The direction cosines of the line passing through `P(2,3,-1)` and the origin are

Options

A. `2/(sqrt(14)),3/(sqrt(14)),1/(sqrt(14))`
B. `2/(sqrt(14)),-3/(sqrt(14)),1/(sqrt(14))`
C. `-2/(sqrt(14)),-3/(sqrt(14)),1/(sqrt(14))`
D. `-2/(sqrt(14)),-3/(sqrt(14)),-1/(sqrt(14))`

Explanation

To find the direction cosines of the line passing through the point \( P(2, 3, -1) \) and the origin \( O(0, 0, 0) \), we can follow these steps:

### Step 1: Determine the position vector of point P and the origin O
The position vector of point \( P \) is given by:
\[ \vec{OP} = 2\hat{i} + 3\hat{j} - 1\hat{k} \]
The position vector of the origin \( O \) is:
\[ \vec{O} = 0\hat{i} + 0\hat{j} + 0\hat{k} \]

### Step 2: Calculate the vector \( \vec{OP} \)
To find the vector \( \vec{OP} \), we subtract the position vector of \( O \) from that of \( P \):
\[ \vec{OP} = \vec{P} - \vec{O} = (2\hat{i} + 3\hat{j} - 1\hat{k}) - (0\hat{i} + 0\hat{j} + 0\hat{k}) = 2\hat{i} + 3\hat{j} - 1\hat{k} \]

### Step 3: Calculate the magnitude of vector \( \vec{OP} \)
The magnitude of vector \( \vec{OP} \) is calculated using the formula:
\[ |\vec{OP}| = \sqrt{x^2 + y^2 + z^2} \]
where \( x = 2 \), \( y = 3 \), and \( z = -1 \):
\[ |\vec{OP}| = \sqrt{2^2 + 3^2 + (-1)^2} = \sqrt{4 + 9 + 1} = \sqrt{14} \]

### Step 4: Find the direction cosines
The direction cosines \( l, m, n \) are given by:
\[ l = \frac{x}{|\vec{OP}|}, \quad m = \frac{y}{|\vec{OP}|}, \quad n = \frac{z}{|\vec{OP}|} \]
Substituting the values we have:
\[ l = \frac{2}{\sqrt{14}}, \quad m = \frac{3}{\sqrt{14}}, \quad n = \frac{-1}{\sqrt{14}} \]

### Final Answer
Thus, the direction cosines of the line passing through \( P(2, 3, -1) \) and the origin are:
\[ \left( \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}, \frac{-1}{\sqrt{14}} \right) \]

---