7. Find a vector with magnitude 3 and direction angles \( \frac{\pi}{4}, \frac{\pi}{4}, \frac{\pi}{2} \).

To find a vector with a specified magnitude and direction angles, we can use the formula that relates the direction angles (α, β, γ) to the components of the vector \( \mathbf{v} \): \[ \mathbf{v} = |\mathbf{v}| \begin{pmatrix} \sin(\alpha) \sin(\beta) \\ \sin(\beta) \cos(\alpha) \\ \cos(\beta) \end{pmatrix} \] Here, the magnitude \( |\mathbf{v}| = 3 \), and the direction angles given are \( \alpha = \frac{\pi}{4} \), \( \beta = \frac{\pi}{4} \), and \( \gamma = \frac{\pi}{2} \). Now we can substitute these values into our formula: 1. Calculate the components: - For \( v_x \): \( v_x = 3 \sin\left(\frac{\pi}{4}\right) \sin\left(\frac{\pi}{4}\right) = 3 \cdot \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} = 3 \cdot \frac{1}{2} = \frac{3}{2} \) - For \( v_y \): \( v_y = 3 \sin\left(\frac{\pi}{4}\right) \cos\left(\frac{\pi}{4}\right) = 3 \cdot \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} = 3 \cdot \frac{1}{2} = \frac{3}{2} \) - For \( v_z \): \( v_z = 3 \cos\left(\frac{\pi}{4}\right) = 3 \cdot \frac{\sqrt{2}}{2} = \frac{3\sqrt{2}}{2} \) Putting it all together, the vector \( \mathbf{v} \) is: \[ \mathbf{v} = \begin{pmatrix} \frac{3}{2} \\ \frac{3}{2} \\ \frac{3\sqrt{2}}{2} \end{pmatrix} \]