Q.Find a unit vector perpendicular to each of the vector $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$, where $\vec{a}=3\hat{i}+2\hat{j}+2\hat{k}$ and $\vec{b}=\hat{i}+2\hat{j}-2\hat{k}$.

Concept and Intuition

When you're asked to find a vector that is perpendicular to two other vectors simultaneously, the most direct and fundamental tool in vector algebra is the cross product.

Imagine two non-parallel vectors originating from the same point. They define a unique plane in space. The cross product of these two vectors yields a new vector that is perpendicular to this entire plane. This means the resulting vector is perpendicular to both of the original vectors.

Once we have a vector that is perpendicular to both, the problem asks for a unit vector. A unit vector is simply a vector with a magnitude of 1, pointing in the same direction as the original vector. To convert any non-zero vector $\vec{V}$ into a unit vector $\hat{V}$, we divide it by its own magnitude: $\hat{V} = \frac{\vec{V}}{|\vec{V}|}$.

Step-by-step Solution

1. Identify the given vectors.

   Let the two given vectors be $\vec{A}$ and $\vec{B}$.

   $\vec{A} = \hat{i} + \hat{k}$

   $\vec{B} = \hat{i} - \hat{k}$

   We can write these in component form as:

   $\vec{A} = 1\hat{i} + 0\hat{j} + 1\hat{k}$

   $\vec{B} = 1\hat{i} + 0\hat{j} - 1\hat{k}$

2. Calculate the cross product $\vec{A} \times \vec{B}$.

   This will give us a vector perpendicular to both $\vec{A}$ and $\vec{B}$.

Expand the determinant:
$= \hat{i}((0)(-1) - (1)(0)) - \hat{j}((1)(-1) - (1)(1)) + \hat{k}((1)(0) - (0)(1))$
$= \hat{i}(0 - 0) - \hat{j}(-1 - 1) + \hat{k}(0 - 0)$
$= 0\hat{i} - \hat{j}(-2) + 0\hat{k}$
$= 2\hat{j}$

Let's call this resulting vector $\vec{P} = 2\hat{j}$. This vector $\vec{P}$ is perpendicular to both $\hat{i} + \hat{k}$ and $\hat{i} - \hat{k}$.

> [!TIP]
> You can quickly verify perpendicularity by checking the dot product. If $\vec{P} \cdot \vec{A} = 0$ and $\vec{P} \cdot \vec{B} = 0$, then $\vec{P}$ is indeed perpendicular to both.
> $\vec{P} \cdot \vec{A} = (0\hat{i} + 2\hat{j} + 0\hat{k}) \cdot (1\hat{i} + 0\hat{j} + 1\hat{k}) = (0)(1) + (2)(0) + (0)(1) = 0$.
> $\vec{P} \cdot \vec{B} = (0\hat{i} + 2\hat{j} + 0\hat{k}) \cdot (1\hat{i} + 0\hat{j} - 1\hat{k}) = (0)(1) + (2)(0) + (0)(-1) = 0$.
> Both dot products are zero, confirming $\vec{P}$ is perpendicular to both $\vec{A}$ and $\vec{B}$.

3. Normalize the vector $\vec{P}$ to find the unit vector.

The vector we found is $\vec{P} = 2\hat{j}$. To make it a unit vector, we need to divide it by its magnitude.

First, calculate the magnitude of $\vec{P}$:

$|\vec{P}| = |2\hat{j}| = \sqrt{0^2 + 2^2 + 0^2} = \sqrt{4} = 2$.

Now, divide $\vec{P}$ by its magnitude:
$\hat{P} = \frac{\vec{P}}{|\vec{P}|} = \frac{2\hat{j}}{2} = \hat{j}$.

This is the unit vector perpendicular to both $\hat{i} + \hat{k}$ and $\hat{i} - \hat{k}$.

> [!NOTE]
> The cross product $\vec{B} \times \vec{A}$ would yield $-2\hat{j}$, which is also perpendicular to both vectors but points in the opposite direction. Normalizing it would give $-\hat{j}$. Both $\hat{j}$ and $-\hat{j}$ are valid unit vectors perpendicular to the given plane. Since $\hat{j}$ is an option, we select it.

The correct option is (B).