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

Question

If \(\vec{u}, \vec{v}, \vec{w}\) are non-coplanar vectors and \(p, q\) are real numbers, then the equality \([3 \vec{u} p \vec{v} \quad p \vec{w}]-[p \vec{v} \vec{w} q \vec{u}]\) \(-\left[\begin{array}{lll}2 \vec{w} & q \vec{v} & q \vec{u}\end{array}\right]=0\) 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

Option (B): exactly two values of \((p, q)\).

Step 1: Understand the Given Problem

The expression given involves the scalar triple product. The problem asks under what conditions the combination of these triple products equals zero, where \(\vec{u}, \vec{v}, \vec{w}\) are non-coplanar vectors and \ p, q\ are real numbers. The scalar triple product \[ \vec{a} \, \vec{b} \, \vec{c} \] is equal to the determinant of the matrix formed by vectors \vec{a}, \vec{b}, \vec{c}\.

Step 2: Express the Scalar Triple Products

The given expression involves the scalar triple products: \[ [3 \vec{u}, p \vec{v}, p \vec{w}] - [p \vec{v}, \vec{w}, q \vec{u}] - [2 \vec{w}, q \vec{v}, q \vec{u}] = 0. \] Expand each triple product using determinants. For example, \[ [3 \vec{u}, p \vec{v}, p \vec{w}] = 3p^2 (\vec{u} \cdot (\vec{v} \times \vec{w})). \]

Step 3: Calculate each Scalar Triple Product

Compute each part: \[ [3 \vec{u}, p \vec{v}, p \vec{w}] = 3p^2 (\vec{u} \cdot (\vec{v} \times \vec{w})), \ [p \vec{v}, \vec{w}, q \vec{u}] = pq (\vec{v} \cdot (\vec{w} \times \vec{u})), \ [2 \vec{w}, q \vec{v}, q \vec{u}] = 2q^2 (\vec{w} \cdot (\vec{v} \times \vec{u})). \] Note the cyclic nature and switch of order in determinants, which introduces sign changes.

Step 4: Substitute and Simplify

Substitute these expressions back into the condition: \[ 3p^2 (\vec{u} \cdot (\vec{v} \times \vec{w})) - pq (\vec{v} \cdot (\vec{w} \times \vec{u})) - 2q^2 (\vec{w} \cdot (\vec{v} \times \vec{u})) = 0. \] Simplifying, since \ (\vec{a} \cdot (\vec{b} \times \vec{c})) = (\vec{b} \cdot (\vec{c} \times \vec{a})) = (\vec{c} \cdot (\vec{a} \times \vec{b})), \ we have: \[ 3p^2 - pq - 2q^2 = 0. \]

Step 5: Solve the Quadratic Equation

The equation \[ 3p^2 - pq - 2q^2 = 0 \] is a quadratic in terms of \(p\) for a fixed \(q\). Solving this quadratic using the quadratic formula, \ p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \, with \ a = 3, b = -q, \ and \ c = -2q^2 \, the discriminant \[ q^2 + 24q^2 = 25q^2 \] is always positive. Therefore, there are always two real roots for \(p\) given \(q eq 0\).

Step 6: Conclusion

Since the quadratic equation in \(p\) has two solutions for each nonzero \ \ \(q\), this means that the original vector condition is true for exactly two pairs of \(p, q\) where \(q eq 0\). Therefore, the correct option is (B) exactly two values of \(p, q\).

Key Concepts

Non-Coplanar Vectors

Vectors are considered non-coplanar if they do not lie on the same plane. In simpler terms, think of non-coplanar vectors as directions in space that form a three-dimensional structure, where no two vectors can be contained within a single plane along with a third vector. This is an essential property because it ensures that volume can be described exactly without collapsing into a flat shape.
Why are non-coplanar vectors important? They allow you to calculate volumes and solve spatial problems in vector mathematics. In our given problem, the vectors \(\vec{u}, \vec{v}, \vec{w}\) being non-coplanar helps unequivocally define a new space using these vectors. This sort of arrangement is crucial when working with the scalar triple product or any problem involving space dimensions beyond a flat surface.
Some key points about non-coplanar vectors:

• They form the basis of three-dimensional vector spaces.
• They allow for the representation of volume through the scalar triple product.
• If the scalar triple product of three vectors is non-zero, then the vectors are non-coplanar.

Determinant

The determinant is a special number that can be calculated from a square matrix. It holds significant meaning in linear algebra, notably in understanding whether a system of equations has solutions, and how those solutions might be oriented.
When dealing with vectors, the scalar triple product is a specific application of the determinant, where it evaluates the volume of the parallelepiped spanned by three vectors. In the exercise problem, the determinant helps to simplify the expression of several scalar triple products.
To calculate the determinant of a 3x3 matrix \[\begin{bmatrix}a & b & c \d & e & f \g & h & i \\end{bmatrix}\]use the formula:\[a(ei - fh) - b(di - fg) + c(dh - eg)\]
Understanding the determinant:

• If the determinant is non-zero, it means that the vectors are non-coplanar.
• A zero determinant indicates linear dependence, meaning the vectors lie within the same plane.
• The determinant sign can indicate orientation, with positive and negative values reflecting different directions.

Quadratic Equation

Quadratic equations are a form of polynomial equation of degree 2, commonly written as \(ax^2 + bx + c = 0\). The solutions to this equation are found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In our specific exercise, we derive a quadratic equation in the context of solving for the variable \(p\) with a fixed \(q\). Here it appears as:\[ 3p^2 - pq - 2q^2 = 0 \]
Why are quadratic equations important here? They offer a structured way to find possible values of our unknowns (\(p\) in this case), giving two real solutions when the discriminant (\(b^2 - 4ac\)) is positive.
Key features of quadratic equations:

• The discriminant determines the nature of roots (two distinct, one repeated, or no real roots).
• Quadratic equations are easily solved using the quadratic formula when in their standard form.
• These equations frequently appear in physics, engineering, and many real-world problems.

Solving the quadratic equation in the exercise tells us that there are two possible values of \(p\) for each nonzero \(q\), hence why only two \((p, q)\) pairs satisfy the given condition in the problem.