1. **Problem 1: Expand and approximate $(1.03)^3$ using the Binomial Theorem up to four terms.** The Binomial Theorem states: $$ (x + y)^n = \sum_{k=0}^n \binom{n}{k} x^{n-k} y^k $$ Here, let $x=1$, $y=0.03$, and $n=3$. 2. Write the expansion up to four terms (which is all terms for $n=3$): $$ (1 + 0.03)^3 = \binom{3}{0}1^3(0.03)^0 + \binom{3}{1}1^2(0.03)^1 + \binom{3}{2}1^1(0.03)^2 + \binom{3}{3}1^0(0.03)^3 $$ 3. Calculate each term: - $\binom{3}{0} = 1$, so first term is $1 \times 1 \times 1 = 1$ - $\binom{3}{1} = 3$, so second term is $3 \times 1 \times 0.03 = 0.09$ - $\binom{3}{2} = 3$, so third term is $3 \times 1 \times (0.03)^2 = 3 \times 0.0009 = 0.0027$ - $\binom{3}{3} = 1$, so fourth term is $1 \times 1 \times (0.03)^3 = 0.000027$ 4. Sum the terms: $$ 1 + 0.09 + 0.0027 + 0.000027 = 1.092727 $$ 5. Approximate to 3 decimal places: $$ 1.093 $$ --- 6. **Problem 2: Find the projection of vector $\mathbf{a} = 5\mathbf{i} - 2\mathbf{j} + \mathbf{k}$ along $\mathbf{b} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k}$ and projection of $\mathbf{b}$ along $\mathbf{a}$.** 7. Recall the formula for projection of $\mathbf{u}$ along $\mathbf{v}$: $$ \text{proj}_{\mathbf{v}} \mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{v}\|^2} \mathbf{v} $$ 8. Compute dot products: $$ \mathbf{a} \cdot \mathbf{b} = 5 \times 1 + (-2) \times 2 + 1 \times 2 = 5 - 4 + 2 = 3 $$ $$ \mathbf{b} \cdot \mathbf{a} = \mathbf{a} \cdot \mathbf{b} = 3 $$ 9. Compute magnitudes squared: $$ \|\mathbf{b}\|^2 = 1^2 + 2^2 + 2^2 = 1 + 4 + 4 = 9 $$ $$ \|\mathbf{a}\|^2 = 5^2 + (-2)^2 + 1^2 = 25 + 4 + 1 = 30 $$ 10. Calculate projections: $$ \text{proj}_{\mathbf{b}} \mathbf{a} = \frac{3}{9} \mathbf{b} = \frac{1}{3} (\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}) = \frac{1}{3} \mathbf{i} + \frac{2}{3} \mathbf{j} + \frac{2}{3} \mathbf{k} $$ $$ \text{proj}_{\mathbf{a}} \mathbf{b} = \frac{3}{30} \mathbf{a} = \frac{1}{10} (5\mathbf{i} - 2\mathbf{j} + \mathbf{k}) = 0.5 \mathbf{i} - 0.2 \mathbf{j} + 0.1 \mathbf{k} $$ --- **Final answers:** - Approximate value of $(1.03)^3$ is **1.093**. - Projection of $\mathbf{a}$ along $\mathbf{b}$ is $\frac{1}{3} \mathbf{i} + \frac{2}{3} \mathbf{j} + \frac{2}{3} \mathbf{k}$. - Projection of $\mathbf{b}$ along $\mathbf{a}$ is $0.5 \mathbf{i} - 0.2 \mathbf{j} + 0.1 \mathbf{k}$.