Find the vector equation of the line which is parallel to the vector `3hati-2hatj+6hatk` and which passes through the point `(1,-2,3)`.

Answer

To find the vector equation of the line that is parallel to the vector \( \mathbf{b} = 3\hat{i} - 2\hat{j} + 6\hat{k} \) and passes through the point \( (1, -2, 3) \), we can follow these steps:

### Step 1: Identify the point and the direction vector
The point through which the line passes is given as \( (1, -2, 3) \). We can denote this point as a position vector:
\[ \mathbf{a} = 1\hat{i} - 2\hat{j} + 3\hat{k} \]
The direction vector of the line, which is parallel to the given vector, is:
\[ \mathbf{b} = 3\hat{i} - 2\hat{j} + 6\hat{k} \]

### Step 2: Write the vector equation of the line
The vector equation of a line can be expressed in the form:
\[ \mathbf{r} = \mathbf{a} + \lambda \mathbf{b} \]
where \( \lambda \) is a scalar parameter.

Substituting the values of \( \mathbf{a} \) and \( \mathbf{b} \):
\[ \mathbf{r} = (1\hat{i} - 2\hat{j} + 3\hat{k}) + \lambda (3\hat{i} - 2\hat{j} + 6\hat{k}) \]

### Step 3: Simplify the equation
Now, we can simplify the equation:
\[ \mathbf{r} = 1\hat{i} - 2\hat{j} + 3\hat{k} + \lambda (3\hat{i} - 2\hat{j} + 6\hat{k}) \]
Distributing \( \lambda \):
\[ \mathbf{r} = 1\hat{i} - 2\hat{j} + 3\hat{k} + 3\lambda\hat{i} - 2\lambda\hat{j} + 6\lambda\hat{k} \]
Combining like terms:
\[ \mathbf{r} = (1 + 3\lambda)\hat{i} + (-2 - 2\lambda)\hat{j} + (3 + 6\lambda)\hat{k} \]

### Final Vector Equation
Thus, the vector equation of the line is:
\[ \mathbf{r} = (1 + 3\lambda)\hat{i} + (-2 - 2\lambda)\hat{j} + (3 + 6\lambda)\hat{k} \]