Question

$$\begin{gathered} 23.If\stackrel{\rightharpoonup }{a}\&\stackrel{\rightharpoonup }{b}areunitvectors,thenanglebetween\stackrel{\rightharpoonup }{a}\&\stackrel{\rightharpoonup }{b}for\sqrt{3}\stackrel{\rightharpoonup }{a}-\stackrel{\rightharpoonup }{b}tobeaunitvectoris \\ \left(A\right){90}^0\left(B\right){30}^0 \\ \left(C\right){45}^0\left(D\right){60}^0 \end{gathered}$$

Answer 1

Dear Student

$$\begin{gathered} Given \\ \overrightarrow{a}and\overrightarrow{b}areunitvectors \\ \Rightarrow \left|\overrightarrow{a}\right|=1 \\ and\left|\overrightarrow{b}\right|=1 \\ weknow \\ \overrightarrow{a}.\overrightarrow{b}=\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \theta where\theta istheanglebetween\overrightarrow{a}and\overrightarrow{b} \\ alsogiven \\ \left|\sqrt{3}\overrightarrow{a}-\overrightarrow{b}\right|=1 \\ Squaringonbothsides \\ {\left|\sqrt{3}\overrightarrow{a}-\overrightarrow{b}\right|}^2=1 \\ \Rightarrow 3{\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2-2\sqrt{3}\overrightarrow{a}.\overrightarrow{b}=1 \\ \Rightarrow 3+1-2\sqrt{3}\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \theta =1 \\ \Rightarrow 3+1-2\sqrt{3}\cos \theta =1 \\ \Rightarrow 2\sqrt{3}\cos \theta =3 \\ \Rightarrow \cos \theta =\frac{3}{2\sqrt{3}}=\frac{\sqrt{3}}{2}=\cos {30}^0 \\ \Rightarrow \theta ={30}^0 \\ \therefore Theanglebetween\overrightarrow{a}and\overrightarrow{b}is{30}^0 \end{gathered}$$

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