27. Let \( \boldsymbol{u}=\boldsymbol{i}+m \boldsymbol{j}+\boldsymbol{k} \) and \( \boldsymbol{v}=2 \boldsymbol{i}-\boldsymbol{j}+n \boldsymbol{k} \). Compute all values of \( m \) and \( n \) for which \( \boldsymbol{u} \perp \boldsymbol{v} \) and \( |\boldsymbol{u}|=|\boldsymbol{v}| \).

To find the values of \( m \) and \( n \) such that \( \boldsymbol{u} \perp \boldsymbol{v} \) and \( |\boldsymbol{u}|=|\boldsymbol{v}| \), we start with the conditions given. **Step 1: Perpendicularity Condition** For two vectors to be perpendicular, their dot product must be zero: \[ \boldsymbol{u} \cdot \boldsymbol{v} = 0 \] Calculating the dot product: \[ (1)(2) + (m)(-1) + (1)(n) = 2 - m + n = 0 \] This leads to the equation: \[ m - n = 2 \quad \text{(1)} \] **Step 2: Equal Magnitude Condition** Next, we find the magnitudes of both vectors: \[ |\boldsymbol{u}| = \sqrt{1^2 + m^2 + 1^2} = \sqrt{2 + m^2} \] \[ |\boldsymbol{v}| = \sqrt{2^2 + (-1)^2 + n^2} = \sqrt{4 + 1 + n^2} = \sqrt{5 + n^2} \] Setting the magnitudes equal: \[ \sqrt{2 + m^2} = \sqrt{5 + n^2} \] Squaring both sides gives: \[ 2 + m^2 = 5 + n^2 \] Rearranging leads to: \[ m^2 - n^2 = 3 \quad \text{(2)} \] **Step 3: Solving the System of Equations** We now have two equations: 1. \( m - n = 2 \) 2. \( m^2 - n^2 = 3 \) We can express \( m \) in terms of \( n \) from equation (1): \[ m = n + 2 \] Substituting this into equation (2): \[ (n + 2)^2 - n^2 = 3 \] Expanding the left side: \[ n^2 + 4n + 4 - n^2 = 3 \] This simplifies to: \[ 4n + 4 = 3 \] Subtracting 4 from both sides: \[ 4n = -1 \] Thus, we find: \[ n = -\frac{1}{4} \] Substituting back to find \( m \): \[ m = -\frac{1}{4} + 2 = \frac{7}{4} \] **Final Values:** The values of \( m \) and \( n \) for which \( \boldsymbol{u} \perp \boldsymbol{v} \) and \( |\boldsymbol{u}| = |\boldsymbol{v}| \) are: \[ m = \frac{7}{4}, \quad n = -\frac{1}{4} \]