABC,

By the triangle law of inequality

\vec{BC}+\vec{CA}=\vec{BA}

\vec{CA}=\vec{BA}-\vec{BC}(1)

Again in \Delta ADC,

By triangle law of inequality

\vec{CA}=\vec{CD}+\vec{DA}.......(2)

From (1) and (2)

\begin{aligned} &\overrightarrow{C D}+\overrightarrow{D A}=\overrightarrow{B A}-\overrightarrow{B C} \\ &\overrightarrow{B A}+\overrightarrow{B C}+\overrightarrow{C D}+\overrightarrow{D A}=\overrightarrow{B A}+\overrightarrow{B A} \\ &\overrightarrow{B A}+\overrightarrow{B C}+\overrightarrow{C D}+\overrightarrow{D A}=2 \overrightarrow{B A} \end{aligned}

 

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