Question

Find the unit vector perpendicular to each of the vectors (\( \vec{a} + \vec{b} \)) and (\( \vec{a} - \vec{b} \)) where \[\vec{a} = \hat{i} + \hat{j} + \hat{k}, \, \vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}.\]

Hint:

{Key Points:}

• Vector perpendicular to two vectors = their cross product
• Unit vector = \( \frac{\vec{n}}{|\vec{n}|} \)
• Both \( \hat{n} \) and \( -\hat{n} \) are perpendicular unit vectors

Answer 1

Step 1: Find \( \vec{a} + \vec{b} \) and \( \vec{a} - \vec{b} \)

Given: \[ \vec{a} = \hat{i} + \hat{j} + \hat{k} \] \[ \vec{b} = \hat{i} + 2\hat{j} + 3\hat{k} \] \[ \vec{a} + \vec{b} = 2\hat{i} + 3\hat{j} + 4\hat{k} \] \[ \vec{a} - \vec{b} = -\hat{j} - 2\hat{k} \] 
Step 2: Find a vector perpendicular to both

\[ \vec{n} = (\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}) \] \[ \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 0 & -1 & -2 \end{vmatrix} \] 
Step 3: Compute the cross product

\[ \vec{n} = \hat{i} \begin{vmatrix} 3 & 4 \\ -1 & -2 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 4 \\ 0 & -2 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 3 \\ 0 & -1 \end{vmatrix} \] \[ \vec{n} = \hat{i}(-6 + 4) - \hat{j}(-4) + \hat{k}(-2) \] \[ \vec{n} = -2\hat{i} + 4\hat{j} - 2\hat{k} \] 
Step 4: Find the magnitude of \( \vec{n} \)

\[ |\vec{n}| = \sqrt{(-2)^2 + 4^2 + (-2)^2} = \sqrt{4 + 16 + 4} = \sqrt{24} = 2\sqrt{6} \] 
Step 5: Find the unit vector

\[ \hat{n} = \frac{\vec{n}}{|\vec{n}|} = \frac{-2\hat{i} + 4\hat{j} - 2\hat{k}}{2\sqrt{6}} \] \[ \hat{n} = \frac{-\hat{i} + 2\hat{j} - \hat{k}}{\sqrt{6}} \] 
Step 6: Final Answer

\[ \boxed{\frac{-\hat{i} + 2\hat{j} - \hat{k}}{\sqrt{6}}} \] Note: The negative of this vector is also a unit vector perpendicular to both given vectors.