Let `Aequiv(1, 2, 3) B equiv (3, 4, 5) ` then

Options

A. Direction ratios of AB line cannot be (2, 2, 2)
B. Direction ratios of `vec(AB)` are (-1, 1, -1)
C. Direction ratios of `vec(AB)` are (1, 1, 1)
D. Direction cosine of AB line may be
`(1/sqrt(3), -1/sqrt(3), - 1/sqrt(3))`

Answer

To solve the problem, we need to find the direction ratios and direction cosines of the line segment joining the points A(1, 2, 3) and B(3, 4, 5).

### Step-by-Step Solution:

1. **Identify the Points**:
We have two points A and B given as:
- A = (1, 2, 3)
- B = (3, 4, 5)

2. **Calculate the Direction Ratios**:
The direction ratios of the line segment AB can be calculated using the formula:
\[ \text{Direction Ratios} = (x_2 - x_1, y_2 - y_1, z_2 - z_1) \]
Substituting the coordinates of points A and B:
\[ \text{Direction Ratios} = (3 - 1, 4 - 2, 5 - 3) = (2, 2, 2) \]

3. **Simplify the Direction Ratios**:
The direction ratios (2, 2, 2) can be simplified by taking out the common factor:
\[ (2, 2, 2) = 2(1, 1, 1) \]
Thus, the direction ratios can also be expressed as (1, 1, 1).

4. **Calculate the Magnitude of AB**:
The magnitude of the vector AB is calculated as follows:
\[ |AB| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} = \sqrt{(3 - 1)^2 + (4 - 2)^2 + (5 - 3)^2} \]
\[ = \sqrt{2^2 + 2^2 + 2^2} = \sqrt{4 + 4 + 4} = \sqrt{12} = 2\sqrt{3} \]

5. **Calculate the Direction Cosines**:
The direction cosines are calculated by dividing each direction ratio by the magnitude of AB:
\[ \text{Direction Cosines} = \left(\frac{2}{2\sqrt{3}}, \frac{2}{2\sqrt{3}}, \frac{2}{2\sqrt{3}}\right) = \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) \]

6. **Conclusion**:
The direction ratios of line segment AB are (2, 2, 2) or (1, 1, 1) when simplified. The direction cosines are \(\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)\).