Consider the plane `(x,y,z)= (0,1,1) + lamda(1,-1,1)+mu(2,-1,0)` The distance of this plane from the origin is

Options

A. a) `(1)/(3)`
B. b)`(sqrt(3))/(2)`
C. c) `sqrt((3)/(2))`
D. d) `(2)/(sqrt(3))`

Answer

To find the distance of the given plane from the origin, we will follow these steps:

### Step 1: Rewrite the equation of the plane
The equation of the plane is given as:
\[ (x, y, z) = (0, 1, 1) + \lambda(1, -1, 1) + \mu(2, -1, 0) \]
This can be rewritten in parametric form:
\[ x = 0 + \lambda \cdot 1 + \mu \cdot 2 \]
\[ y = 1 + \lambda \cdot (-1) + \mu \cdot (-1) \]
\[ z = 1 + \lambda \cdot 1 + \mu \cdot 0 \]

### Step 2: Express the equations in terms of λ and μ
From the above equations, we can express:
\[ x = \lambda + 2\mu \]
\[ y = 1 - \lambda - \mu \]
\[ z = 1 + \lambda \]

### Step 3: Eliminate λ and μ
We can rearrange the equations to eliminate λ and μ. From the third equation:
\[ \lambda = z - 1 \]
Substituting this into the first equation:
\[ x = (z - 1) + 2\mu \implies 2\mu = x - z + 1 \implies \mu = \frac{x - z + 1}{2} \]
Now substituting λ and μ into the second equation:
\[ y = 1 - (z - 1) - \frac{x - z + 1}{2} \]
Simplifying this gives:
\[ y = 1 - z + 1 - \frac{x - z + 1}{2} \]
\[ y = 2 - z - \frac{x}{2} + \frac{z}{2} - \frac{1}{2} \]
\[ y = \frac{-x + 2z + 3}{2} \]
Multiplying through by 2 gives:
\[ 2y = -x + 2z + 3 \implies x + 2y - 2z + 3 = 0 \]

### Step 4: Identify the normal vector and calculate the distance
The equation of the plane is now:
\[ x + 2y - 2z + 3 = 0 \]
The normal vector \( \mathbf{n} \) of the plane is \( (1, 2, -2) \).

The distance \( D \) of a plane \( Ax + By + Cz + D = 0 \) from the origin \( (0, 0, 0) \) is given by:
\[ D = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \]
Substituting \( A = 1, B = 2, C = -2, D = 3 \):
\[ D = \frac{|1(0) + 2(0) - 2(0) + 3|}{\sqrt{1^2 + 2^2 + (-2)^2}} = \frac{|3|}{\sqrt{1 + 4 + 4}} = \frac{3}{\sqrt{9}} = \frac{3}{3} = 1 \]

### Final Result
The distance of the plane from the origin is:
\[ D = 1 \]