Let's solve each part step by step: --- ### **1.1** Given: \[ \vec{a} = 2\vec{i} + 3\vec{j} + 4\vec{k} \] \[ \vec{b} = \vec{i} - \vec{j} + \vec{k} \] --- #### **(i) \(\vec{a} \times \vec{b}\) [5]** The cross product formula: \[ \vec{a} \times \vec{b} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 2 & 3 & 4 \\ 1 & -1 & 1 \\ \end{vmatrix} \] Expanding the determinant: \[ = \vec{i} \left(3 \cdot 1 - 4 \cdot (-1)\right) - \vec{j} \left(2 \cdot 1 - 4 \cdot 1\right) + \vec{k} \left(2 \cdot (-1) - 3 \cdot 1\right) \] \[ = \vec{i}(3 + 4) - \vec{j}(2 - 4) + \vec{k}(-2 - 3) \] \[ = 7\vec{i} + 2\vec{j} - 5\vec{k} \] **Final Answer (i):** \[ \boxed{\vec{a} \times \vec{b} = 7\vec{i} + 2\vec{j} - 5\vec{k}} \] --- #### **(ii) Sine of the angle between these vectors [6]** We know: \[ |\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta \] First, calculate magnitudes: \[ |\vec{a}| = \sqrt{2^2 + 3^2 + 4^2} = \sqrt{4 + 9 + 16} = \sqrt{29} \] \[ |\vec{b}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \] Magnitude of cross product: \[ |\vec{a} \times \vec{b}| = \sqrt{7^2 + 2^2 + (-5)^2} = \sqrt{49 + 4 + 25} = \sqrt{78} \] Now, \[ |\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta \] \[ \sqrt{78} = \sqrt{29} \times \sqrt{3} \times \sin\theta \] \[ \sin\theta = \frac{\sqrt{78}}{\sqrt{29} \times \sqrt{3}} = \frac{\sqrt{78}}{\sqrt{87}} \] \[ = \frac{\sqrt{78}}{ \sqrt{87}} \] **Final Answer (ii):** \[ \boxed{\sin\theta = \frac{\sqrt{78}}{\sqrt{87}}} \] --- #### **(iii) Unit vector perpendicular to each vector [2]** The unit vector perpendicular to \(\vec{a}\) and \(\vec{b}\) is in the direction of \(\vec{a} \times \vec{b}\): \[ \text{Unit vector} = \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} \] \[ = \frac{7\vec{i} + 2\vec{j} - 5\vec{k}}{\sqrt{78}} \] **Final Answer (iii):** \[ \boxed{\frac{7\vec{i} + 2\vec{j} - 5\vec{k}}{\sqrt{78}}} \] --- ### **1.2** Given: \[ \vec{a} = \vec{i} - \vec{j} + \vec{k} \] \[ \vec{b} = 2\vec{j} - 3\vec{k} \] Area of parallelogram = \( |\vec{a} \times \vec{b}| \) \[ \vec{a} \times \vec{b} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & -1 & 1 \\ 0 & 2 & -3 \\ \end{vmatrix} \] Expand: \[ = \vec{i}((-1)(-3) - (1)(2)) - \vec{j}(1 \cdot (-3) - 1 \cdot 0) + \vec{k}(1 \cdot 2 - (-1) \cdot 0) \] \[ = \vec{i}(3 - 2) - \vec{j}(-3 - 0) + \vec{k}(2 - 0) \] \[ = 1\vec{i} + 3\vec{j} + 2\vec{k} \] Magnitude: \[ |\vec{a} \times \vec{b}| = \sqrt{1^2 + 3^2 + 2^2} = \sqrt{1 + 9 + 4} = \sqrt{14} \] **Final Answer 1.2:** \[ \boxed{\text{Area} = \sqrt{14}} \] --- ## **Summary of Final Answers:** ### 1.1 (i) \(\boxed{7\vec{i} + 2\vec{j} - 5\vec{k}}\) (ii) \(\boxed{\sin\theta = \frac{\sqrt{78}}{\sqrt{87}}}\) (iii) \(\boxed{\frac{7\vec{i} + 2\vec{j} - 5\vec{k}}{\sqrt{78}}}\) ### 1.2 \(\boxed{\sqrt{14}}\)