6. Is there a vector whose direction angles are \( \frac{\pi}{4}, \frac{\pi}{4}, \frac{\pi}{3} \) ?

To determine if there is a vector whose direction angles are \( \frac{\pi}{4}, \frac{\pi}{4}, \frac{\pi}{3} \), we can use the relationship between direction angles and the components of the vector. The direction angles \( \alpha, \beta, \gamma \) of a vector \( \mathbf{v} = (v_1, v_2, v_3) \) are related to the components of the vector by the following equations: \[ \cos(\alpha) = \frac{v_1}{|\mathbf{v}|}, \quad \cos(\beta) = \frac{v_2}{|\mathbf{v}|}, \quad \cos(\gamma) = \frac{v_3}{|\mathbf{v}|} \] where \( |\mathbf{v}| \) is the magnitude of the vector given by: \[ |\mathbf{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2} \] The direction angles must satisfy the following condition: \[ \cos^2(\alpha) + \cos^2(\beta) + \cos^2(\gamma) = 1 \] Let's calculate \( \cos(\alpha) \), \( \cos(\beta) \), and \( \cos(\gamma) \) for the given angles: 1. \( \alpha = \frac{\pi}{4} \) \[ \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \] 2. \( \beta = \frac{\pi}{4} \) \[ \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \] 3. \( \gamma = \frac{\pi}{3} \) \[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \] Now, we can substitute these values into the condition: \[ \cos^2\left(\frac{\pi}{4}\right) + \cos^2\left(\frac{\pi}{4}\right) + \cos^2\left(\frac{\pi}{3}\right) = \left(\frac{\sqrt{2}}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2 + \left(\frac{1}{2}\right)^2 \] Calculating each term: \[ \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{2}{4} = \frac{1}{2} \] \[ \left(\frac{1}{2}\right)^2 = \frac{1}{4} \] Now, summing these: \[ \frac{1}{2} + \frac{1}{2} + \frac{1}{4} = 1 + \frac{1}{4} = \frac{5}{4} \] Since \( \frac{5}{4} \neq 1 \), the condition is not satisfied. Therefore, there is no vector whose direction angles are \( \frac{\pi}{4}, \frac{\pi}{4}, \frac{\pi}{3} \).