Write line `vecr =(hati -hatj)+lambda(2hatj-hatk)` in to cartesian form

Answer

To convert the given vector equation of the line \( \vec{r} = (\hat{i} - \hat{j}) + \lambda(2\hat{j} - \hat{k}) \) into Cartesian form, we can follow these steps:

### Step 1: Identify the position vector and direction vector
The vector equation is in the form:
\[ \vec{r} = \vec{a} + \lambda \vec{b} \]
where \( \vec{a} \) is the position vector of a point on the line, and \( \vec{b} \) is the direction vector of the line.

From the given equation:
- The position vector \( \vec{a} = \hat{i} - \hat{j} \) can be expressed as:
\[ \vec{a} = 1\hat{i} - 1\hat{j} + 0\hat{k} \]
This gives us the point \( (x_1, y_1, z_1) = (1, -1, 0) \).

- The direction vector \( \vec{b} = 2\hat{j} - \hat{k} \) can be expressed as:
\[ \vec{b} = 0\hat{i} + 2\hat{j} - 1\hat{k} \]
This gives us the direction ratios \( (a_1, b_1, c_1) = (0, 2, -1) \).

### Step 2: Write the parametric equations
Using the position vector and direction ratios, we can write the parametric equations of the line:
\[ x = x_1 + \lambda a_1 = 1 + \lambda \cdot 0 = 1 \]
\[ y = y_1 + \lambda b_1 = -1 + \lambda \cdot 2 = -1 + 2\lambda \]
\[ z = z_1 + \lambda c_1 = 0 + \lambda \cdot (-1) = -\lambda \]

### Step 3: Eliminate the parameter \( \lambda \)
From the equation for \( x \):
\[ x = 1 \implies \lambda \text{ is not involved in } x \]
From the equation for \( y \):
\[ y = -1 + 2\lambda \implies 2\lambda = y + 1 \implies \lambda = \frac{y + 1}{2} \]
From the equation for \( z \):
\[ z = -\lambda \implies \lambda = -z \]

### Step 4: Set the expressions for \( \lambda \) equal to each other
Equating the two expressions for \( \lambda \):
\[ \frac{y + 1}{2} = -z \]
Multiplying through by 2:
\[ y + 1 = -2z \]
Rearranging gives:
\[ y + 2z + 1 = 0 \]

### Final Cartesian Form
Thus, the Cartesian form of the line is:
\[ y + 2z + 1 = 0 \]