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  • Need solution for RD Sharma maths class 12 chapter 23 Scaler and Dot Product exercise Multiple choice question 28

Need solution for RD Sharma maths class 12 chapter 23 Scaler and Dot Product exercise Multiple choice question 28

Answers (1)

Answer:

Option (c) -19

Hint:

Apply the identity(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a \cdot b+2 b . c+2 c . a

Given:

|\vec{a}|=2,|\vec{b}|=3,|\vec{c}|=5 \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})

                                                                            \left[\because(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(a b+b c+c a)\right]

0=(2)^{2}+(3)^{2}+(5)^{2}+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}) \quad\left[\begin{array}{l} \because \vec{a}+\vec{b}+\vec{c}=0 \\ \because|\vec{a}|=2,|\vec{b}|=3,|\vec{c}|=5 \end{array}\right]

\begin{aligned} &0=4+9+25+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}) \\\\ &\frac{-38}{2}=\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a} \\\\ &\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}=-19 \end{aligned}

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