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

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

Answers (1)

Answer :-   \theta = \frac{\pi}{2}

Hint :- You must know how to find the angle between two vectors

Given :-  |\vec{a}|=|\vec{b}|=1

Solution :- Let the angle between

\begin{aligned} &\Rightarrow \cos \theta=\frac{(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})}{|\vec{a}+\vec{b}| \cdot|\vec{a}-\vec{b}|} \\ & \end{aligned}

\Rightarrow \cos \theta=\frac{|\vec{a}|^{2}-|\vec{b}|^{2}}{|\vec{a}+\vec{b}| \cdot|\vec{a}-\vec{b}|}\qquad\qquad\left[\because(\mathrm{a}+\mathrm{b}) \cdot(\mathrm{a}-\mathrm{b})=a^{2}-b^{2}\right]

\Rightarrow \cos \theta=\frac{1-1}{|\vec{a}+\vec{b}| \cdot|\vec{a}-\vec{b}|}\qquad[\because|\vec{a}|=|\vec{b}|=1]

\begin{aligned} \Rightarrow &\cos \theta=\frac{0}{|\vec{a}+\vec{b}| \cdot|\vec{a}-\vec{b}|} \\ & \end{aligned}

\Rightarrow \cos \theta=0 \\

\Rightarrow \cos \theta=\cos \frac{\pi}{2} \\

\Rightarrow \theta=\frac{\pi}{2}

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