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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)

best_answer

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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