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  • Provide solution for RD Sharma maths class 12 chapter 23 Scaler and Dot Product exercise Multiple choice question 14

Provide solution for RD Sharma maths class 12 chapter 23 Scaler and Dot Product exercise Multiple choice question 14

Answers (1)

Answer:

Option (c) \sqrt{3} a

Hint:

Use the formula off(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} are mutually perpendicular and |\vec{a}|=|\vec{b}|=|\vec{c}|=a

Solution:

|\vec{a}|=|\vec{b}|=|\vec{c}|=a \text { and } \vec{a}, \vec{b}, \vec{c}are mutually perpendicular.

\therefore \vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{c}=\vec{c} \cdot \vec{a}=0

Now,

\begin{aligned} &(\vec{a}+\vec{b}+\vec{c})^{2}=(\vec{a})^{2}+(\vec{b})^{2}+(\vec{c})^{2}+2 \vec{a} \cdot \vec{b}+2 \vec{b} \cdot \vec{c}+2 \vec{c} \cdot \vec{a} \\\\ &(\vec{a}+\vec{b}+\vec{c})^{2}=a^{2}+a^{2}+a^{2} \end{aligned}

\begin{aligned} &(\vec{a}+\vec{b}+\vec{c})^{2}=3 a^{2} \\\\ &|\vec{a}+\vec{b}+\vec{c}|=\sqrt{3} a \end{aligned}

Hence, option (c) is correct

 

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