1. Problem: Find the magnitude and slope of vectors $\mathbf{A}$ and $\mathbf{B}$ where $\mathbf{A}=6+ j8$ and $\mathbf{B}=3- j4$. 2. Magnitude of a vector $\mathbf{v}=x+jy$ is given by: $$ \|\mathbf{v}\|=\sqrt{x^2+y^2} $$ 3. For $\mathbf{A}=6+j8$: $$ \|\mathbf{A}\|=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10 $$ 4. For $\mathbf{B}=3-j4$: $$ \|\mathbf{B}\|=\sqrt{3^2+(-4)^2}=\sqrt{9+16}=\sqrt{25}=5 $$ 5. Slope of a vector $\mathbf{v}=x+jy$ in the $xy$-plane is: $$ m=\frac{y}{x} $$ 6. Slope of $\mathbf{A}$: $$ m_A=\frac{8}{6}=\frac{4}{3} $$ 7. Slope of $\mathbf{B}$: $$ m_B=\frac{-4}{3} $$ Final Answer: - Magnitude: $\|\mathbf{A}\|=10$, $\|\mathbf{B}\|=5$ - Slope: $m_A=\frac{4}{3}$, $m_B=-\frac{4}{3}$