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. \( -\frac{1}{\sqrt{6}} \hat{i} + \frac{2}{\sqrt{6}} \hat{j} - \frac{1}{\sqrt{6}} \hat{k} \)

Hint:

When finding a unit vector perpendicular to two vectors, first compute the cross product of the given vectors. After that, normalize the resulting vector by dividing each component by its magnitude. This process is critical when solving for the direction of vectors in 3D space. Also, be careful with determinant calculations to avoid errors in the cross product.

Answer 1

The Correct Option is D

Given vectors: \( \vec{a} = \hat{i} + \hat{j} + \hat{k} \) and \( \vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}. \) We require the unit vector perpendicular to \( \vec{a} + \vec{b} \) and \( \vec{a} - \vec{b}. \) A vector perpendicular to both is obtained via their cross product. We proceed by calculating \( (\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}). \)

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

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

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

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

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

To obtain the unit vector, divide this resulting vector by its magnitude.

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

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

This result corresponds to Option (4), making it the correct answer.
(4)