Key Concepts and Formulas

1. Equation of a Plane in Point-Normal Form: If a plane passes through a point $(x_1, y_1, z_1)$ and has a normal vector $\vec{n} = (A, B, C)$, its equation can be written as $A(x-x_1) + B(y-y_1) + C(z-z_1) = 0$. This form directly uses the plane's orientation (normal vector) and its position (a point it contains).
2. Normal Vector from Perpendicularity: If a plane is perpendicular to a given line, then the direction vector of that line serves as the normal vector to the plane. The direction vector of a line joining two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by $(x_2-x_1, y_2-y_1, z_2-z_1)$.
3. Distance of a Point from a Plane: The perpendicular distance of a point $(x_0, y_0, z_0)$ from a plane $Ax + By + Cz + D = 0$ is given by the formula: $\text{Distance} = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$

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Step-by-Step Solution

Step 1: Determine the Normal Vector of the Plane $P$

• What we are doing: We need to find the normal vector to the plane $P$.
• Why this is important: The normal vector $(A, B, C)$ defines the orientation of the plane and is crucial for writing its equation.
• Applying the concept: The problem states that plane $P$ is perpendicular to the line joining the points $L_1(4,1,-3)$ and $L_2(2,4,3)$. According to our key concepts, if a plane is perpendicular to a line, the direction vector of that line can be taken as the normal vector to the plane.
• Calculation: Let $\vec{n}$ be the normal vector to plane $P$. We calculate the direction vector of the line $L_1L_2$: $\vec{n} = \vec{L_2} - \vec{L_1} = (2-4, 4-1, 3-(-3)) = (-2, 3, 6)$ Alternatively, we could have calculated $\vec{L_1} - \vec{L_2} = (4-2, 1-4, -3-3) = (2, -3, -6)$. Both vectors are valid normal vectors for the plane, as they are scalar multiples of each other and represent the same direction (or its opposite). For convenience in calculations, we will use $\vec{n} = (2, -3, -6)$. Thus, the components of the normal vector are $(A, B, C) = (2, -3, -6)$.

Step 2: Formulate the Equation of the Plane $P$

• What we are doing: We will now write the equation of plane $P$ using the normal vector found in Step 1 and the given point it passes through.
• Why this is important: To find the distance of a point from the plane, we first need the explicit algebraic equation of the plane in the general form $Ax+By+Cz+D=0$.
• Applying the concept: We use the point-normal form of the plane equation: $A(x-x_1) + B(y-y_1) + C(z-z_1) = 0$.
• Given Information:
  • Normal vector $\vec{n} = (A, B, C) = (2, -3, -6)$.
  • The plane passes through the point $(x_1, y_1, z_1) = (1, -1, -5)$.
• Calculation: Substitute these values into the point-normal form: $2(x-1) - 3(y-(-1)) - 6(z-(-5)) = 0$ $2(x-1) - 3(y+1) - 6(z+5) = 0$ Now, we expand and simplify this equation to the general form $Ax + By + Cz + D = 0$: $2x - 2 - 3y - 3 - 6z - 30 = 0$ $2x - 3y - 6z - 35 = 0$ This is the equation of plane $P$. Here, $A=2$, $B=-3$, $C=-6$, and $D=-35$.

Step 3: Calculate the Distance of Plane $P$ from the Point $(3,-2,2)$

• What we are doing: We will compute the perpendicular distance from the given point $(3,-2,2)$ to the plane $P$ whose equation we just found.
• Why this is important: This is the final objective of the problem and involves a direct application of the distance formula.
• Applying the concept: We use the formula for the perpendicular distance of a point from a plane.
• Given Information:
  • Equation of plane $P$: $2x - 3y - 6z - 35 = 0$. (So, $A=2, B=-3, C=-6, D=-35$).
  • The point is $(x_0, y_0, z_0) = (3, -2, 2)$.
• Calculation: Substitute these values into the distance formula: $\text{Distance} = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$ $\text{Distance} = \frac{|2(3) - 3(-2) - 6(2) - 35|}{\sqrt{(2)^2 + (-3)^2 + (-6)^2}}$ $\text{Distance} = \frac{|6 + 6 - 12 - 35|}{\sqrt{4 + 9 + 36}}$ $\text{Distance} = \frac{|12 - 12 - 35|}{\sqrt{49}}$ $\text{Distance} = \frac{|-35|}{7}$ $\text{Distance} = \frac{35}{7}$ $\text{Distance} = 5$

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Common Mistakes & Tips

• Direction Vector vs. Normal Vector: Always remember that if a plane is perpendicular to a line, the line's direction vector is the plane's normal vector. If the plane were parallel to a line, their normal vector and direction vector would be perpendicular (dot product is zero).
• Absolute Value in Distance Formula: Distance is a scalar and always non-negative. Do not forget the absolute value in the numerator of the distance formula to ensure a positive result.
• Simplifying Plane Equation: Ensure the plane equation is in the standard general form $Ax+By+Cz+D=0$ before applying the distance formula. Any constant terms not combined into 'D' will lead to incorrect results.
• Arithmetic Precision: Be careful with signs and basic arithmetic, especially when squaring negative numbers (e.g., $(-3)^2 = 9$, not $-9$) and summing terms under the square root.

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Summary

To find the distance of a point from a plane, we first determined the normal vector of the plane using the property that it is perpendicular to a given line. Then, we used this normal vector along with a point lying on the plane to derive the plane's equation in the general form. Finally, we applied the standard formula for the perpendicular distance of a point from a plane. All calculations led to a distance of 5 units.

The final answer is $\boxed{5}$, which corresponds to option (A).