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

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).

Answer 2

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).