मान लीजिए hat{a} और hat{b} दो इकाई सदिश हैं ज | vedclass

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(A) दिया गया है कि $\hat{a} \cdot \hat{b} = |\hat{a}||\hat{b}| \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$.मान लीजिए $\vec{u} = \hat{a} + \hat{b}$ और $\

Vedclass · Accepted Answer

$90+27 \sqrt{2}$

Vedclass · Answer

(A) दिया गया है कि $\hat{a} \cdot \hat{b} = |\hat{a}||\hat{b}| \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$.मान लीजिए $\vec{u} = \hat{a} + \hat{b}$ और $\vec{v} = \hat{a} + 2\hat{b} + 2(\hat{a} 	imes \hat{b})$.$|\vec{u}|^2 = |\hat{a}|^2 + |\hat{b}|^2 + 2(\hat{a} \cdot \hat{b}) = 1 + 1 + 2(\frac{1}{\sqrt{2}}) = 2 + \sqrt{2}$.$|\vec{v}|^2 = |\hat{a} + 2\hat{b} + 2(\hat{a} 	imes \hat{b})|^2 = |\hat{a}|^2 + 4|\hat{b}|^2 + 4|\hat{a} 	imes \hat{b}|^2 + 4(\hat{a} \cdot \hat{b}) + 0 + 0$.चूँकि $|\hat{a} 	imes \hat{b}| = |\hat{a}||\hat{b}| \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$,इसलिए $|\hat{a} 	imes \hat{b}|^2 = \frac{1}{2}$.$|\vec{v}|^2 = 1 + 4(1) + 4(\frac{1}{2}) + 4(\frac{1}{\sqrt{2}}) = 1 + 4 + 2 + 2\sqrt{2} = 7 + 2\sqrt{2}$.अब,$\vec{u} \cdot \vec{v} = (\hat{a} + \hat{b}) \cdot (\hat{a} + 2\hat{b} + 2(\hat{a} 	imes \hat{b})) = |\hat{a}|^2 + 2(\hat{a} \cdot \hat{b}) + 0 + (\hat{b} \cdot \hat{a}) + 2|\hat{b}|^2 + 0 = 1 + 2(\frac{1}{\sqrt{2}}) + \frac{1}{\sqrt{2}} + 2 = 3 + \frac{3}{\sqrt{2}}$.$\cos 	heta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}||\vec{v}|} = \frac{3 + \frac{3}{\sqrt{2}}}{\sqrt{2+\sqrt{2}}\sqrt{7+2\sqrt{2}}}$.$\cos^2 	heta = \frac{9(1 + \frac{1}{\sqrt{2}})^2}{(2+\sqrt{2})(7+2\sqrt{2})} = \frac{9(\frac{\sqrt{2}+1}{\sqrt{2}})^2}{14 + 4\sqrt{2} + 7\sqrt{2} + 4} = \frac{9(\frac{3+2\sqrt{2}}{2})}{18 + 11\sqrt{2}} = \frac{9(3+2\sqrt{2})}{2(18+11\sqrt{2})}$.$164$ से गुणा करने पर: $164 \cos^2 	heta = 164 	imes \frac{9(3+2\sqrt{2})}{2(18+11\sqrt{2})} = 82 	imes \frac{9(3+2\sqrt{2})}{18+11\sqrt{2}} 	imes \frac{18-11\sqrt{2}}{18-11\sqrt{2}} = 738 	imes \frac{54 - 33\sqrt{2} + 36\sqrt{2} - 44}{324 - 242} = 738 	imes \frac{10 + 3\sqrt{2}}{82} = 9(10 + 3\sqrt{2}) = 90 + 27\sqrt{2}$.

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