Find the directional cosines of vector `(5hati+2hatj+6hatk)`. Also write the value of sum of squares of directional cosines of this vector.

Explanation

To find the directional cosines of the vector \( \vec{A} = 5\hat{i} + 2\hat{j} + 6\hat{k} \), we will follow these steps:

### Step 1: Calculate the Magnitude of the Vector
The magnitude \( |\vec{A}| \) of the vector \( \vec{A} \) is given by the formula:
\[ |\vec{A}| = \sqrt{a_x^2 + a_y^2 + a_z^2} \]
where \( a_x, a_y, a_z \) are the components of the vector along the \( x, y, z \) axes respectively.

For our vector:
- \( a_x = 5 \)
- \( a_y = 2 \)
- \( a_z = 6 \)

Thus, the magnitude is:
\[ |\vec{A}| = \sqrt{5^2 + 2^2 + 6^2} = \sqrt{25 + 4 + 36} = \sqrt{65} \]

### Step 2: Calculate the Directional Cosines
The directional cosines \( \cos \alpha, \cos \beta, \cos \gamma \) are defined as:
\[ \cos \alpha = \frac{a_x}{|\vec{A}|}, \quad \cos \beta = \frac{a_y}{|\vec{A}|}, \quad \cos \gamma = \frac{a_z}{|\vec{A}|} \]

Now we can calculate each directional cosine:

1. **For \( \cos \alpha \)**:
\[ \cos \alpha = \frac{5}{\sqrt{65}} \]

2. **For \( \cos \beta \)**:
\[ \cos \beta = \frac{2}{\sqrt{65}} \]

3. **For \( \cos \gamma \)**:
\[ \cos \gamma = \frac{6}{\sqrt{65}} \]

### Step 3: Sum of Squares of Directional Cosines
The sum of the squares of the directional cosines is given by:
\[ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma \]

Calculating this:
\[ \cos^2 \alpha = \left(\frac{5}{\sqrt{65}}\right)^2 = \frac{25}{65} \]
\[ \cos^2 \beta = \left(\frac{2}{\sqrt{65}}\right)^2 = \frac{4}{65} \]
\[ \cos^2 \gamma = \left(\frac{6}{\sqrt{65}}\right)^2 = \frac{36}{65} \]

Now, summing these:
\[ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = \frac{25}{65} + \frac{4}{65} + \frac{36}{65} = \frac{65}{65} = 1 \]

### Final Answers
- The directional cosines are:
- \( \cos \alpha = \frac{5}{\sqrt{65}} \)
- \( \cos \beta = \frac{2}{\sqrt{65}} \)
- \( \cos \gamma = \frac{6}{\sqrt{65}} \)

- The sum of squares of the directional cosines is \( 1 \).