The direction cosines of the vector `3hati-4hatj+5hatk` are

Options

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

Explanation

To find the direction cosines of the vector \( \vec{A} = 3\hat{i} - 4\hat{j} + 5\hat{k} \), we will follow these steps:

### Step 1: Calculate the magnitude of the vector
The magnitude of the vector \( \vec{A} \) is given by the formula:
\[ |\vec{A}| = \sqrt{a^2 + b^2 + c^2} \]
where \( a, b, c \) are the coefficients of \( \hat{i}, \hat{j}, \hat{k} \) respectively.

For our vector:
- \( a = 3 \)
- \( b = -4 \)
- \( c = 5 \)

Calculating the magnitude:
\[ |\vec{A}| = \sqrt{3^2 + (-4)^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2} \]

### Step 2: Find the unit vector
The unit vector \( \hat{A} \) in the direction of \( \vec{A} \) is given by:
\[ \hat{A} = \frac{\vec{A}}{|\vec{A}|} \]

Substituting the values:
\[ \hat{A} = \frac{3\hat{i} - 4\hat{j} + 5\hat{k}}{5\sqrt{2}} = \frac{3}{5\sqrt{2}}\hat{i} - \frac{4}{5\sqrt{2}}\hat{j} + \frac{5}{5\sqrt{2}}\hat{k} \]

This simplifies to:
\[ \hat{A} = \frac{3}{5\sqrt{2}}\hat{i} - \frac{4}{5\sqrt{2}}\hat{j} + \frac{1}{\sqrt{2}}\hat{k} \]

### Step 3: Identify the direction cosines
The direction cosines \( \cos \alpha, \cos \beta, \cos \gamma \) are the components of the unit vector:
- \( \cos \alpha = \frac{3}{5\sqrt{2}} \)
- \( \cos \beta = -\frac{4}{5\sqrt{2}} \)
- \( \cos \gamma = \frac{1}{\sqrt{2}} \)

### Final Result
The direction cosines of the vector \( 3\hat{i} - 4\hat{j} + 5\hat{k} \) are:
\[ \left(\frac{3}{5\sqrt{2}}, -\frac{4}{5\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \]