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  • Need solution for RD sharma maths class 12 chapter 23 Scalar or dot product exercise Fill in the blanks question 12

Need solution for RD sharma maths class 12 chapter 23 Scalar or dot product exercise Fill in the blanks question 12

Answers (1)

best_answer

\sqrt{a^{2}+b^{2}+c^{2}}

Hint:

               Use a.b=0

Given:

               If a,b,c are mutually perpendicular vector, then \left | \vec{a}+\vec{b}+\vec{c} \right |

Solution:

\begin{aligned} &|\vec{a}|=1,|\vec{b}|=1,|\vec{c}|=1 \\ &\vec{a} \cdot \vec{b}=0, \vec{b} \cdot \vec{c}=0, \vec{c} \cdot \vec{a}=0 \\ &\vec{b} \cdot \vec{a}=0, \vec{c} \cdot \vec{b}=0, \vec{a} \cdot \vec{c}=0 \\ \end{aligned}

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

\begin{aligned} &=\vec{a} \cdot \vec{a}+\vec{a} \cdot \vec{b}+\vec{c} \cdot \vec{c}+\vec{b} \cdot \vec{a}+\vec{b} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}+\vec{c} \cdot \vec{b}+\vec{c} \cdot \vec{c} \\ \end{aligned}

\begin{aligned} &=|\vec{a}|^{2}+0+0+0+|\vec{b}|^{2}+0+0+0+|\vec{c}|^{2} \\ &=|\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2} \\ \end{aligned}

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

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