The foot of the perpendicular drawn from the origin to the plane `x+y+z=3` is

Options

A. `(0,1,2)`
B. `(0,0,3)`
C. `(1,1,1)`
D. `(-1,1,3)`

Answer

To find the foot of the perpendicular drawn from the origin to the plane given by the equation \( x + y + z = 3 \), we can follow these steps:

### Step 1: Identify the equation of the plane
The equation of the plane is given as:
\[ x + y + z = 3 \]

### Step 2: Identify the coordinates of the point from which the perpendicular is drawn
The origin has the coordinates:
\[ (0, 0, 0) \]

### Step 3: Use the formula for the distance from a point to a plane
The distance \( d \) from a point \( (x_0, y_0, z_0) \) to the plane \( ax + by + cz + d = 0 \) is given by:
\[ d = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}} \]
For our plane \( x + y + z - 3 = 0 \), we have:
- \( a = 1 \)
- \( b = 1 \)
- \( c = 1 \)
- \( d = -3 \)

### Step 4: Substitute the coordinates of the origin into the distance formula
Substituting \( (0, 0, 0) \) into the formula:
\[ d = \frac{|1(0) + 1(0) + 1(0) - 3|}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{|-3|}{\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3} \]

### Step 5: Find the direction ratios of the normal to the plane
The normal vector to the plane \( x + y + z = 3 \) is given by the coefficients of \( x, y, z \), which are:
\[ (1, 1, 1) \]

### Step 6: Calculate the coordinates of the foot of the perpendicular
To find the coordinates of the foot of the perpendicular, we can move from the origin in the direction of the normal vector. The coordinates of the foot of the perpendicular \( Q \) can be calculated as:
\[ Q = (0, 0, 0) + t(1, 1, 1) \]
where \( t \) is the distance we found earlier, \( t = \sqrt{3} \).

Thus:
\[ Q = (t, t, t) = (\sqrt{3}, \sqrt{3}, \sqrt{3}) \]

### Step 7: Find the actual coordinates that satisfy the plane equation
However, we need to ensure that these coordinates satisfy the plane equation \( x + y + z = 3 \). Substituting \( Q = (t, t, t) \):
\[ \sqrt{3} + \sqrt{3} + \sqrt{3} = 3\sqrt{3} \]
This does not satisfy the plane equation.

Instead, we can find the foot of the perpendicular by scaling down the normal vector to meet the plane. The coordinates of the foot of the perpendicular will be:
\[ Q = \left(1, 1, 1\right) \]
since \( 1 + 1 + 1 = 3 \).

### Final Answer
The foot of the perpendicular from the origin to the plane \( x + y + z = 3 \) is:
\[ (1, 1, 1) \]