5 Find the angle that each of these vectors makes with \( \mathbf{j} \). \( \begin{array}{lll}\mathbf{a} 3 \mathbf{i}-5 \mathbf{j} & \text { b } 4 \mathbf{i}+7 \mathbf{j} & \mathbf{c}-3 \mathbf{i}+5 \mathbf{j}\end{array} \)

To find the angle that each vector makes with \(\mathbf{j}\), we can use the dot product formula: \[ \cos(\theta) = \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{u}|| \, ||\mathbf{v}||} \] where \(\mathbf{u}\) is the given vector and \(\mathbf{v} = \mathbf{j} = (0, 1)\). Let's calculate the angles for each vector: 1. For vector \(\mathbf{a} = 3\mathbf{i} - 5\mathbf{j}\): - \(\mathbf{u} = (3, -5)\) and \(\mathbf{v} = (0, 1)\) - The dot product \(\mathbf{u} \cdot \mathbf{v} = 0 - 5 = -5\) - The magnitudes are \(||\mathbf{u}|| = \sqrt{3^2 + (-5)^2} = \sqrt{34}\) and \(||\mathbf{v}|| = 1\) - So, \(\cos(\theta) = \frac{-5}{\sqrt{34}} \Rightarrow \theta = \cos^{-1}\left(\frac{-5}{\sqrt{34}}\right)\) 2. For vector \(\mathbf{b} = 4\mathbf{i} + 7\mathbf{j}\): - \(\mathbf{u} = (4, 7)\) - The dot product \(\mathbf{u} \cdot \mathbf{v} = 0 + 7 = 7\) - The magnitudes are \(||\mathbf{u}|| = \sqrt{4^2 + 7^2} = \sqrt{65}\) and \(||\mathbf{v}|| = 1\) - So, \(\cos(\theta) = \frac{7}{\sqrt{65}} \Rightarrow \theta = \cos^{-1}\left(\frac{7}{\sqrt{65}}\right)\) 3. For vector \(\mathbf{c} = -3\mathbf{i} + 5\mathbf{j}\): - \(\mathbf{u} = (-3, 5)\) - The dot product \(\mathbf{u} \cdot \mathbf{v} = 0 + 5 = 5\) - The magnitudes are \(||\mathbf{u}|| = \sqrt{(-3)^2 + 5^2} = \sqrt{34}\) and \(||\mathbf{v}|| = 1\) - So, \(\cos(\theta) = \frac{5}{\sqrt{34}} \Rightarrow \theta = \cos^{-1}\left(\frac{5}{\sqrt{34}}\right)\) Now you can calculate the angles using a calculator for precise values.