If the two directional cosines of a vectors are `1/sqrt(2)` and `1/sqrt(3)` then the value of third directional cosine is

Options

A. `1/sqrt(6)`
B. `1/sqrt(5)`
C. `1/sqrt(7)`
D. `1/sqrt(10)`

Answer

To find the value of the third directional cosine given the first two, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Direction Cosines**:
Direction cosines are the cosines of the angles that a vector makes with the coordinate axes. If the directional cosines are denoted as \( L \), \( M \), and \( N \), then they satisfy the equation:
\[ L^2 + M^2 + N^2 = 1 \]

2. **Given Values**:
We are given:
\[ L = \frac{1}{\sqrt{2}}, \quad M = \frac{1}{\sqrt{3}} \]
We need to find \( N \).

3. **Substituting the Values**:
Substitute \( L \) and \( M \) into the equation:
\[ \left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{3}}\right)^2 + N^2 = 1 \]

4. **Calculating \( L^2 \) and \( M^2 \)**:
Calculate \( L^2 \) and \( M^2 \):
\[ L^2 = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}, \quad M^2 = \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3} \]

5. **Adding \( L^2 \) and \( M^2 \)**:
Now add \( L^2 \) and \( M^2 \):
\[ \frac{1}{2} + \frac{1}{3} \]
To add these fractions, find a common denominator, which is 6:
\[ \frac{1}{2} = \frac{3}{6}, \quad \frac{1}{3} = \frac{2}{6} \]
Therefore,
\[ \frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6} \]

6. **Substituting Back into the Equation**:
Substitute back into the equation:
\[ \frac{5}{6} + N^2 = 1 \]

7. **Solving for \( N^2 \)**:
Rearranging gives:
\[ N^2 = 1 - \frac{5}{6} = \frac{1}{6} \]

8. **Finding \( N \)**:
Taking the square root gives:
\[ N = \sqrt{\frac{1}{6}} = \frac{1}{\sqrt{6}} \]

### Final Answer:
The value of the third directional cosine \( N \) is:
\[ N = \frac{1}{\sqrt{6}} \]