If $$\vec{A} = 2\hat{i} + 3\hat{j} - \hat{k} \text{ m}$$ and $$\vec{B} = \hat{i} + 2\hat{j} + 2\hat{k} \text{ m}$$. The magnitude of component of vector $$\vec{A}$$ along vector $$\vec{B}$$ will be ______ m.

Given $$\vec{A} = 2\hat{i} + 3\hat{j} - \hat{k}$$ and $$\vec{B} = \hat{i} + 2\hat{j} + 2\hat{k}$$, the component of $$\vec{A}$$ along $$\vec{B}$$ is given by $$\text{Component} = \frac{\vec{A} \cdot \vec{B}}{|\vec{B}|}$$.

First, the dot product $$\vec{A} \cdot \vec{B}$$ is calculated as $$\vec{A} \cdot \vec{B} = (2)(1) + (3)(2) + (-1)(2) = 2 + 6 - 2 = 6$$.

Next, the magnitude of $$\vec{B}$$ is determined from $$|\vec{B}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$$.

Therefore, substituting these values into the component formula gives $$\text{Component} = \frac{6}{3} = 2 \text{ m}$$, so the magnitude of the component of $$\vec{A}$$ along $$\vec{B}$$ is $$\textbf{2}$$ m.