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  • Provide solution for RD Sharma maths class 12 chapter 23 Scaler and Dot Product exercise Multiple choice question 27

Provide solution for RD Sharma maths class 12 chapter 23 Scaler and Dot Product exercise Multiple choice question 27

Answers (1)

Answer:

Option (c)  -\frac{3}{2}

Hint:

You must know the formula of (a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a \cdot b+2 b . c+2 c . a

Given:

|\vec{a}|=|\vec{b}|=|\vec{c}|=1 \text { and } \vec{a}+\vec{b}+\vec{c}=0

Solution:

\vec{a}+\vec{b}+\vec{c}=0

We know that,

(\vec{a}+\vec{b}+\vec{c})^{2}=(\vec{a})^{2}+(\vec{b})^{2}+(\vec{c})^{2}+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})

0=1+1+1+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})                    \begin{aligned} &{[\because|\vec{a}|=|\vec{b}|=|\vec{c}|=1]} \\ &{[\because \vec{a}+\vec{b}+\vec{c}=0]} \end{aligned}

\frac{-3}{2}=\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}

Hence, option (c) is correct

Posted by

infoexpert26

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