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Chapter 22: Problem 31

\(P\) is any point on the plane \(l x+m y+n z=p ;\) a point \(Q\) is taken on the line \(O P\) such that \(O P \cdot O Q=p^{2}\), then the locus of \(Q\) is

Short Answer

Expert verified
The locus of Q is the plane equation \(lx + my + nz = p\).

Step by step solution

01

Understand the Problem

We have a plane with the equation \(l x + m y + n z = p\). A point \(P\) lies on this plane, and a point \(Q\) is on the line \(OP\), where \(O\) is the origin. We want to find the locus of \(Q\) such that \(OP \cdot OQ = p^2\).
02

Define the Vectors

Let \(\vec{OP} = \vec{r}\) and \(\vec{OQ} = \lambda \vec{r}\) where \(\lambda\) is a scalar that defines the position of \(Q\) along the line \(OP\). Thus, \(OQ\) is a scaled version of \(OP\).
03

Express the Dot Product Condition

The condition \(OP \cdot OQ = p^2\) can be rewritten as \(\vec{r} \cdot (\lambda \vec{r}) = p^2\). This simplifies to \(\lambda (\vec{r} \cdot \vec{r}) = p^2\).
04

Use Plane Equation for Vector \(\vec{r}\)

Since \(P\) is on the plane \(lx + my + nz = p\), we have \((l, m, n) \cdot (x, y, z) = p\). But, \(\vec{r} = (x, y, z)\), meaning \(\vec{n} \cdot \vec{r} = p\) where \(\vec{n} = (l, m, n)\) is the normal to the plane.
05

Relate \(\lambda\) to Plane Equation

Substitute the plane condition in the dot product equation: \(\lambda (p^2) = p^2\). Solve for \(\lambda\): \(\lambda = 1\).
06

Determine Locus of \(Q\)

Since \(\lambda = 1\), \(Q\) coincides with \(P\), meaning the locus of \(Q\) is simply the plane itself. Therefore, the locus of \(Q\) is given by the plane equation \(lx + my + nz = p\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Plane Equation
The equation of a plane in three-dimensional space can be described as \( l x + m y + n z = p \), where \( l, m, \) and \( n \) are the coefficients that form the normal vector \( \vec{n} = (l, m, n) \). The plane is defined as the set of all points \((x, y, z)\) such that their dot product with the normal vector equals \(p\), which is a constant.
This equation is useful because it directly tells us how the plane is oriented in three-dimensional space.
  • The coefficients \( l, m, \) and \( n \) determine the tilt and direction of the plane.
  • The value \(p\) is essentially the distance from the plane to the origin along the normal vector, assuming \( \vec{n} \) is a unit vector.
Understanding and using the plane equation is crucial for solving geometric problems in vector spaces, such as determining if a point lies on a certain plane.
Dot Product
The dot product, also known as the scalar product, is an operation that takes two equal-length sequences of numbers (usually coordinate vectors), and returns a single number. The dot product of vectors \(\vec{a} = (a_1, a_2, a_3)\) and \(\vec{b} = (b_1, b_2, b_3)\) is calculated as:\[\vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3\]This operation has several qualities:
  • It provides a measure of how parallel two vectors are. If the dot product is zero, the vectors are orthogonal.
  • It gives the projection of one vector onto another.
  • This product results in a scalar rather than a vector.
The dot product is pivotal when computing angles between vectors or understanding plane geometries, as it forms the basis for interpreting geometric conditions.
Vector Algebra
Vector algebra is the branch of mathematics that combines elements of algebra and geometry to study vectors. Vectors are quantities defined by both a magnitude and a direction. In this exercise, understanding vector algebra is essential in manipulating vectors to express positions and equations.
Vectors are usually represented in component form such as \(\vec{r} = (x, y, z)\), where each component is its respective coordinate in space.
  • Vector addition and subtraction are performed by adding or subtracting corresponding components.
  • Scaling a vector involves multiplying each component by a scalar, which stretches or shrinks the vector.
  • The dot product and cross product are crucial operations in vector algebra that help in solving geometric problems and understanding spatial relationships.
By employing vector algebra, we can relate points, lines, and planes in a coherent manner to solve complex geometric challenges involving spatial loci.

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Most popular questions from this chapter

If \(l_{1}, m_{1}, n_{1}\) and \(l_{2}, m_{2}, n_{2}\) are d.c.'s of the two lines inclined to each other at an angle \(\theta\), then the d.c.'s of the internal bisector of the angle between these lines are (A) \(\frac{l_{1}+l_{2}}{2 \sin \theta / 2}, \frac{m_{1}+m_{2}}{2 \sin \theta / 2}, \frac{n_{1}+n_{2}}{2 \sin \theta / 2}\) (B) \(\frac{l_{1}+l_{2}}{2 \cos \theta / 2}, \frac{m_{1}+m_{2}}{2 \cos \theta / 2}, \frac{n_{1}+n_{2}}{2 \cos \theta / 2}\) (C) \(\frac{l_{1}-l_{2}}{2 \sin \theta / 2}, \frac{m_{1}-m_{2}}{2 \sin \theta / 2}, \frac{n_{1}-n_{2}}{2 \sin \theta / 2}\) (D) \(\frac{l_{1}-l_{2}}{2 \cos \theta / 2}, \frac{m_{1}-m_{2}}{2 \cos \theta / 2}, \frac{n_{1}-n_{2}}{2 \cos \theta / 2}\)

If the angle \(Q\) between the line \(\frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2}\) and the plane \(2 x-y+\sqrt{\lambda z}+4=0\) is such that \(\sin \theta=\frac{1}{3}\) the value of \(\lambda\) is \(\quad\) [2005](A) \(\frac{5}{3}\) (B) \(\frac{-3}{5}\) (C) \(\frac{3}{4}\) (D) \(\frac{-4}{3}\)

The plane \(x+y+z=5 \sqrt{3}\) and sphere \(x^{2}+y^{2}+z^{2}=5\) (A) touch each other (B) cut in a circle (C) do not meet (D) none of these

The equation of the plane through the points \((2,3,1)\) and \((4,-5,3)\) and parallel to \(x\)-axis is (A) \(x-z-1=0\) (B) \(4 x+y-11=0\) (C) \(y+4 z-7=0\) (D) none of these

A line with direction cosines proportional to \(2,1,2\) meets each of the lines \(x=y+a=z\) and \(x+a=2 y=2 z\). The co-ordinates of each of the point of intersection are given by [2004] (A) \((3 a, 3 a, 3 a),(a, a, a)\) (B) \((3 a, 2 a, 3 a),(a, a, a)\) (C) \((3 a, 2 a, 3 a),(a, a, 2 a)\) (D) \((2 a, 3 a, 3 a),(2 a, a, a)\)
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