Problem 6 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(\alpha \in \mathrm{R}\) and the three vectors \(\vec{a}=\alpha \hat{i}+\hat{j}+3 \hat{k}\), \(\vec{b}=2 \hat{i}+\hat{j}-\alpha \hat{k}\) and \(\vec{c}=\alpha \hat{i}-2 \hat{j}+3 \hat{k}\). Then the set \(S=(\alpha: \vec{a}, \vec{b}\) and \(\vec{c}\) are coplanar) \(\quad\) [April 12, 2019 (II)] (a) is singleton (b) is empty (c) contains exactly two positive numbers (d) contains exactly two numbers only one of which is positive

Step-by-Step Solution

Answer

(b) is empty.

Step 1: Understand Coplanarity Condition

The vectors \( \vec{a}, \vec{b}, \vec{c} \) are coplanar if their scalar triple product is zero. This means \( \vec{a} \cdot (\vec{b} \times \vec{c}) = 0 \). We need to compute \( \vec{b} \times \vec{c} \), and then take the dot product with \( \vec{a} \).

Step 2: Calculate Cross Product \( \vec{b} \times \vec{c} \)

Calculate the cross product \( \vec{b} \times \vec{c} \) using the determinant formula:\[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 1 & -\alpha \ \alpha & -2 & 3 \end{vmatrix} \ = (1*3-(-\alpha)(-2))\hat{i} - (2*3-(-\alpha)*3)\hat{j} + (2*(-2)-1*\alpha)\hat{k}. \]Solving, we get:\[ \vec{b} \times \vec{c} = (3-2\alpha)\hat{i} - (6-3\alpha)\hat{j} + (-4-\alpha)\hat{k}. \]

Step 3: Dot Product \( \vec{a} \cdot (\vec{b} \times \vec{c}) \)

Compute the dot product:\[ \vec{a} \cdot ((3-2\alpha)\hat{i} - (6-3\alpha)\hat{j} + (-4-\alpha)\hat{k}). \]Calculating step-by-step: \[ \alpha(3-2\alpha) + 1(-6+3\alpha) + 3(-4-\alpha) \]Simplifying:\[ 3\alpha - 2\alpha^2 - 6 + 3\alpha - 12 - 3\alpha \]This equals:\[ -2\alpha^2 + 3\alpha - 18. \]

Step 4: Solve \(-2\alpha^2 + 3\alpha - 18 = 0\)

This is a quadratic equation in the form \( a\alpha^2 + b\alpha + c = 0 \), with \( a = -2 \), \( b = 3 \), \( c = -18 \). We solve it using the quadratic formula: \[ \alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \]Substituting, we have:\[ \alpha = \frac{-3 \pm \sqrt{3^2 - 4(-2)(-18)}}{-4}. \]Calculate:\[ \alpha = \frac{-3 \pm \sqrt{9 - 144}}{-4}. \]Since \( 9 - 144 = -135 \), the discriminant is negative, indicating there are no real solutions.

Key Concepts

Scalar Triple Product

To determine if vectors are coplanar, we use the scalar triple product. This involves three vectors, say \( \vec{a}, \vec{b}, \vec{c} \), and is expressed as \( \vec{a} \cdot (\vec{b} \times \vec{c}) \). If this product is zero, then the vectors are coplanar.
This calculation involves two steps:

• First, calculate the cross product \( \vec{b} \times \vec{c} \).
• Then, compute the dot product of \( \vec{a} \) with the cross product result.

The scalar triple product geometrically represents the volume of the parallelepiped formed by the vectors. When the volume is zero, it implies that the vectors lie in the same plane. This visualization is crucial to grasp why a zero scalar triple product indicates coplanarity.

Cross Product

The cross product, represented as \( \vec{b} \times \vec{c} \), is a vector product that yields a vector perpendicular to the plane containing \( \vec{b} \) and \( \vec{c} \).
To compute the cross product:

• Use the determinant of a matrix constructed from the standard unit vectors \( \hat{i}, \hat{j}, \hat{k} \) and the components of vectors \( \vec{b} \) and \( \vec{c} \).
• This operation helps in finding elements like \( (3-2\alpha)\hat{i}, -(6-3\alpha)\hat{j}, (-4-\alpha)\hat{k} \), showing how components interact.

The result of the cross product is a vector orthogonal to \( \vec{b} \) and \( \vec{c} \), fundamental for calculating the scalar triple product. Understanding this helps in geometry and 3D vector analysis.

Dot Product

The dot product, in this context, \( \vec{a} \cdot (\vec{b} \times \vec{c}) \), involves two vectors and results in a scalar.
This product reflects the magnitude of \( \vec{a} \) in the direction of \( \vec{b} \times \vec{c} \).
It is essential to:

• Take each component of \( \vec{a} \) and multiply it with the respective component of \( \vec{b} \times \vec{c} \).
• Sum these products to get a single scalar quantity.

This scalar essentially tells you how much of one vector aligns with another. In the coplanarity context, a result of zero is key as it means the vectors lie flat in the same plane, contributing no volume.

Quadratic Equation

Quadratic equations appear when solving polynomial equations of the second degree, such as \( -2\alpha^2 + 3\alpha - 18 = 0 \).
To solve this equation, you can use the quadratic formula:\[ \alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Key aspects of solving quadratic equations include:

• Identifying coefficients \( a, b, c \).
• Calculating the discriminant \( b^2 - 4ac \).
• Determining the nature of roots (real, complex, or repeated).

For our problem, a negative discriminant indicates no real roots, hence no real values of \( \alpha \) satisfy coplanarity. Understanding the nature of solutions in quadratic equations helps determine possible real solutions and their implications in geometric scenarios.