Problem 98 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(\vec{u}, \vec{v}, \vec{w}\) are non-coplanar vectors and \(p, q\) are real numbers, then the equality \([3 \vec{u} p \vec{v} p \vec{w}]-[p \vec{v} \vec{w} q \vec{u}]\) \(-\left[\begin{array}{lll}2 \vec{w} & q \vec{v} & q \vec{u}\end{array}\right]=0\) holds for [2009] (A) exactly one value of \((p, q)\) (B) exactly two values of \((p, q)\) (C) more than two but not all values of \((p, q)\) (D) all values of \((p, q)\)

Step-by-Step Solution

Answer

Two values of \((p, q)\) satisfy the equation.

Step 1: Understand the Problem

We are given three non-coplanar vectors \(\vec{u}, \vec{v}, \vec{w}\) and need to find the conditions on the real numbers \(p\) and \(q\) such that the equation \([3 \vec{u} \hspace{1mm} p \vec{v} \hspace{1mm} p \vec{w}]-[p \vec{v} \hspace{1mm} \vec{w} \hspace{1mm} q \vec{u}] -[2 \vec{w} \hspace{1mm} q \vec{v} \hspace{1mm} q \vec{u}]=0\) holds.

Step 2: Apply the Determinant Property for Linear Relations

Recognize that \([\vec{a} \hspace{1mm} \vec{b} \hspace{1mm} \vec{c}]\) represents the scalar triple product (a determinant) of the vectors. Given that the vectors are non-coplanar, analyze how changing the scalar multiples or coefficients affect the product.

Step 3: Expand the Determinants

Identify that each term represents a determinant: \[ [3 \vec{u} \hspace{1mm} p \vec{v} \hspace{1mm} p \vec{w}] = 3 \det(\vec{u}, \vec{v}, \vec{w}) \times p^2, \] \[ [p \vec{v} \hspace{1mm} \vec{w} \hspace{1mm} q \vec{u}] = \det(\vec{v}, \vec{w}, \vec{u}) \times pq, \] \[ [2 \vec{w} \hspace{1mm} q \vec{v} \hspace{1mm} q \vec{u}] = 2 \det(\vec{w}, \vec{v}, \vec{u}) \times q^2. \] Simplify and equate these using their linear properties.

Step 4: Combine and Solve for Values

Combine the scalar products:\[ 3p^2 - pq - 2q^2 = 0. \]Factor the quadratic equation to solve for \(p\) and \(q\).

Step 5: Factor and Solve the Quadratic Equation

Rewrite the equation: \(3p^2 - pq - 2q^2 = 0\) as \((3p + 2q)(p - q) = 0\). This gives potential solutions, either \(3p + 2q = 0\) or \(p = q\).

Step 6: Determine Number of Solutions

Check the conditions that arise from \(3p + 2q = 0\) and \(p = q\) alongside the requirement that non-zero solutions are needed for the determinants to remain non-zero.

Key Concepts

Scalar Triple Product

The scalar triple product is a crucial concept in vector analysis. It is denoted by \([\vec{a} \hspace{1mm} \vec{b} \hspace{1mm} \vec{c}]\) and represents the volume of the parallelepiped formed by the vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\). It is calculated using the determinant of a 3x3 matrix constructed from these vectors. This determinant is expressed as

• \( \vec{a} \cdot ( \vec{b} \times \vec{c} ) \)
• It results in a scalar value.

The scalar triple product is significant because:

• It provides geometric insights: If the result is zero, the vectors are coplanar, meaning they lie in the same plane.
• It informs us about the volume, with the absolute value representing the volume, while the sign indicates orientation.

For the problem at hand, understanding the scalar triple product is key because the problem involves analyzing several of these determinants to deduce conditions for \(p\) and \(q\). By examining how changes in these parameters affect the products, we can determine the necessary relationships.

Non-Coplanar Vectors

Non-coplanar vectors are vectors that do not lie in the same plane, which means they span the three-dimensional space. Generally, you need three vectors to entirely define a three-dimensional space. Here’s why non-coplanarity is important:

• If three vectors are coplanar, they are linearly dependent, meaning one of them can be expressed as a linear combination of the others.
• With non-coplanar vectors, any vector within the space can be expressed uniquely using the given set.
• The scalar triple product for non-coplanar vectors is non-zero, confirming their independence in spanning the space.

In the provided problem, using non-coplanar vectors ensures that the determinants used in calculations are not zero. Thus, the validity of the conditions for \(p\) and \(q\) can be accurately determined from the derived equations.

Quadratic Equations

Quadratic equations are polynomial equations of degree two, generally represented in the form \(ax^2 + bx + c = 0\). The solution to such equations can be found using various methods, such as factoring, completing the square, or the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Quadratic equations can have:

• Two solutions: when the discriminant \((b^2 - 4ac)\) is positive.
• One solution: when the discriminant is zero, indicating a repeated root.
• No real solutions: when the discriminant is negative.

In the context of this problem, the equation \(3p^2 - pq - 2q^2 = 0\) is quadratic. By factoring it into \((3p + 2q)(p - q) = 0\), we effectively reduce the problem to solving simpler linear equations. Each factor equated to zero gives potential relationships between \(p\) and \(q\), crucial for identifying the set of solutions that satisfy the original problem conditions.