Find the vector equations of the line
` (x)/(1) = (y-1)/(2) = (z-2)/(3)`

Answer

To find the vector equations of the line given by the equation

\[ \frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3}, \]

we will follow these steps:

### Step 1: Identify the point and direction ratios
From the equation, we can identify the direction ratios and a point through which the line passes.

- The direction ratios are given by the coefficients of \(x\), \(y\), and \(z\) in the equation. Here, we have:
- \(L = 1\)
- \(M = 2\)
- \(N = 3\)

- To find a point on the line, we can set the parameter equal to 0. This gives us:
- \(x = 1 \cdot 0 = 0\)
- \(y - 1 = 2 \cdot 0 \Rightarrow y = 1\)
- \(z - 2 = 3 \cdot 0 \Rightarrow z = 2\)

Thus, the point \(P(0, 1, 2)\) lies on the line.

### Step 2: Write the position vector of the point
The position vector \(\vec{a}\) of the point \(P(0, 1, 2)\) can be expressed as:
\[ \vec{a} = 0\hat{i} + 1\hat{j} + 2\hat{k} = \hat{j} + 2\hat{k}. \]

### Step 3: Write the direction vector
The direction vector \(\vec{b}\) corresponding to the direction ratios is:
\[ \vec{b} = 1\hat{i} + 2\hat{j} + 3\hat{k}. \]

### Step 4: Write the vector equation of the line
The vector equation of the line can be expressed in the form:
\[ \vec{r} = \vec{a} + \lambda \vec{b}, \]
where \(\lambda\) is a parameter.

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

### Step 5: Simplify the equation
Distributing \(\lambda\):
\[ \vec{r} = \hat{j} + 2\hat{k} + \lambda\hat{i} + 2\lambda\hat{j} + 3\lambda\hat{k}. \]
Combining like terms:
\[ \vec{r} = \lambda\hat{i} + (1 + 2\lambda)\hat{j} + (2 + 3\lambda)\hat{k}. \]

### Final Answer
The vector equation of the line is:
\[ \vec{r} = \lambda\hat{i} + (1 + 2\lambda)\hat{j} + (2 + 3\lambda)\hat{k}. \]

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