(ক) দেওয়া আছে, $\mathbf{a} = 2 \hat{i} - 2 \hat{j} + \hat{k}$ এবং $\mathbf{b} = 6 \hat{i} - 2 \hat{j} + 3 \hat{k}$। $$\begin{aligned} 2\mathbf{a} + 3\mathbf{b} &= 2(2 \hat{i} - 2 \hat{j} + \hat{k}) + 3(6 \hat{i} - 2 \hat{j} + 3 \hat{k}) \\ &= (4 \hat{i} - 4 \hat{j} + 2 \hat{k}) + (18 \hat{i} - 6 \hat{j} + 9 \hat{k}) \\ &= 22 \hat{i} - 10 \hat{j} + 11 \hat{k} \end{aligned}$$ মান নির্ণয়: $$|2\mathbf{a} + 3\mathbf{b}| = \sqrt{22^2 + (-10)^2 + 11^2} = \sqrt{484 + 100 + 121} = \sqrt{705}$$ (খ) চিত্রে $\mathbf{a}$ ও $\mathbf{b}$ এর মধ্যবর্তী কোণ $\beta$। আমরা জানি, $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \beta$। $|\mathbf{a}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = 3$ $|\mathbf{b}| = \sqrt{6^2 + (-2)^2 + 3^2} = \sqrt{36 + 4 + 9} = 7$ $\mathbf{a} \cdot \mathbf{b} = (2)(6) + (-2)(-2) + (1)(3) = 12 + 4 + 3 = 19$ সুতরাং, $$19 = 3 \times 7 \times \cos \beta \Rightarrow \cos \beta = \frac{19}{21} \Rightarrow \beta = \cos^{-1}(\frac{19}{21})$$ (গ) দেওয়া আছে, $|\mathbf{c}| = 12$ এবং $\mathbf{c} = p\mathbf{a}$। $|\mathbf{c}| = |p\mathbf{a}| = |p| |\mathbf{a}| = 12$ যেহেতু $|\mathbf{a}| = 3$, তাই $|p| \times 3 = 12 \Rightarrow |p| = 4$। ধরি, $p = 4$, তাহলে $\mathbf{c} = 4(2 \hat{i} - 2 \hat{j} + \hat{k}) = 8 \hat{i} - 8 \hat{j} + 4 \hat{k}$। আবার, $\mathbf{d} = 2\mathbf{b} = 2(6 \hat{i} - 2 \hat{j} + 3 \hat{k}) = 12 \hat{i} - 4 \hat{j} + 6 \hat{k}$। $$\begin{aligned} \mathbf{d} - \mathbf{c} &= (12 - 8) \hat{i} + (-4 - (-8)) \hat{j} + (6 - 4) \hat{k} \\ &= 4 \hat{i} + 4 \hat{j} + 2 \hat{k} \end{aligned}$$ মান নির্ণয়: $$|\mathbf{d} - \mathbf{c}| = \sqrt{4^2 + 4^2 + 2^2} = \sqrt{16 + 16 + 4} = \sqrt{36} = 6$$