The projection vector of the vector $2\hat i+ 3\hat j+\hat k$ on $2\hat i+\hat j-2\hat k$ is

$\frac{5}{9}(2\hat i+\hat j-2\hat k)$

The correct answer is Option (2) → $\frac{5}{9}(2\hat i+\hat j-2\hat k)$

Given vectors:

$\vec{a} = 2\hat{i} + 3\hat{j} + \hat{k}$

$\vec{b} = 2\hat{i} + \hat{j} - 2\hat{k}$

Formula for projection vector of $\vec{a}$ on $\vec{b}$:

$\text{Proj}_{\vec{b}} \vec{a} = \left( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \right) \vec{b}$

Step 1: Dot product

$\vec{a} \cdot \vec{b} = (2)(2) + (3)(1) + (1)(-2) = 4 + 3 - 2 = 5$

Step 2: Magnitude squared of $\vec{b}$

$|\vec{b}|^2 = 2^2 + 1^2 + (-2)^2 = 4 + 1 + 4 = 9$

Step 3: Projection vector

$\text{Proj}_{\vec{b}} \vec{a} = \left( \frac{5}{9} \right)(2\hat{i} + \hat{j} - 2\hat{k})$

$= \frac{10}{9}\hat{i} + \frac{5}{9}\hat{j} - \frac{10}{9}\hat{k}$