ভেক্টর দুইটির অন্তর্ভুক্ত কোণ-
A=sqrt2 cosalphai-sqrt2 sinalphaj, B=-sqrt2cosalphaj+sqrt2sinalphak(0<alpha<pi/2)
cos^-1(sinalphacosalpha)
ভেক্টর দুইটির অন্তর্ভুক্ত কোণ নির্ণয়
ধরি, \( \vec{A} = \sqrt{2} \cos{\alpha} \hat{i} - \sqrt{2} \sin{\alpha} \hat{j} \) এবং \( \vec{B} = -\sqrt{2} \cos{\alpha} \hat{j} + \sqrt{2} \sin{\alpha} \hat{k} \)। এই দুইটি ভেক্টরের অন্তর্ভুক্ত কোণ \( \theta \) হলে, আমরা জানি:
\( \cos{\theta} = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|} \)
প্রথমে, \( \vec{A} \cdot \vec{B} \) নির্ণয় করি:
\( \vec{A} \cdot \vec{B} = (\sqrt{2} \cos{\alpha} \hat{i} - \sqrt{2} \sin{\alpha} \hat{j}) \cdot (-\sqrt{2} \cos{\alpha} \hat{j} + \sqrt{2} \sin{\alpha} \hat{k}) \)
\( = (\sqrt{2} \cos{\alpha})(0) + (-\sqrt{2} \sin{\alpha})(-\sqrt{2} \cos{\alpha}) + (0)(\sqrt{2} \sin{\alpha}) \)
\( = 0 + 2 \sin{\alpha} \cos{\alpha} + 0 \)
\( = 2 \sin{\alpha} \cos{\alpha} \)
এখন, \( |\vec{A}| \) এবং \( |\vec{B}| \) নির্ণয় করি:
\( |\vec{A}| = \sqrt{(\sqrt{2} \cos{\alpha})^2 + (-\sqrt{2} \sin{\alpha})^2} \)
\( = \sqrt{2 \cos^2{\alpha} + 2 \sin^2{\alpha}} \)
\( = \sqrt{2(\cos^2{\alpha} + \sin^2{\alpha})} = \sqrt{2} \)
\( |\vec{B}| = \sqrt{(-\sqrt{2} \cos{\alpha})^2 + (\sqrt{2} \sin{\alpha})^2} \)
\( = \sqrt{2 \cos^2{\alpha} + 2 \sin^2{\alpha}} \)
\( = \sqrt{2(\cos^2{\alpha} + \sin^2{\alpha})} = \sqrt{2} \)
সুতরাং,
\( \cos{\theta} = \frac{2 \sin{\alpha} \cos{\alpha}}{\sqrt{2} \cdot \sqrt{2}} = \frac{2 \sin{\alpha} \cos{\alpha}}{2} = \sin{\alpha} \cos{\alpha} \)
অতএব,
\( \theta = \cos^{-1}(\sin{\alpha} \cos{\alpha}) \) 🥳🎉
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