2hati-hatj+3hatk এবং4hati-2hatj+6hatk ভেক্টরদ্বয়ের মধ্যবর্তী কোণ-

Explanation:

Another Explanation (5): vector দুটি হল : \(\vec{a} = 2\hat{i} - \hat{j} + 3\hat{k}\) এবং \(\vec{b} = 4\hat{i} - 2\hat{j} + 6\hat{k}\) দুটি ভেক্টরের মধ্যবর্তী কোণ \(\theta\) হলে, \(\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}\) \(\vec{a} \cdot \vec{b} = (2 \times 4) + (-1 \times -2) + (3 \times 6) = 8 + 2 + 18 = 28\) \(|\vec{a}| = \sqrt{2^2 + (-1)^2 + 3^2} = \sqrt{4 + 1 + 9} = \sqrt{14}\) \(|\vec{b}| = \sqrt{4^2 + (-2)^2 + 6^2} = \sqrt{16 + 4 + 36} = \sqrt{56} = 2\sqrt{14}\) সুতরাং, \(\cos \theta = \frac{28}{\sqrt{14} \times 2\sqrt{14}} = \frac{28}{2 \times 14} = \frac{28}{28} = 1\) \(\cos \theta = 1\) \(\theta = \cos^{-1}(1) = 0^\circ\) অতএব, ভেক্টরদ্বয়ের মধ্যবর্তী কোণ \(0^\circ\)।😊🎉