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 the vectors \( \vec{a} + \vec{b} \) and \( \vec{a} - \vec{b} \).
Given: \[ \vec{a} = \hat{i} + \hat{j} + \hat{k}, \quad \vec{b} = \hat{i} + 2\hat{j} + 3\hat{k} \] First, find \( \vec{a} + \vec{b} \): \[ \vec{a} + \vec{b} = (\hat{i} + \hat{j} + \hat{k}) + (\hat{i} + 2\hat{j} + 3\hat{k}) = 2\hat{i} + 3\hat{j} + 4\hat{k} \] Next, find \( \vec{a} - \vec{b} \): \[ \vec{a} - \vec{b} = (\hat{i} + \hat{j} + \hat{k}) - (\hat{i} + 2\hat{j} + 3\hat{k}) = 0\hat{i} - \hat{j} - 2\hat{k} = -\hat{j} - 2\hat{k} \] Step 2: Find the cross product of \( (\vec{a} + \vec{b}) \) and \( (\vec{a} - \vec{b}) \).
To find the vector perpendicular to both \( \vec{a} + \vec{b} \) and \( \vec{a} - \vec{b} \), we take the cross product of these two vectors: \[ (\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}) = (2\hat{i} + 3\hat{j} + 4\hat{k}) \times (-\hat{j} - 2\hat{k}) \] Using the determinant form of the cross product: \[ \vec{u} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 0 & -1 & -2 \end{vmatrix} \] Expanding the determinant: \[ = \hat{i} \begin{vmatrix} 3 & 4 \\ -1 & -2 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 4 \\ 0 & -2 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 3 \\ 0 & -1 \end{vmatrix} \] Calculating the 2x2 determinants: \[ \hat{i} \left( 3(-2) - 4(-1) \right) = \hat{i} \left( -6 + 4 \right) = -2\hat{i} \] \[ - \hat{j} \left( 2(-2) - 4(0) \right) = - \hat{j} \left( -4 \right) = 4\hat{j} \] \[ + \hat{k} \left( 2(-1) - 3(0) \right) = \hat{k} \left( -2 \right) = -2\hat{k} \] Thus, the cross product is: \[ (\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}) = -2\hat{i} + 4\hat{j} - 2\hat{k} \] Step 3: Find the unit vector.
The magnitude of the cross product is: \[ | -2\hat{i} + 4\hat{j} - 2\hat{k} | = \sqrt{(-2)^2 + 4^2 + (-2)^2} = \sqrt{4 + 16 + 4} = \sqrt{24} = 2\sqrt{6} \] Now, the unit vector in the direction of the cross product is: \[ \hat{u} = \frac{-2\hat{i} + 4\hat{j} - 2\hat{k}}{2\sqrt{6}} = \frac{-\hat{i} + 2\hat{j} - \hat{k}}{\sqrt{6}} \] Final Answer: The unit vector perpendicular to both \( (\vec{a} + \vec{b}) \) and \( (\vec{a} - \vec{b}) \) is: \[ \hat{u} = \frac{-\hat{i} + 2\hat{j} - \hat{k}}{\sqrt{6}} \]