A plane passes through the point `(0,-1,0)` and `(0,0,1)` and makes an angle of `(pi)/(4)` with the plane `y-z=0` then the point which satisfies the desired plane is

Options

A. `(-sqrt(2), 1, -4)`
B. `(sqrt(2),1,4)`
C. `(sqrt(2),-1,4)`
D. `(-sqrt(2),-1,-4)`

Explanation

To solve the problem, we need to find the equation of a plane that passes through the points (0, -1, 0) and (0, 0, 1) and makes an angle of π/4 with the plane defined by the equation y - z = 0.

### Step 1: Determine the Normal Vector of the Given Plane
The equation of the plane y - z = 0 can be rewritten as 0x + 1y - 1z = 0. From this, we can identify the normal vector of this plane:
- Normal vector of the plane y - z = 0 is **N1 = (0, 1, -1)**.

### Step 2: Find the Direction Vector of the Line through the Given Points
The two points through which the desired plane passes are (0, -1, 0) and (0, 0, 1). We can find the direction vector between these two points:
- Direction vector **D = (0 - 0, 0 - (-1), 1 - 0) = (0, 1, 1)**.

### Step 3: Determine the Normal Vector of the Desired Plane
Let the normal vector of the desired plane be **N2 = (A, B, C)**. Since the plane passes through the two points, we can express the equation of the plane as:
- **Ax + By + Cz = D**.

### Step 4: Use the Angle Condition
The angle θ between two planes can be determined using the dot product of their normal vectors:
\[ \cos(\theta) = \frac{N1 \cdot N2}{\|N1\| \|N2\|} \]
Given that θ = π/4, we have:
\[ \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]
Thus, we can set up the equation:
\[ \frac{(0)(A) + (1)(B) + (-1)(C)}{\sqrt{0^2 + 1^2 + (-1)^2} \sqrt{A^2 + B^2 + C^2}} = \frac{1}{\sqrt{2}} \]
This simplifies to:
\[ \frac{B - C}{\sqrt{2} \sqrt{A^2 + B^2 + C^2}} = \frac{1}{\sqrt{2}} \]
Cross-multiplying gives:
\[ B - C = \sqrt{A^2 + B^2 + C^2} \]

### Step 5: Substitute the Points into the Plane Equation
Since the plane passes through the points (0, -1, 0) and (0, 0, 1), we can substitute these points into the plane equation:
1. For the point (0, -1, 0):
\[ A(0) + B(-1) + C(0) = D \implies -B = D \implies B = -D \]
2. For the point (0, 0, 1):
\[ A(0) + B(0) + C(1) = D \implies C = D \]

### Step 6: Solve for A, B, C
From the above, we have:
- \(B = -D\)
- \(C = D\)

Substituting these into the angle condition:
\[ -D - D = \sqrt{A^2 + (-D)^2 + D^2} \]
This simplifies to:
\[ -2D = \sqrt{A^2 + 2D^2} \]
Squaring both sides:
\[ 4D^2 = A^2 + 2D^2 \implies A^2 = 2D^2 \implies A = \pm \sqrt{2}D \]

### Step 7: Write the Equation of the Plane
Substituting back, we can write the equation of the plane as:
\[ \pm \sqrt{2}D \cdot x - D \cdot y + D \cdot z = D \]
Dividing through by D (assuming D ≠ 0):
\[ \pm \sqrt{2} x - y + z = 1 \]

### Conclusion
The equation of the desired plane is:
\[ \sqrt{2} x - y + z = 1 \quad \text{or} \quad -\sqrt{2} x - y + z = 1 \]