Key Concepts and Formulas

1. Equation of a Plane: The equation of a plane passing through a point $(x_1, y_1, z_1)$ and having a normal vector with direction ratios $(a, b, c)$ is given by: $a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$ The coefficients $(a, b, c)$ represent the direction ratios of any vector perpendicular (normal) to the plane.
2. Condition for a Point on a Plane: If a plane passes through a point, the coordinates of that point must satisfy the plane's equation.
3. Angle Between Two Planes: If $\theta$ is the acute angle between two planes with normal vectors $\vec{n_1} = (a_1, b_1, c_1)$ and $\vec{n_2} = (a_2, b_2, c_2)$, then: $\cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{||\vec{n_1}|| \cdot ||\vec{n_2}||} = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$ The absolute value ensures that $\theta$ is the acute angle. For a plane $Ax + By + Cz + D = 0$, its normal vector has direction ratios $(A, B, C)$.

Step-by-Step Solution

Step 1: Formulate the Equation of the Required Plane Let the direction ratios of the normal vector to the required plane be $(a, b, c)$. The plane passes through the point $(0, -1, 0)$. Why this step? We use the point-normal form of the plane equation. By using one of the given points, we introduce the unknown direction ratios $(a, b, c)$ and establish a general equation for the plane.

Using the point-normal form with $(x_1, y_1, z_1) = (0, -1, 0)$: $a(x - 0) + b(y - (-1)) + c(z - 0) = 0$ $ax + b(y + 1) + cz = 0 \quad \text{(Equation 1)}$

Step 2: Incorporate the Second Given Point The required plane also passes through the point $(0, 0, 1)$. Why this step? Since the second point lies on the plane, its coordinates must satisfy the plane's equation. Substituting these coordinates into Equation 1 will give us a linear relationship between $a, b, c$.

Substitute $(x, y, z) = (0, 0, 1)$ into Equation 1: $a(0) + b(0 + 1) + c(1) = 0$ $b + c = 0$ $c = -b \quad \text{(Equation 2)}$

Step 3: Apply the Angle Condition The required plane makes an angle of $\theta = \frac{\pi}{4}$ with the plane $y - z + 5 = 0$. Why this step? The angle between two planes is determined by the angle between their respective normal vectors. This condition allows us to establish a second relationship between $a, b, c$ using the given angle.

The normal vector to our required plane is $\vec{n_1} = (a, b, c)$. The normal vector to the given plane $y - z + 5 = 0$ is $\vec{n_2} = (0, 1, -1)$.

We are given $\theta = \frac{\pi}{4}$, so $\cos \theta = \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$.

Now, we calculate the dot product and magnitudes of the normal vectors:

• Dot product: $\vec{n_1} \cdot \vec{n_2} = (a)(0) + (b)(1) + (c)(-1) = b - c$
• Magnitude of $\vec{n_1}$: $||\vec{n_1}|| = \sqrt{a^2 + b^2 + c^2}$
• Magnitude of $\vec{n_2}$: $||\vec{n_2}|| = \sqrt{0^2 + 1^2 + (-1)^2} = \sqrt{0 + 1 + 1} = \sqrt{2}$

Substitute these into the angle formula: $\frac{1}{\sqrt{2}} = \frac{|b - c|}{\sqrt{a^2 + b^2 + c^2} \cdot \sqrt{2}}$ Multiply both sides by $\sqrt{2}$: $1 = \frac{|b - c|}{\sqrt{a^2 + b^2 + c^2}}$ Rearrange and square both sides to eliminate the square root and absolute value: $\sqrt{a^2 + b^2 + c^2} = |b - c|$ $a^2 + b^2 + c^2 = (b - c)^2$ $a^2 + b^2 + c^2 = b^2 - 2bc + c^2$ $a^2 = -2bc \quad \text{(Equation 3)}$

Step 4: Solve for the Direction Ratios We now have a system of two equations relating $a, b, c$:

1. $c = -b$ (from Equation 2)
2. $a^2 = -2bc$ (from Equation 3)

Why this step? By solving this system, we can find the proportional relationships between $a, b, c$, which represent the direction ratios of the normal vector.

Substitute Equation 2 ($c = -b$) into Equation 3: $a^2 = -2b(-b)$ $a^2 = 2b^2$ Taking the square root of both sides gives: $a = \pm \sqrt{2} b$

So, we have the relationships: $a = \pm \sqrt{2} b$ and $c = -b$. The direction ratios $(a, b, c)$ are therefore proportional to $(\pm \sqrt{2}b, b, -b)$. Since direction ratios are proportional, we can choose a convenient non-zero value for $b$. Let's choose $b = 1$. Then $c = -1$. And $a = \pm \sqrt{2}(1) = \pm \sqrt{2}$.

This gives two possible sets of direction ratios:

1. $(\sqrt{2}, 1, -1)$
2. $(-\sqrt{2}, 1, -1)$

If we choose $b = -1$: Then $c = -(-1) = 1$. And $a = \pm \sqrt{2}(-1) = \mp \sqrt{2}$. This gives $(-\sqrt{2}, -1, 1)$ or $(\sqrt{2}, -1, 1)$.

All these sets of direction ratios are proportional to $(\sqrt{2}, 1, -1)$. Comparing these with the given options, option (C) is $(\sqrt{2}, 1, -1)$, which is one of the derived possibilities.

Common Mistakes & Tips

• Absolute Value in Angle Formula: Always use the absolute value in the numerator of the angle between planes formula ($|\vec{n_1} \cdot \vec{n_2}|$) to ensure you calculate the acute angle, as is standard unless specified otherwise.
• Proportionality of Direction Ratios: Remember that direction ratios are proportional. If $(a, b, c)$ are direction ratios, then $(ka, kb, kc)$ for any non-zero scalar $k$ are also valid direction ratios. This means your derived ratios might be a multiple of an option, but the ratios themselves must match.
• Algebraic Simplification: Be careful when squaring both sides of an equation to remove square roots and absolute values. For example, $\sqrt{X} = |Y|$ simplifies to $X = Y^2$.

Summary

This problem required us to find the direction ratios of the normal to a plane given two points it passes through and the angle it makes with another plane. We first used the two points to establish a relationship between the direction ratios $(a,b,c)$. Then, we used the angle between planes formula to establish a second relationship. Solving the system of these two equations ($c = -b$ and $a^2 = -2bc$) yielded $a = \pm \sqrt{2}b$. By choosing a convenient value for $b$ (e.g., $b=1$), we found the direction ratios to be proportional to $(\sqrt{2}, 1, -1)$.

The final answer is $\boxed{\text{(C)}}$.