Find the direction cosines of the vector `vec(F) = 4hat(i) + 3hat(j) + 2 hat(k)`

Explanation

To find the direction cosines of the vector \(\vec{F} = 4\hat{i} + 3\hat{j} + 2\hat{k}\), we can follow these steps:

### Step 1: Identify the components of the vector
The vector \(\vec{F}\) can be expressed in terms of its components:
- \(F_x = 4\) (coefficient of \(\hat{i}\))
- \(F_y = 3\) (coefficient of \(\hat{j}\))
- \(F_z = 2\) (coefficient of \(\hat{k}\))

### Step 2: Calculate the magnitude of the vector
The magnitude (or modulus) of the vector \(\vec{F}\) is given by the formula:
\[ |\vec{F}| = \sqrt{F_x^2 + F_y^2 + F_z^2} \]
Substituting the values:
\[ |\vec{F}| = \sqrt{4^2 + 3^2 + 2^2} = \sqrt{16 + 9 + 4} = \sqrt{29} \]

### Step 3: Calculate the direction cosines
The direction cosines are defined as:
- \(\cos \alpha = \frac{F_x}{|\vec{F}|}\)
- \(\cos \beta = \frac{F_y}{|\vec{F}|}\)
- \(\cos \gamma = \frac{F_z}{|\vec{F}|}\)

Now, substituting the values we calculated:
1. For \(\cos \alpha\):
\[ \cos \alpha = \frac{F_x}{|\vec{F}|} = \frac{4}{\sqrt{29}} \]
2. For \(\cos \beta\):
\[ \cos \beta = \frac{F_y}{|\vec{F}|} = \frac{3}{\sqrt{29}} \]
3. For \(\cos \gamma\):
\[ \cos \gamma = \frac{F_z}{|\vec{F}|} = \frac{2}{\sqrt{29}} \]

### Final Result
The direction cosines of the vector \(\vec{F}\) are:
- \(\cos \alpha = \frac{4}{\sqrt{29}}\)
- \(\cos \beta = \frac{3}{\sqrt{29}}\)
- \(\cos \gamma = \frac{2}{\sqrt{29}}\)