1. The problem gives two vectors $\mathbf{a} = 11 \mathbf{i} + 9 \mathbf{j} + 0 \mathbf{k}$ and $\mathbf{b} = x \mathbf{i} + 7 \mathbf{j} + 0 \mathbf{k}$. 2. Find $x$ such that $\mathbf{a}$ and $\mathbf{b}$ are parallel. - Two vectors are parallel if one is a scalar multiple of the other. - So $\frac{x}{11} = \frac{7}{9} = \frac{0}{0}$ (ignoring the zero component). - From $\frac{x}{11} = \frac{7}{9}$, solving for $x$ gives: $$x = \frac{11 \times 7}{9} = \frac{77}{9}$$ 3. Find $x$ such that $\mathbf{a}$ and $\mathbf{b}$ are perpendicular. - Two vectors are perpendicular if their dot product is zero. - The dot product: $$\mathbf{a} \cdot \mathbf{b} = 11x + 9 \times 7 + 0 \times 0 = 11x + 63$$ - Set equal to zero and solve for $x$: $$11x + 63 = 0$$ $$11x = -63$$ $$x = \frac{-63}{11}$$ Final answers: - (a) $x = \frac{77}{9}$ - (b) $x = \frac{-63}{11}$