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  • Please Solve RD Sharma Class 12 Chapter 23 Scalar or Dot Product Exercise Very Short Answer Question 13 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter 23 Scalar or Dot Product Exercise Very Short Answer Question 13 Maths Textbook Solution.

Answers (1)

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ANSWER:  \sqrt{3}

GIVEN:  |\vec{a}|=|\vec{b}|=1 \quad \&|\vec{a}+\vec{b}|=1

HINTS: Use the identity of  (a+b)^{2} \&(a-b)^{2}

SOLUTION:

\begin{aligned} &|\vec{a}+\vec{b}|=1 \\ & \end{aligned}

|\vec{a}+\vec{b}|^{2}=1^{2}\qquad[\text { Squaring both sides }] \\

\Rightarrow|\vec{a}|^{2}+|\vec{b}|^{2}+2 \vec{a} \cdot \vec{b}=1\qquad\left[\because(a+b)^{2}=a^{2}+b^{2}+2 a b\right] \\

\Rightarrow 1+1+2 \vec{a} \cdot \vec{b}=1 \qquad[\because|\vec{a}|=|\vec{b}|=1] \\

\Rightarrow 2 \vec{a} \cdot \vec{b}=-1 \Rightarrow \vec{a} \cdot \vec{b}=-1

Now

\begin{aligned} &|\vec{a}-\vec{b}|^{2}=|\vec{a}|^{2}+|\vec{b}|^{2}-2 \vec{a} \cdot \vec{b}\qquad\left[\because(\mathrm{a}-\mathrm{b})^{2}=\mathrm{a}^{2}+b^{2}-2 a b\right] \\ & \end{aligned}

\Rightarrow|\vec{a}-\vec{b}|^{2}=1+1-2 \frac{1}{2}=1 \qquad\left[\because|\vec{a}|=|\vec{b}|=-\frac{1}{2}\right] \\

\Rightarrow|\vec{a}-\vec{b}|=\sqrt{3}

 

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