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  • Provide solution for RD sharma maths class 12 chapter 23 Scalar or dot product exercise Fill in the blanks question 10

Provide solution for RD sharma maths class 12 chapter 23 Scalar or dot product exercise Fill in the blanks question 10

Answers (1)

\frac{\pi}{4}

Hint:

               Angle between \vec{a} and \vec{b}

Given:

Let \vec{a},\vec{b} are unit vector such that, \vec{a}-\sqrt{2}\vec{b} is also a unit vector, then angle between \vec{a} and \vec{b}

Solution:

\begin{aligned} &|\vec{a}|=|\vec{b}|=1 \\ &|\vec{a}-\sqrt{2} \vec{b}|=1 \\ &|\vec{a}-\sqrt{2} \vec{b}|^{2}=1^{2} \\ \end{aligned}     ….squaring on both sides

\begin{aligned} &|\vec{a}|^{2}+|\sqrt{2} \vec{b}|^{2}-2|\vec{a}| \cdot|\sqrt{2} \vec{b}| \cos \theta=1 \\ &1+2(1)^{2}-2(1)(\sqrt{2}(1)) \cos \theta=1 \end{aligned}

\begin{aligned} &1+2-2 \sqrt{2} \cos \theta=1 \\ &3-2 \sqrt{2} \cos \theta=1 \\ \end{aligned}

\begin{aligned} &3-1=2 \sqrt{2} \cos \theta \\ &2=2 \sqrt{2} \cos \theta \\ \end{aligned}

\begin{aligned} &\cos \theta=\frac{2}{2 \sqrt{2}} \\ &\cos \theta=\frac{1}{\sqrt{2}} \\ &\theta=\frac{\pi}{4} \end{aligned}

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