1. **Problem statement:** Given three displacement vectors $ \mathbf{B} = -\mathbf{i} + 2\mathbf{j} $, $ \mathbf{C} = 3\mathbf{i} - 2\mathbf{j} $, and $ \mathbf{D} $ unknown, find the magnitude and direction of $ \mathbf{D} $ such that $ 3\mathbf{B} - \mathbf{C} + \mathbf{D} = 0 $. 2. **Formula and rules:** The equation can be rearranged to find $ \mathbf{D} $: $$ \mathbf{D} = -3\mathbf{B} + \mathbf{C} $$ We will calculate $ \mathbf{D} $ by substituting $ \mathbf{B} $ and $ \mathbf{C} $, then find its magnitude and direction. 3. **Calculate $ 3\mathbf{B} $:** $$ 3\mathbf{B} = 3(-\mathbf{i} + 2\mathbf{j}) = -3\mathbf{i} + 6\mathbf{j} $$ 4. **Calculate $ -3\mathbf{B} + \mathbf{C} $:** $$ -3\mathbf{B} + \mathbf{C} = -(-3\mathbf{i} + 6\mathbf{j}) + (3\mathbf{i} - 2\mathbf{j}) = 3\mathbf{i} - 6\mathbf{j} + 3\mathbf{i} - 2\mathbf{j} $$ $$ = (3 + 3)\mathbf{i} + (-6 - 2)\mathbf{j} = 6\mathbf{i} - 8\mathbf{j} $$ So, $$ \mathbf{D} = 6\mathbf{i} - 8\mathbf{j} $$ 5. **Magnitude of $ \mathbf{D} $:** $$ |\mathbf{D}| = \sqrt{6^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 $$ 6. **Direction of $ \mathbf{D} $:** The direction angle $ \theta $ with respect to the positive x-axis is given by: $$ \theta = \tan^{-1} \left( \frac{y}{x} \right) = \tan^{-1} \left( \frac{-8}{6} \right) = \tan^{-1}(-\frac{4}{3}) $$ Since $ x = 6 > 0 $ and $ y = -8 < 0 $, $ \mathbf{D} $ lies in the fourth quadrant. Calculate: $$ \theta = -53.13^\circ $$ or equivalently, $$ \theta = 360^\circ - 53.13^\circ = 306.87^\circ $$ **Final answer:** Magnitude of $ \mathbf{D} $ is 10 units. Direction of $ \mathbf{D} $ is approximately $ 306.87^\circ $ measured counterclockwise from the positive x-axis.