Key Concepts and Formulas

To solve this problem, we will utilize the following fundamental concepts from 3D Geometry:

1. Equation of a Plane: The general equation of a plane is $Ax + By + Cz + D = 0$. Here, $A, B, C$ are the direction ratios of the normal vector to the plane.
2. Normal Vector to a Plane: For a plane $Ax + By + Cz + D = 0$, its normal vector is $\vec{n} = A\hat{i} + B\hat{j} + C\hat{k}$.
3. Plane Passing Through Points: If a plane passes through a point $(x_0, y_0, z_0)$, then substituting these coordinates into the plane's equation must satisfy the equation. This helps in determining the coefficients of the plane.
4. Angle Between Two Planes: The angle $\theta$ between two planes $P_1: A_1x + B_1y + C_1z + D_1 = 0$ and $P_2: A_2x + B_2y + C_2z + D_2 = 0$ is the angle between their normal vectors $\vec{n_1} = A_1\hat{i} + B_1\hat{j} + C_1\hat{k}$ and $\vec{n_2} = A_2\hat{i} + B_2\hat{j} + C_2\hat{k}$. The formula is: $\cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{||\vec{n_1}|| \cdot ||\vec{n_2}||}$ The absolute value ensures we find the acute angle ($0 \le \theta \le \frac{\pi}{2}$).

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Step-by-Step Solution

Step 1: Formulating the Equation of the Required Plane

Let the general equation of the required plane be $ax + by + cz + d = 0$. Our goal is to find the values of $a, b, c, d$.

1. Using point $(0, -1, 0)$: The plane passes through $(0, -1, 0)$. Substituting these coordinates into the plane equation: $a(0) + b(-1) + c(0) + d = 0$ $-b + d = 0 \quad \Rightarrow \quad b = d$ Explanation: Any point lying on the plane must satisfy its equation. This provides a relationship between the coefficients.

2. Using point $(0, 0, 1)$: The plane also passes through $(0, 0, 1)$. Substituting these coordinates: $a(0) + b(0) + c(1) + d = 0$ $c + d = 0 \quad \Rightarrow \quad c = -d$ Explanation: Similar to the first point, this gives us another relationship.

Now, substitute $b = d$ and $c = -d$ back into the general equation $ax + by + cz + d = 0$: $ax + (d)y + (-d)z + d = 0$ $ax + dy - dz + d = 0$ Explanation: By expressing $b$ and $c$ in terms of $d$, we reduce the number of independent unknowns.

Important Note on $d$: We must check if $d$ can be zero. If $d=0$, then $b=0$ and $c=0$. The plane equation would simplify to $ax=0$. For this to represent a plane, $a$ must be non-zero (otherwise $0=0$, which is not a plane). If $a \neq 0$, $ax=0$ implies $x=0$, which is the YZ-plane. This plane passes through $(0,-1,0)$ and $(0,0,1)$. The normal vector for $x=0$ is $\vec{n_1} = (1,0,0)$. The given plane is $y-z+5=0$, with normal $\vec{n_2} = (0,1,-1)$. The cosine of the angle between them would be $\cos\theta = \frac{|(1)(0) + (0)(1) + (0)(-1)|}{\sqrt{1^2}\sqrt{0^2+1^2+(-1)^2}} = \frac{0}{1 \cdot \sqrt{2}} = 0$. This implies $\theta = \frac{\pi}{2}$. However, the problem states the angle is $\frac{\pi}{4}$. Therefore, $d \neq 0$.

Since $d \neq 0$, we can divide the entire equation $ax + dy - dz + d = 0$ by $d$: $\frac{a}{d}x + \frac{d}{d}y - \frac{d}{d}z + \frac{d}{d} = 0$ $\frac{a}{d}x + y - z + 1 = 0$ Let $k = \frac{a}{d}$. This substitution simplifies the equation of the required plane to: $P_1: kx + y - z + 1 = 0$ Explanation: Dividing by $d$ normalizes the equation, which is a common practice to reduce the number of independent variables without losing generality, as $d$ cannot be zero.

Step 2: Using the Angle Condition Between Planes

The problem states that plane $P_1$ makes an angle of $\theta = \frac{\pi}{4}$ with the plane $P_2: y - z + 5 = 0$.

1. Identify Normal Vectors:

   • For $P_1: kx + y - z + 1 = 0$, the normal vector is $\vec{n_1} = k\hat{i} + 1\hat{j} - 1\hat{k}$.
   • For $P_2: y - z + 5 = 0$, the normal vector is $\vec{n_2} = 0\hat{i} + 1\hat{j} - 1\hat{k}$. Explanation: The coefficients of $x, y, z$ in the plane equation directly represent the components of its normal vector.

2. Calculate Dot Product and Magnitudes:

   • Dot Product: $\vec{n_1} \cdot \vec{n_2} = (k)(0) + (1)(1) + (-1)(-1) = 0 + 1 + 1 = 2$
   • Magnitudes: $||\vec{n_1}|| = \sqrt{k^2 + (1)^2 + (-1)^2} = \sqrt{k^2 + 1 + 1} = \sqrt{k^2 + 2}$ $||\vec{n_2}|| = \sqrt{(0)^2 + (1)^2 + (-1)^2} = \sqrt{0 + 1 + 1} = \sqrt{2}$

3. Apply Angle Formula: Given $\theta = \frac{\pi}{4}$, we know $\cos \left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$. Using the formula $\cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{||\vec{n_1}|| \cdot ||\vec{n_2}||}$: $\frac{1}{\sqrt{2}} = \frac{|2|}{\sqrt{k^2 + 2} \cdot \sqrt{2}}$ Explanation: We use the absolute value of the dot product to ensure we are considering the acute angle between the planes.

4. Solve for $k$: $\frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2(k^2 + 2)}}$ Cancel $\sqrt{2}$ from both denominators: $1 = \frac{2}{\sqrt{k^2 + 2}}$ Rearrange the equation: $\sqrt{k^2 + 2} = 2$ Square both sides: $(\sqrt{k^2 + 2})^2 = 2^2$ $k^2 + 2 = 4$ $k^2 = 2$ $k = \pm \sqrt{2}$ Explanation: Squaring both sides is a standard algebraic method to remove square roots. The $\pm$ sign indicates two possible values for $k$, meaning there are two distinct planes satisfying the given conditions.

Step 3: Determining the Equations of the Possible Planes

We have two possible values for $k$, leading to two possible plane equations:

1. If $k = \sqrt{2}$, the equation of the plane is $P_A: \sqrt{2}x + y - z + 1 = 0$.
2. If $k = -\sqrt{2}$, the equation of the plane is $P_B: -\sqrt{2}x + y - z + 1 = 0$.

Step 4: Checking the Options

The question asks which of the given points the plane also passes through. We test each option by substituting its coordinates into the derived plane equations.

• Option (A): $(\sqrt{2}, 1, 4)$

  • Test with $P_A: \sqrt{2}x + y - z + 1 = 0$: Substitute $x=\sqrt{2}, y=1, z=4$: $\sqrt{2}(\sqrt{2}) + (1) - (4) + 1 = 2 + 1 - 4 + 1 = 0$. Since $0 = 0$, point (A) lies on plane $P_A$.
  • Test with $P_B: -\sqrt{2}x + y - z + 1 = 0$: Substitute $x=\sqrt{2}, y=1, z=4$: $-\sqrt{2}(\sqrt{2}) + (1) - (4) + 1 = -2 + 1 - 4 + 1 = -4 \neq 0$. Since point (A) satisfies the equation for $P_A$, it is a point on one of the planes.

• Option (B): $(-\sqrt{2}, 1, 4)$

  • Test with $P_A: \sqrt{2}x + y - z + 1 = 0$: $\sqrt{2}(-\sqrt{2}) + (1) - (4) + 1 = -2 + 1 - 4 + 1 = -4 \neq 0$.
  • Test with $P_B: -\sqrt{2}x + y - z + 1 = 0$: $-\sqrt{2}(-\sqrt{2}) + (1) - (4) + 1 = 2 + 1 - 4 + 1 = 0$. Since $0 = 0$, point (B) lies on plane $P_B$.

Since the question asks for a point the plane passes through, and Option (A) is satisfied by one of the derived planes ($P_A$), it is the correct answer. (Both (A) and (B) are valid points for the two distinct planes, but only one option can be chosen).

• Option (C): $(-\sqrt{2}, -1, -4)$

  • Testing with $P_A$: $\sqrt{2}(-\sqrt{2}) + (-1) - (-4) + 1 = -2 - 1 + 4 + 1 = 2 \neq 0$.
  • Testing with $P_B$: $-\sqrt{2}(-\sqrt{2}) + (-1) - (-4) + 1 = 2 - 1 + 4 + 1 = 6 \neq 0$. Point (C) does not lie on either plane.

• Option (D): $(\sqrt{2}, -1, 4)$

  • Testing with $P_A$: $\sqrt{2}(\sqrt{2}) + (-1) - (4) + 1 = 2 - 1 - 4 + 1 = -2 \neq 0$.
  • Testing with $P_B$: $-\sqrt{2}(\sqrt{2}) + (-1) - (4) + 1 = -2 - 1 - 4 + 1 = -6 \neq 0$. Point (D) does not lie on either plane.

Thus, Option (A) is the only point that satisfies one of the derived plane equations.

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Common Mistakes & Tips

• Normal Vector Signs: Be careful when extracting the components of the normal vector from the plane equation. For $y-z+5=0$, the normal is $(0, 1, -1)$.
• Absolute Value in Angle Formula: Always use the absolute value of the dot product in the $\cos \theta$ formula for the angle between planes to ensure you calculate the acute angle.
• Considering Both Roots: When solving for a squared variable (e.g., $k^2 = 2$), remember that there are two possible roots ($k = \pm \sqrt{2}$). Both solutions represent valid planes that satisfy the initial conditions.
• Systematic Option Checking: If multiple planes are possible, systematically check each given option against all derived plane equations.

Summary

This problem required us to first determine the general equation of a plane using two given points, simplifying it by expressing coefficients in terms of a single parameter. Then, we utilized the condition of the angle between this plane and a given second plane. By calculating the dot product and magnitudes of their normal vectors and applying the angle formula, we solved for the unknown parameter, which yielded two possible plane equations. Finally, we checked which of the given options satisfied either of these derived plane equations to find the correct point.

The final answer is $\boxed{\left( {\sqrt 2 ,1,4} \right)}$, which corresponds to option (A).