Problem 37 - A Complete Resource Book in Mathematics for JEE Main 2017

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) \(l x+m y+n z=p\left(x^{2}+y^{2}+z^{2}\right)\) (C) \(p(b x+m y+n z)=x^{2}+y^{2}+z^{2}\) (D) none of these

Step-by-Step Solution

Answer

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

Step 1: Understand the Problem

The exercise gives a plane equation where a point \( P \) lies, and a point \( Q \) lies on the line connecting the origin \( O \) and \( P \). The condition \( OP \cdot OQ = p^2 \) is given, and we need to find the equation, or locus, for \( Q \).

Step 2: Define Points O, P, and Q

Since \( P(x_1, y_1, z_1) \) lies on the plane \( l x + m y + n z = p \), and \( Q \) is on the line \( OP \), parameterize \( Q \) as \( (kx_1, ky_1, kz_1) \), where \( k \) is a scalar multiplier.

Step 3: Calculate OP and OQ

Calculate the magnitudes of \( OP \) and \( OQ \): - \( OP = \sqrt{x_1^2 + y_1^2 + z_1^2} \) - \( OQ = \sqrt{(kx_1)^2 + (ky_1)^2 + (kz_1)^2} = |k| \cdot OP \). Using the condition \( OP \cdot OQ = p^2 \), substitute these values.

Step 4: Apply the OP.OQ Condition

Substitute the expressions of \( OP \) and \( OQ \) into the condition: \( OP \cdot OQ = (x_1^2 + y_1^2 + z_1^2) \cdot |k| = p^2 \). This implies \( |k| = \frac{p}{\sqrt{x_1^2 + y_1^2 + z_1^2}} \).

Step 5: Find Locus Equation

Find a relation between \( x_1, y_1, z_1 \) and \( k \). Since \( l x_1 + m y_1 + n z_1 = p \) and \( Q = (kx_1, ky_1, kz_1) \), use \( k^2(x_1^2 + y_1^2 + z_1^2) = p^2 \). Substitute back to get \( lx + my + nz = \frac{p}{k} = \sqrt{x_1^2 + y_1^2 + z_1^2} \).

Step 6: Simplify the Locus Equation

We derive from previous steps that \( lx + my + nz = x^2 + y^2 + z^2 \). Hence, the correct option comparing with given choices is \( \text{(A) } lx + my + nz = x^2 + y^2 + z^2 \).

Key Concepts

Plane Equations

A plane equation is a mathematical expression that represents a flat surface in three-dimensional space. It is generally written in the form \( l x + m y + n z = p \). Here, \( l, m, \) and \( n \) are the coefficients that determine the orientation of the plane, while \( p \) is a constant that helps locate the exact position of the plane in space.
To understand the concept, imagine a sheet of paper extending infinitely. This sheet, being perfectly flat, can be described using a plane equation.
The point \( P \) identified in our exercise, which lies on this plane, is a solution to the equation \( l x + m y + n z = p \). It implies that plugging the coordinates of \( P \) into the equation will satisfy it perfectly. This is because every plane can be thought of as a set of infinite points that makes the equation true when plugged into it.
Understanding plane equations is crucial, especially in geometric loci problems, as these equations form the foundational base from which you can derive further properties or solutions related to spaces and shapes.

Parametric Equations

Parametric equations are expressions that use parameters to uniquely define a curve or line in space. Instead of describing points using traditional \( (x, y, z) \) placements, parametric equations use a parameter \( k \) to describe a whole series of positions on a line.
In the context of our exercise, the point \( Q \) is defined parametrically as \( (kx_1, ky_1, kz_1) \), where \( k \) serves as a scalar that stretches or shrinks the position of \( Q \) relative to \( P \). In essence, parametric equations allow us to describe any point on the line having some relation to \( O \) and \( P \), by manipulating the value of \( k \).
This becomes particularly useful because it simplifies the calculation and understanding of geometric problems by translating a potentially complicated system into a more manageable form. In our example, it allows us to translate the condition \( OP \cdot OQ = p^2 \) into a straightforward solvable form.

Geometric Loci

A geometric locus is a collection of points that satisfy a certain condition or set of conditions. When we seek to identify the locus of a point such as \( Q \) in our exercise, we are essentially looking for all possible locations \( Q \) could occupy given the defined conditions.
The problem at hand provides a direct relation condition \( OP \cdot OQ = p^2 \) alongside the plane equation, allowing us to derive a specific equation that \( Q \) must satisfy. With sufficient understanding of plane and parametric equations, we execute mathematical manipulations to find that the condition simplifies down to \( lx + my + nz = x^2 + y^2 + z^2 \).
This result means that wherever \( Q \) moves, it must satisfy this equation precisely, making it the geometric locus. In simpler terms, it's like saying "Q can exist anywhere, as long as it stays on this "geometric path" defined by the conditions."

• Geometric loci help solve complex spatial problems by reducing them to identifiable paths or curves.
• They are used extensively in architecture, robotics, and various other engineering fields.

Recognizing these paths allows one to understand and predict the behavior of systems within defined constraints.