Find the vector equation of line passing through `A(3, 4, -7) and B(1, -1, 6)`. Also, find its cartesian equations.

Answer

To find the vector equation of the line passing through the points \( A(3, 4, -7) \) and \( B(1, -1, 6) \), we will follow these steps:

### Step 1: Identify the position vectors of points A and B
The position vector of point \( A \) is given by:
\[ \vec{A} = 3\hat{i} + 4\hat{j} - 7\hat{k} \]
The position vector of point \( B \) is given by:
\[ \vec{B} = 1\hat{i} - 1\hat{j} + 6\hat{k} \]

### Step 2: Calculate the direction vector \( \vec{B} - \vec{A} \)
To find the direction vector of the line, we subtract the position vector of \( A \) from the position vector of \( B \):
\[ \vec{B} - \vec{A} = (1 - 3)\hat{i} + (-1 - 4)\hat{j} + (6 - (-7))\hat{k} \]
Calculating this gives:
\[ \vec{B} - \vec{A} = -2\hat{i} - 5\hat{j} + 13\hat{k} \]

### Step 3: Write the vector equation of the line
The vector equation of the line can be expressed as:
\[ \vec{r} = \vec{A} + \lambda(\vec{B} - \vec{A}) \]
Substituting the values we have:
\[ \vec{r} = (3\hat{i} + 4\hat{j} - 7\hat{k}) + \lambda(-2\hat{i} - 5\hat{j} + 13\hat{k}) \]
This can be simplified to:
\[ \vec{r} = (3 - 2\lambda)\hat{i} + (4 - 5\lambda)\hat{j} + (-7 + 13\lambda)\hat{k} \]

### Step 4: Find the Cartesian equations of the line
The general form for the Cartesian equations of a line is:
\[ \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} \]
Where \( (x_1, y_1, z_1) \) is a point on the line and \( (a, b, c) \) are the direction ratios.

From our calculations:
- Point \( A(3, 4, -7) \) gives us \( x_1 = 3, y_1 = 4, z_1 = -7 \)
- The direction ratios are \( a = -2, b = -5, c = 13 \)

Thus, the Cartesian equations can be written as:
\[ \frac{x - 3}{-2} = \frac{y - 4}{-5} = \frac{z + 7}{13} \]

### Final Result
The vector equation of the line is:
\[ \vec{r} = (3 - 2\lambda)\hat{i} + (4 - 5\lambda)\hat{j} + (-7 + 13\lambda)\hat{k} \]
And the Cartesian equations are:
\[ \frac{x - 3}{-2} = \frac{y - 4}{-5} = \frac{z + 7}{13} \]