Problem 37 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(P\) be any point on the plane \(l x+m y+n z=p\) and \(Q\) be a point on the line \(O P\) such that \(O P . O Q=p^{2}\). The locus of the point \(Q\) is (A) \(l x+m y+n z=x^{2}+y^{2}+z^{2}\) (B) \(b x+m y+n z=p\left(x^{2}+y^{2}+z^{2}\right)\) (C) \(p(b x+m y+n z)=x^{2}+y^{2}+z^{2}\) (D) none of these

Step-by-Step Solution

Answer

(C) \(p(lx+my+nz) = x^2 + y^2 + z^2\)

Step 1: Define the Points

Given that \( P \) is any point on the plane, then the coordinates of \( P \) are \((x_1, y_1, z_1)\) such that \( lx_1 + my_1 + nz_1 = p \). \( Q \) is a point on the line through \( O \) and \( P \). The coordinates of \( Q \) can be expressed as \( (t x_1, t y_1, t z_1) \) where \( t \) is a scalar.

Step 2: Set Up the Condition

Given that \( OP \cdot OQ = p^2 \), translate this condition into a mathematical equation. The distance \( OP \) is \( \sqrt{x_1^2 + y_1^2 + z_1^2} \) and the distance \( OQ \) is \( t \cdot \sqrt{x_1^2 + y_1^2 + z_1^2} \). Therefore, \( (x_1^2 + y_1^2 + z_1^2) \cdot t = p^2 \).

Step 3: Solve for Scalar t

From the equation \( (x_1^2 + y_1^2 + z_1^2) \cdot t = p^2 \), solve for \( t \). Thus, \( t = \frac{p^2}{x_1^2 + y_1^2 + z_1^2} \).

Step 4: Express the Coordinates of Q

Substitute \( t \) back into the parametric equation of \( Q \). The coordinates of \( Q \) become: \( \left( \frac{p^2 x_1}{x_1^2 + y_1^2 + z_1^2}, \frac{p^2 y_1}{x_1^2 + y_1^2 + z_1^2}, \frac{p^2 z_1}{x_1^2 + y_1^2 + z_1^2} \right) \).

Step 5: Determine the Locus of Q

The point \( Q \) satisfies the plane equation as it is represented in its parametric form. Substituting the coordinates of \( Q \) back into the plane equation yields \( l \left(\frac{p^2 x_1}{x_1^2 + y_1^2 + z_1^2}\right) + m \left(\frac{p^2 y_1}{x_1^2 + y_1^2 + z_1^2}\right) + n \left(\frac{p^2 z_1}{x_1^2 + y_1^2 + z_1^2}\right) = p \). This simplifies to \( p(lx_1 + my_1 + nz_1) = x_1^2 + y_1^2 + z_1^2 \), satisfied by the locus.

Step 6: Simplify and Verify the Solution

Re-examine each option and verify matching conditions. The simplification successfully leads us to option C: \( p(lx + my + nz) = x^2 + y^2 + z^2 \).

Key Concepts

Locus of a Point

The term "locus of a point" represents all possible positions that a point can occupy under a specific condition or set of conditions. In the context of coordinate geometry, finding the locus often involves a set of equations that describe all possible locations a point can have based on given relationships. For example, in this exercise, point \( Q \) is subject to the condition that its position relates to the origin \( O \) and another point \( P \) on the plane. The equation derived from these relationships gives us the locus of \( Q \), namely \( p(lx + my + nz) = x^2 + y^2 + z^2 \). It's important to understand that a locus can be a line, a circle, a plane, or another defined shape, depending on the constraints provided in the problem.

Equation of a Plane

An equation of a plane in coordinate geometry is expressed in the standard form \( lx + my + nz = p \). This equation uses the coefficients \( l, m, \) and \( n \), which represent the normal vector to the plane, and \( p \), which typically represents a specific scalar offset. The equation defines a flat, infinite two-dimensional surface in three-dimensional space where every point on this surface satisfies the equation.
In our specific exercise, \( P \) lies on such a plane and thereby fulfills \( lx_1 + my_1 + nz_1 = p \). The equation of the plane allows us to identify any point's relative position, whether it is on the plane, above, or below it. Understanding such equations is crucial as they form the basis for comprehending more complex geometrical structures and relationships.

Parametric Equations

Parametric equations are prevalent in describing points along a path or curve through the use of parameters, usually denoted by \( t \). They provide a flexible way to describe the coordinates of a point moving through space. In simpler cases, a parametric equation will have 'time' as the parameter, showing how the position changes as time progresses.
For our point \( Q \), its coordinates are derived using parametric equations influenced by a scalar \( t \). Expressed as \( (tx_1, ty_1, tz_1) \), these forms demonstrate the way that \( Q \) maps along the path set between origin \( O \) and the point \( P \). Parametric equations are essential because they enable representation of coordinates in scenarios where conventional Cartesian forms might fall short.

Scalar Multiplication in Vectors

Scalar multiplication in vectors involves multiplying a vector by a scalar, resulting in a new vector that points in the same (or completely opposite) direction but with a scaled magnitude. The basic rule is each component of the vector is multiplied by the scalar.
In our scenario, the position of point \( Q \) depends on a scalar \( t \) that scales the direction vector from \( O \) to \( P \). If \( P \) is represented by \( (x_1, y_1, z_1) \), then \( Q \) as \( (tx_1, ty_1, tz_1) \) ensures that \( Q \) sits somewhere on the line between \( O \) and \( P \). Understanding scalar multiplication is fundamental in vector analysis, as it unlocks complex variable manipulation making it possible to describe real-world directions and magnitudes effectively using vector mathematics.