Question

\(\text{The unit vector perpendicular to each of the vectors } \vec{a} + \vec{b} \text{ and } \vec{a} - \vec{b}, \text{ where } \vec{a} = \hat{i} + \hat{j} + \hat{k} \text{ and } \vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}, \text{ is:}\)

• A. \( \frac{1}{\sqrt{6}} \hat{i} + \frac{2}{\sqrt{6}} \hat{j} + \frac{1}{\sqrt{6}} \hat{k} \)
• B. \( -\frac{1}{\sqrt{6}} \hat{i} + \frac{1}{\sqrt{6}} \hat{j} - \frac{1}{\sqrt{6}} \hat{k} \)
• C. \( -\frac{1}{\sqrt{6}} \hat{i} + \frac{2}{\sqrt{6}} \hat{j} + \frac{2}{\sqrt{6}} \hat{k} \)
• D. \( -\frac{1}{\sqrt{6}} \hat{i} + \frac{2}{\sqrt{6}} \hat{j} - \frac{1}{\sqrt{6}} \hat{k} \)

Answer 1

The Correct Option is D

Let the given vectors be:
\( \vec{a} = \hat{i} + \hat{j} + \hat{k}, \quad \vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}. \) We need to find the unit vector perpendicular to both \( \vec{a} + \vec{b} \) and \( \vec{a} - \vec{b}. \) The cross product of these two vectors will give a vector perpendicular to both. Let’s first compute the cross product of \( \vec{a} + \vec{b} \) and \( \vec{a} - \vec{b}. \)

First, compute the two vectors:
\( \vec{a} + \vec{b} = (\hat{i} + \hat{j} + \hat{k}) + (\hat{i} + 2\hat{j} + 3\hat{k}) = 2\hat{i} + 3\hat{j} + 4\hat{k}, \)
\( \vec{a} - \vec{b} = (\hat{i} + \hat{j} + \hat{k}) - (\hat{i} + 2\hat{j} + 3\hat{k}) = 0\hat{i} - \hat{j} - 2\hat{k}. \)

Now compute the cross product:
\( (\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 0 & -1 & -2 \end{vmatrix}. \)

Expand the determinant:
\( (\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}) = \hat{i} \begin{vmatrix} 3 & 4 \\ -1 & -2 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 4 \\ 0 & -2 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 3 \\ 0 & -1 \end{vmatrix}. \)

Calculate the 2x2 determinants:
\( \begin{vmatrix} 3 & 4 \\ -1 & -2 \end{vmatrix} = (3)(-2) - (4)(-1) = -6 + 4 = -2, \)
\( \begin{vmatrix} 2 & 4 \\ 0 & -2 \end{vmatrix} = (2)(-2) - (4)(0) = -4, \)
\( \begin{vmatrix} 2 & 3 \\ 0 & -1 \end{vmatrix} = (2)(-1) - (3)(0) = -2. \)

So the cross product is:
\( (\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}) = -2\hat{i} + 4\hat{j} - 2\hat{k}. \)

Now, to find the unit vector, we need to divide this vector by its magnitude.

The magnitude of the vector is:
\( ||\vec{v}|| = \sqrt{(-2)^2 + 4^2 + (-2)^2} = \sqrt{4 + 16 + 4} = \sqrt{24} = 2\sqrt{6}. \)

Thus, the unit vector is:
\( \hat{v} = \frac{-2\hat{i} + 4\hat{j} - 2\hat{k}}{2\sqrt{6}} = \frac{-1}{\sqrt{6}} \hat{i} + \frac{2}{\sqrt{6}} \hat{j} - \frac{1}{\sqrt{6}} \hat{k}. \)

This matches Option (4), so the correct answer is:
(4).