Find the equation of the plane through the point (0, 1, 2) and parpendicular to the line lt brgt `vecr=(hati+hatj)+lambda (2hati-hatj +hatk).` (Vector form )

Answer

To find the equation of the plane that passes through the point (0, 1, 2) and is perpendicular to the given line, we can follow these steps:

### Step 1: Identify the point and the direction vector of the line
The point through which the plane passes is given as \( P(0, 1, 2) \). The line is represented in vector form as:
\[ \vec{r} = \hat{i} + \hat{j} + \lambda(2\hat{i} - \hat{j} + \hat{k}) \]
From this equation, we can identify the direction vector of the line, which is:
\[ \vec{d} = 2\hat{i} - \hat{j} + \hat{k} \]

### Step 2: Determine the normal vector of the plane
Since the plane is perpendicular to the line, the direction vector of the line \( \vec{d} \) will serve as the normal vector \( \vec{n} \) of the plane. Thus:
\[ \vec{n} = 2\hat{i} - \hat{j} + \hat{k} \]

### Step 3: Use the equation of the plane
The general equation of a plane in vector form can be expressed as:
\[ \vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n} \]
where \( \vec{a} \) is the position vector of the point through which the plane passes.

### Step 4: Write the position vector of the point
The position vector \( \vec{a} \) corresponding to the point \( (0, 1, 2) \) is:
\[ \vec{a} = 0\hat{i} + 1\hat{j} + 2\hat{k} = \hat{j} + 2\hat{k} \]

### Step 5: Calculate \( \vec{a} \cdot \vec{n} \)
Now, we need to calculate the dot product \( \vec{a} \cdot \vec{n} \):
\[ \vec{a} \cdot \vec{n} = (0\hat{i} + 1\hat{j} + 2\hat{k}) \cdot (2\hat{i} - \hat{j} + \hat{k}) \]
Calculating this gives:
\[ = 0 \cdot 2 + 1 \cdot (-1) + 2 \cdot 1 = 0 - 1 + 2 = 1 \]

### Step 6: Write the equation of the plane
Now substituting back into the plane equation:
\[ \vec{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) = 1 \]
This can be expressed in component form as:
\[ (x, y, z) \cdot (2, -1, 1) = 1 \]
Expanding this gives:
\[ 2x - y + z = 1 \]

### Final Answer
The equation of the plane is:
\[ 2x - y + z = 1 \]

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