If `A equiv (2, 3, 4), B equiv (3, 4, 5).` The direction cosine of a line are `(1/sqrt(3), 1/sqrt(3), 1/sqrt(3))` Now integral value of the
projection of AB on the given line is _______.

Answer

To find the integral value of the projection of the vector \( \vec{AB} \) on the given line with direction cosines, we can follow these steps:

### Step 1: Find the vector \( \vec{AB} \)

Given points:
- \( A(2, 3, 4) \)
- \( B(3, 4, 5) \)

The vector \( \vec{AB} \) can be calculated as:
\[ \vec{AB} = B - A = (3 - 2, 4 - 3, 5 - 4) = (1, 1, 1) \]

**Hint:** To find the vector between two points, subtract the coordinates of the initial point from the coordinates of the terminal point.

### Step 2: Calculate the magnitude of \( \vec{AB} \)

The magnitude of \( \vec{AB} \) is given by:
\[ |\vec{AB}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \]

**Hint:** The magnitude of a vector \( (x, y, z) \) is calculated using the formula \( \sqrt{x^2 + y^2 + z^2} \).

### Step 3: Identify the direction cosines of the line

The direction cosines of the line are given as:
\[ l = \frac{1}{\sqrt{3}}, \quad m = \frac{1}{\sqrt{3}}, \quad n = \frac{1}{\sqrt{3}} \]

This can be represented as a vector \( \vec{p} = \left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right) \).

**Hint:** Direction cosines represent the cosines of the angles that a line makes with the coordinate axes.

### Step 4: Calculate the dot product \( \vec{AB} \cdot \vec{p} \)

The dot product is calculated as:
\[ \vec{AB} \cdot \vec{p} = (1, 1, 1) \cdot \left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right) = 1 \cdot \frac{1}{\sqrt{3}} + 1 \cdot \frac{1}{\sqrt{3}} + 1 \cdot \frac{1}{\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3} \]

**Hint:** The dot product of two vectors \( (a, b, c) \) and \( (x, y, z) \) is calculated as \( ax + by + cz \).

### Step 5: Calculate the projection of \( \vec{AB} \) on the line

The projection of \( \vec{AB} \) on the line is given by:
\[ \text{Projection} = \frac{\vec{AB} \cdot \vec{p}}{|\vec{p}|} \]
Since \( |\vec{p}| = 1 \) (as it is a direction cosine vector), we have:
\[ \text{Projection} = \vec{AB} \cdot \vec{p} = \sqrt{3} \]

**Hint:** The projection of a vector on another vector can be simplified if the second vector is a unit vector (magnitude 1).

### Step 6: Find the integral value

The integral value of the projection is:
\[ \text{Integral value} = \lfloor \sqrt{3} \rfloor = 1 \]

Thus, the final answer is:
\[ \text{The integral value of the projection of } \vec{AB} \text{ on the given line is } 1. \]