If `A(6,3,2), B(5,1,4), C(3,-4,7), D(0,2,5)` be from points, the projection of the sement CD on the line AB is

Options

A. `-13/3`
B. `-13/7`
C. `-3/13`
D. `-7/13`

Answer

To find the projection of the segment \( CD \) on the line \( AB \), we will follow these steps:

### Step 1: Find the vector \( \overrightarrow{CD} \)
The vector \( \overrightarrow{CD} \) can be calculated using the coordinates of points \( C(3, -4, 7) \) and \( D(0, 2, 5) \).

\[ \overrightarrow{CD} = D - C = (0 - 3, 2 - (-4), 5 - 7) = (-3, 6, -2) \]

### Step 2: Find the vector \( \overrightarrow{AB} \)
Next, we calculate the vector \( \overrightarrow{AB} \) using the coordinates of points \( A(6, 3, 2) \) and \( B(5, 1, 4) \).

\[ \overrightarrow{AB} = B - A = (5 - 6, 1 - 3, 4 - 2) = (-1, -2, 2) \]

### Step 3: Find the magnitude of vector \( \overrightarrow{AB} \)
To find the magnitude of vector \( \overrightarrow{AB} \):

\[ |\overrightarrow{AB}| = \sqrt{(-1)^2 + (-2)^2 + (2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]

### Step 4: Find the unit vector along \( AB \)
The unit vector \( \hat{n} \) along the direction of \( \overrightarrow{AB} \) is given by:

\[ \hat{n} = \frac{\overrightarrow{AB}}{|\overrightarrow{AB}|} = \frac{(-1, -2, 2)}{3} = \left(-\frac{1}{3}, -\frac{2}{3}, \frac{2}{3}\right) \]

### Step 5: Calculate the projection of \( \overrightarrow{CD} \) on \( \overrightarrow{AB} \)
The projection of vector \( \overrightarrow{CD} \) on vector \( \overrightarrow{AB} \) is given by the formula:

\[ \text{Projection} = \frac{\overrightarrow{CD} \cdot \overrightarrow{AB}}{|\overrightarrow{AB}|} \]

First, we calculate the dot product \( \overrightarrow{CD} \cdot \overrightarrow{AB} \):

\[ \overrightarrow{CD} \cdot \overrightarrow{AB} = (-3)(-1) + (6)(-2) + (-2)(2) = 3 - 12 - 4 = -13 \]

Now, substituting into the projection formula:

\[ \text{Projection} = \frac{-13}{3} \]

### Final Answer
The projection of segment \( CD \) on line \( AB \) is:

\[ \text{Projection} = -\frac{13}{3} \]

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