Let's analyze the problem step by. --- # Data: -(\vec{} = \vec{i - 3\{j} 2\{k}\) \(\vecN} = 4\vec{i} + 5\vec{j} - 2\vec{k}\) --- ## a. Find \(\vec{M} \times \vec{N}\) ### Step 1: Write the vectors \[ \vec{M} = \begin{bmatrix} 2 \\ -3 \\ 2 \end{bmatrix}, \quad \vec{N} = \begin{bmatrix} 4 \\ 5 \\ -2 \end{bmatrix} \] ### Step 2: Use the cross product formula \[ \vec{M} \times \vec{N} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 2 & -3 & 2 \\ 4 & 5 & -2 \\ \end{vmatrix} \] ### Step 3: Calculate the determinant \[ \vec{M} \times \vec{N} = \vec{i}((-3)(-2) - 2 \times 5) - \vec{j}(2 \times (-2) - 2 \times 4) + \vec{k}(2 \times 5 - (-3) \times 4) \] Simplify each component: - \(\vec{i}\): \((-3)(-2) - 2 \times 5 = 6 - 10 = -4\) - \(\vec{j}\): \(2 \times (-2) - 2 \times 4 = -4 - 8 = -12\) - \(\vec{k}\): \(2 \times 5 - (-3) \times 4 = 10 + 12 = 22\) ### **Result:** \[ \boxed{ \vec{M} \times \vec{N} = -4 \vec{i} + 12 \vec{j} + 22 \vec{k} } \] --- ## b. Show that \(\vec{M} \times \vec{N}\) is perpendicular to both \(\vec{M}\) and \(\vec{N}\). ### Step 1: Recall the dot product property \[ \vec{A} \cdot (\vec{A} \times \vec{B}) = \] for any vectors \(\vec{A}\) and \(\vec{B}\), because the cross product is perpendicular to both. ### Step 2: Verify for \(\vec{M}\) and \(\vec{N}\) - **Check with \(\vec{M}\):** \[ \vec{M} \cdot (\vec{M} \times \vec{N}) = (2)(-4) + (-3)(12) + (2)(22) = -8 - 36 + 44 = \] - **Check with \(\vec{N}\):** \[ \vec{N} \cdot (\vec{M} \times \vec{N}) = (4)(-4) + (5)(12) + (-2)(22) = -16 + 60 - 44 = \] ### **Conclusion:** \(\vec{M} \times \vec{N}\) is perpendicular to both \(\vec{M}\) and \(\vec{N}\). --- ## Summary: - **Cross product:** \[ \boxed{ \vec{M} \times \vec{N} = -4 \vec{i} + 12 \vec{j} + 22 \vec{k} } \] - **Perpendicularity verified via dot product.**