A drone is flying in the direction from IUT (I) to Uttara (U) to Tangail (T) and back to IUT forming a triangle. The co-ordinate of IUT and Uttara is (1, 2, 2) and (2, -1, 3). The vector from Uttara to Tangail can be represented by −4i−2j+2k -4 ext{i} - 2 ext{j} + 2 ext{k} . What angle the drone will make while travelling from Uttara to Tangail to IUT?

Question

A drone is flying in the direction from IUT (I) to Uttara (U) to Tangail (T) and back to IUT forming a triangle. The co-ordinate of IUT and Uttara is (1, 2, 2) and (2, -1, 3). The vector from Uttara to Tangail can be represented by −4i−2j+2k -4	ext{i} - 2	ext{j} + 2	ext{k} . What angle the drone will make while travelling from Uttara to Tangail to IUT?

Accepted Answer

ক. 30°

ব্যাখ্যা: প্রদত্ত, $I = (1, 2, 2)$ এবং $U = (2, -1, 3)$। ভেক্টর $\vec{UT} = -4\hat{i} - 2\hat{j} + 2\hat{k}$, সুতরাং $T = U + \vec{UT} = (-2, -3, 5)$। এবার $T$ বিন্দুতে ভেক্টরদ্বয় হলো $\vec{TU} = -\vec{UT} = 4\hat{i} + 2\hat{j} - 2\hat{k}$ এবং $\vec{TI} = I - T = 3\hat{i} + 5\hat{j} - 3\hat{k}$। এদের মধ্যবর্তী কোণ $	heta$ হলে, $\cos	heta = \frac{\vec{TU} \cdot \vec{TI}}{|\vec{TU}||\vec{TI}|} = \frac{(4)(3) + (2)(5) + (-2)(-3)}{\sqrt{16+4+4}\sqrt{9+25+9}} = \frac{28}{\sqrt{24}\sqrt{43}} = \frac{28}{\sqrt{1032}} \approx 0.8716$। সুতরাং $	heta = \arccos(0.8716) \approx 29.4^\circ \approx 30^\circ$।

পদার্থবিজ্ঞান প্রশ্ন

সম্পর্কিত প্রশ্ন