If `vecP=3hati+4hatj+12hatk` then find magnitude and the direction cosines of the `vecP`.

Explanation

To solve the problem of finding the magnitude and direction cosines of the vector \(\vec{P} = 3\hat{i} + 4\hat{j} + 12\hat{k}\), we will follow these steps:

### Step 1: Find the Magnitude of the Vector
The magnitude of a vector \(\vec{P} = a\hat{i} + b\hat{j} + c\hat{k}\) is given by the formula:
\[ |\vec{P}| = \sqrt{a^2 + b^2 + c^2} \]
For our vector, \(a = 3\), \(b = 4\), and \(c = 12\). Therefore, we calculate:
\[ |\vec{P}| = \sqrt{3^2 + 4^2 + 12^2} \]
Calculating each term:
\[ 3^2 = 9, \quad 4^2 = 16, \quad 12^2 = 144 \]
Adding these values together:
\[ |\vec{P}| = \sqrt{9 + 16 + 144} = \sqrt{169} \]
Finally, taking the square root:
\[ |\vec{P}| = 13 \text{ units} \]

### Step 2: Find the Direction Cosines
The direction cosines of a vector are given by the ratios of its components to its magnitude. The direction cosines \(l\), \(m\), and \(n\) are defined as:
\[ l = \frac{a}{|\vec{P}|}, \quad m = \frac{b}{|\vec{P}|}, \quad n = \frac{c}{|\vec{P}|} \]
Substituting the values we have:
\[ l = \frac{3}{13}, \quad m = \frac{4}{13}, \quad n = \frac{12}{13} \]
Thus, the direction cosines of the vector \(\vec{P}\) are:
\[ \left( \frac{3}{13}, \frac{4}{13}, \frac{12}{13} \right) \]

### Final Answer
- **Magnitude of \(\vec{P}\)**: \(13 \text{ units}\)
- **Direction Cosines**: \(\left( \frac{3}{13}, \frac{4}{13}, \frac{12}{13} \right)\)