If `vec(A)= 2hat(i)+4hat(j)-5hat(k)` then the direction of cosins of the vector `vec(A)` are

Options

A. `2/(sqrt(45)),4/(sqrt(45)) and (-5)/(sqrt(45))`
B. `1/(sqrt(45)),2/(sqrt(45)) and (3)/(sqrt(45))`
C. `4/(sqrt(45)),0 and (4)/(sqrt(45))`
D. `3/(sqrt(45)),2/(sqrt(45)) and (5)/(sqrt(45))`

Explanation

To find the direction cosines of the vector \(\vec{A} = 2\hat{i} + 4\hat{j} - 5\hat{k}\), we will follow these steps:

### Step 1: Identify the components of the vector
The vector \(\vec{A}\) has the following components:
- \(A_x = 2\) (the coefficient of \(\hat{i}\))
- \(A_y = 4\) (the coefficient of \(\hat{j}\))
- \(A_z = -5\) (the coefficient of \(\hat{k}\))

### Step 2: Calculate the magnitude of the vector
The magnitude (or modulus) of the vector \(\vec{A}\) is calculated using the formula:
\[ |\vec{A}| = \sqrt{A_x^2 + A_y^2 + A_z^2} \]
Substituting the values:
\[ |\vec{A}| = \sqrt{2^2 + 4^2 + (-5)^2} = \sqrt{4 + 16 + 25} = \sqrt{45} \]

### Step 3: Calculate the direction cosines
The direction cosines are given by:
\[ \cos \alpha = \frac{A_x}{|\vec{A}|}, \quad \cos \beta = \frac{A_y}{|\vec{A}|}, \quad \cos \gamma = \frac{A_z}{|\vec{A}|} \]
Substituting the values we found:
- For \(\cos \alpha\):
\[ \cos \alpha = \frac{2}{\sqrt{45}} \]
- For \(\cos \beta\):
\[ \cos \beta = \frac{4}{\sqrt{45}} \]
- For \(\cos \gamma\):
\[ \cos \gamma = \frac{-5}{\sqrt{45}} \]

### Step 4: Final results
Thus, the direction cosines of the vector \(\vec{A}\) are:
\[ \cos \alpha = \frac{2}{\sqrt{45}}, \quad \cos \beta = \frac{4}{\sqrt{45}}, \quad \cos \gamma = \frac{-5}{\sqrt{45}} \]

### Summary
The direction cosines of the vector \(\vec{A} = 2\hat{i} + 4\hat{j} - 5\hat{k}\) are:
- \(\cos \alpha = \frac{2}{\sqrt{45}}\)
- \(\cos \beta = \frac{4}{\sqrt{45}}\)
- \(\cos \gamma = \frac{-5}{\sqrt{45}}\)