If vec{a}=hat{i}+(tan theta) hat{j}+left(frac | vedclass

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(B) Given $\vec{a} \cdot \vec{b} = 0$ as they are orthogonal.$\vec{a} \cdot \vec{b} = (1)(-	an 	heta) + (	an 	heta)(	an 	heta) + \left(\frac{3}{

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$n \pi-	an ^{-1} 2, n \in Z$

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(B) Given $\vec{a} \cdot \vec{b} = 0$ as they are orthogonal.$\vec{a} \cdot \vec{b} = (1)(-	an 	heta) + (	an 	heta)(	an 	heta) + \left(\frac{3}{\sqrt{\sin \frac{	heta}{2}}}ight)(-2 \sqrt{\sin \frac{	heta}{2}}) = 0$$-	an 	heta + 	an^2 	heta - 6 = 0$Let $x = 	an 	heta$, then $x^2 - x - 6 = 0 \Rightarrow (x-3)(x+2) = 0$So, $	an 	heta = 3$ or $	an 	heta = -2$.Vector $\vec{c} = (\sin 2 	heta) \hat{i} - 2 \hat{j} + 2 \hat{k}$ makes an obtuse angle with the $X$-axis, which means the projection of $\vec{c}$ on the $X$-axis must be negative.$\vec{c} \cdot \hat{i} If $	an 	heta = 3$, then $\sin 2 	heta = \frac{2 	an 	heta}{1 + 	an^2 	heta} = \frac{6}{10} > 0$ (Rejected).If $	an 	heta = -2$, then $\sin 2 	heta = \frac{2(-2)}{1 + (-2)^2} = \frac{-4}{5} Thus, $	an 	heta = -2 \Rightarrow 	heta = n \pi - 	an^{-1} 2, n \in Z$.

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