Problem 119 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The values of a for which the two points \((1, a, 1)\) and \((-3,0, a)\) lie on the opposite sides of the plane \(3 x+4 y-\) \(12 z+13=0\), satisfy \(\quad\) [Online May 7, 2012] (a) \(0

Step-by-Step Solution

Answer

The correct answer is not listed among given options. Correct interpretation would be: either \(a < -1\) or \(a > \frac{1}{3}\).

Step 1: Determine the Plane Equation Substitution for the Point (1, a, 1)

Substitute the point \((1, a, 1)\) into the plane equation \(3x + 4y - 12z + 13 = 0\). The substitution gives: \[ 3(1) + 4(a) - 12(1) + 13 = 0 \]Simplify this: \[ 3 + 4a - 12 + 13 = 4a + 4 \] Therefore, the value derived from the substitution is \(4a + 4\). This determines on which side of the plane the point lies.

Step 2: Determine Plane Equation Substitution for the Point (-3, 0, a)

Now, substitute the point \((-3, 0, a)\) into the plane equation: \[ 3(-3) + 4(0) - 12(a) + 13 = 0 \]Simplify it to get: \[ -9 - 12a + 13 = 4 - 12a \]This result \(4 - 12a\) informs us about the position of the point relative to the plane.

Step 3: Compare the Signs of the Two Expressions

For the points to be on opposite sides of the plane, the product of their respective plane equation results must be negative: \[ (4a + 4)(4 - 12a) < 0 \] Expand and simplify the inequality: \[ 16 - 48a + 16a - 48a^2 < 0 \]\[ -48a^2 - 32a + 16 < 0 \]Divide the entire inequality by -16 to simplify and change the inequality sign:\[ 3a^2 + 2a - 1 > 0 \]

Step 4: Solve the Quadratic Inequality

Factor the quadratic inequality: \[ 3a^2 + 2a - 1 = (3a - 1)(a + 1) > 0 \]To solve \((3a - 1)(a + 1) > 0\), determine the interval of solutions. The roots of the corresponding equation are \(a = \frac{1}{3}\) and \(a = -1\). Test intervals around these roots: Recall, the inequality holds when either both terms are positive or both are negative.

Step 5: Determine the Valid Solution Range

Analyze intervals between the roots:- For \(a < -1\), both factors are negative, making the product positive.- For \(-1 < a < \frac{1}{3}\), the product becomes negative.- For \(a > \frac{1}{3}\), both factors are positive, making the product positive.Thus, the inequality solution is \(a < -1\) or \(a > \frac{1}{3}\).

Key Concepts

Plane Equation

Understanding the plane equation is fundamental in 3D Geometry. A plane in 3D space can be described using a linear equation, usually in the form \(Ax + By + Cz + D = 0\).
Here, \(A, B,\) and \(C\) are the coefficients that determine the orientation of the plane, and \(D\) adjusts its position relative to the origin.When a point \((x, y, z)\) is substituted into the equation, it indicates whether the point lies on the plane or which side of the plane it is on.
If the result of substitution equals zero, the point is on the plane; if positive or negative, it reveals the side.
In the exercise, points \( (1, a, 1) \) and \((-3, 0, a)\) were substituted into the plane equation resulting in expressions that tell on which side each point lies.

Quadratic Inequality

Quadratic inequalities, compared to simple linear inequalities, involve higher levels of complexity as they have a second-degree variable (like \(a^2\)).
In this case, the exercise leads to an inequality \(3a^2 + 2a - 1 > 0\).
To solve it, factor the inequality to determine when the expression is greater than zero.After factoring, we test the intervals defined by the roots of the equation to determine when the inequality holds.
The roots are \(a = \frac{1}{3}\) and \(a = -1\). This process involves:

• Dealing with intervals created between these roots.
• Determining where the product of the factors is positive based on the sign changes.

By properly analyzing these intervals, we find the solutions \(a < -1\) or \(a > \frac{1}{3}\), reflecting intervals where the original inequality is satisfied.

Opposite Sides

Determining if two points lie on opposite sides of a plane involves observing the signs of the substituted results.
If two points \( (x_1, y_1, z_1)\) and \( (x_2, y_2, z_2)\) yield values \(V_1\) and \(V_2\) upon substitution into the plane equation, opposite sides are indicated by a change in sign.This means that if \(V_1 imes V_2 < 0\), the points lie on opposite sides.
For the exercise:

• The result from the first substitution was \(4a + 4\).
• The second was \(4 - 12a\).

The product of these values needs to be negative to confirm the opposite side condition, which is reflected by solving the inequality \((4a + 4)(4 - 12a) < 0\).

Substitution Method

The substitution method is crucial in this exercise for finding if points lie on or which side of a plane they are.
Simply substitute the coordinates of the point into the plane equation.
It's a quick way to test the positioning of points relative to the plane.Each step involves calculating a simple numerical value by substituting the \(x, y, z\) coordinates into the plane equation and simplifying.

• If the result is zero, the point lies on the plane.
• If the result is negative or positive, it reveals which side the point lies on.

This technique enables determining spatial relationships in a systematic manner, ideal for complex 3D geometrical scenarios where opposite sides need to be evaluated.