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  • Need solution for RD Sharma Maths Class 12 Chapter 23 Scalar or Dot Products Excercise Very Short Answer Question 24

Need solution for RD Sharma Maths Class 12 Chapter 23 Scalar or Dot Products Excercise Very Short Answer Question 24

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best_answer

Answer :-  \sqrt{2}

Hint :- Use the identity  \left ( a+b\right )^{2}

Given :-   |\vec{a}|=|\vec{b}|=1 \& \vec{a} \cdot \vec{b}=0

Solution :-  |\vec{a}+\vec{b}|^{2}=|\vec{a}|^{2}+|\vec{b}|^{2}+2 \vec{a} \cdot \vec{b} \qquad\qquad\left[\because(a+b)^{2}=a^{2}+b^{2}+2 \mathrm{ab}\right]

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

\Rightarrow|\vec{a}+\vec{b}|^{2}=2 \\

\Rightarrow|\vec{a}+\vec{b}|=\sqrt{2}

 

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