The co-ordiantes of the foot of perpendicular from origin to a plane are `(3,-2,1)`. Find the equation of the plane.

Answer

To find the equation of the plane given that the coordinates of the foot of the perpendicular from the origin to the plane are \( (3, -2, 1) \), we can follow these steps:

### Step 1: Understand the Problem
We know that the coordinates of the foot of the perpendicular from the origin \( O(0, 0, 0) \) to the plane is the point \( P(3, -2, 1) \). The line segment \( OP \) is perpendicular to the plane.

### Step 2: Identify the Normal Vector
Since \( OP \) is perpendicular to the plane, the vector \( \vec{n} \) (normal vector of the plane) is the same as the vector \( \vec{OP} \). We can calculate \( \vec{OP} \) as follows:
\[ \vec{OP} = P - O = (3 - 0, -2 - 0, 1 - 0) = (3, -2, 1) \]
Thus, the normal vector \( \vec{n} \) is \( \vec{n} = 3\hat{i} - 2\hat{j} + 1\hat{k} \).

### Step 3: Use the Standard Equation of the Plane
The standard form of the equation of a plane is given by:
\[ \vec{r} - \vec{a} \cdot \vec{n} = 0 \]
where \( \vec{a} \) is a point on the plane (which is \( P(3, -2, 1) \)) and \( \vec{n} \) is the normal vector.

### Step 4: Substitute Values into the Equation
Let \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \) and \( \vec{a} = 3\hat{i} - 2\hat{j} + 1\hat{k} \). The equation becomes:
\[ (x\hat{i} + y\hat{j} + z\hat{k}) - (3\hat{i} - 2\hat{j} + 1\hat{k}) \cdot (3\hat{i} - 2\hat{j} + 1\hat{k}) = 0 \]

### Step 5: Calculate the Dot Product
Calculating the dot product:
\[ (3\hat{i} - 2\hat{j} + 1\hat{k}) \cdot (3\hat{i} - 2\hat{j} + 1\hat{k}) = 3^2 + (-2)^2 + 1^2 = 9 + 4 + 1 = 14 \]

### Step 6: Write the Final Equation
Now substituting back into the equation:
\[ 3x - 2y + z = 14 \]
This is the required equation of the plane.

### Final Answer
The equation of the plane is:
\[ 3x - 2y + z = 14 \]