10. The points \( P(1,2,3) \) and \( Q(2,3,4) \) are given. Find the norm and di- rection cosines of the vector \( \overrightarrow{P Q} \).

To find the norm and direction cosines of the vector \( \overrightarrow{PQ} \), we will follow these steps: 1. **Calculate the vector \( \overrightarrow{PQ} \)**: The vector \( \overrightarrow{PQ} \) can be calculated using the coordinates of points \( P \) and \( Q \): \[ \overrightarrow{PQ} = Q - P = (2 - 1, 3 - 2, 4 - 3) = (1, 1, 1) \] 2. **Calculate the norm of the vector \( \overrightarrow{PQ} \)**: The norm (or magnitude) of a vector \( \overrightarrow{v} = (x, y, z) \) is given by the formula: \[ ||\overrightarrow{v}|| = \sqrt{x^2 + y^2 + z^2} \] For our vector \( \overrightarrow{PQ} = (1, 1, 1) \): \[ ||\overrightarrow{PQ}|| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \] 3. **Calculate the direction cosines**: The direction cosines \( l, m, n \) of a vector \( \overrightarrow{v} = (x, y, z) \) are given by: \[ l = \frac{x}{||\overrightarrow{v}||}, \quad m = \frac{y}{||\overrightarrow{v}||}, \quad n = \frac{z}{||\overrightarrow{v}||} \] For our vector \( \overrightarrow{PQ} = (1, 1, 1) \) and its norm \( ||\overrightarrow{PQ}|| = \sqrt{3} \): \[ l = \frac{1}{\sqrt{3}}, \quad m = \frac{1}{\sqrt{3}}, \quad n = \frac{1}{\sqrt{3}} \] Now, we can summarize the results: - The norm of the vector \( \overrightarrow{PQ} \) is \( \sqrt{3} \). - The direction cosines are \( l = \frac{1}{\sqrt{3}}, m = \frac{1}{\sqrt{3}}, n = \frac{1}{\sqrt{3}} \).