Problem 67 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

A plane passing through the points \((0,-1,0)\) and \((0,0,1)\) and making an angle \(\frac{\pi}{4}\) with the plane \(\mathrm{y}-\mathrm{z}+5=0\), also passes through the point: [April 09 2019I] (a) \((-\sqrt{2}, 1,-4)\) (b) \((\sqrt{2},-1,4)\) (c) \((-\sqrt{2},-1,-4)\) (d) \((\sqrt{2}, 1,4)\)

Step-by-Step Solution

Answer

Option (b) \((\sqrt{2},-1,4)\) satisfies the plane equation.

Step 1: Determine the line from two points

The two points \( A = (0, -1, 0) \) and \( B = (0, 0, 1) \) lie on the plane. The direction vector of the line passing through these points can be determined as \( \mathbf{BA} = B - A = (0, 0, 1) - (0, -1, 0) = (0, 1, 1) \).

Step 2: Identify the normal to the given plane

The equation of the given plane is \( ext{y} - ext{z} + 5 = 0\). This plane has a normal vector \((0, 1, -1)\).

Step 3: Apply angle condition between planes

The angle between the desired plane and the given plane is \(rac{\pi}{4}\). If \(\mathbf{n_1} = (a, b, c)\) is the normal of the unknown plane, we use the formula for the angle between two planes: \( ext{cos} \theta = \frac{|\mathbf{n_1} \cdot \mathbf{n_2}|}{||\mathbf{n_1}|| ||\mathbf{n_2}||}\). Hence, \[\frac{| a \cdot 0 + b \cdot 1 + c \cdot (-1) |}{\sqrt{a^2 + b^2 + c^2} \sqrt{1 + 1}} = \frac{1}{\sqrt{2}}\] This simplifies to \( |b - c | = \frac{1}{\sqrt{2}} \sqrt{a^2 + b^2 + c^2} \).

Step 4: Determine the equation of the plane

The plane must pass through both given points and have normal perpendicular to the direction vector of the line. Let the normal vector be \(\mathbf{n} = (a, b, c) = (1, b, c)\) so that \(\mathbf{n} \cdot \mathbf{BA} = 0 \), thus \(1\cdot 0 + b\cdot 1 + c\cdot 1 = 0\) leads to \(b + c = 0\). Substituting in the condition from the angle gives \( |b + (-b)| = \frac{1}{\sqrt{2}} ||(1, b, -b)|| \). Solving the two equations, we derive \( b = \frac{1}{\sqrt{2}} \) and \( c = -\frac{1}{\sqrt{2}} \).

Step 5: Plug in point to test true equation

Test each point to find one that satisfies the equation of the plane \(x + \frac{1}{\sqrt{2}}y - \frac{1}{\sqrt{2}}z = \text{constant} \). Use point \( (0, -1, 0) \) to find \( x - \frac{1}{\sqrt{2}}\cdot1 = \frac{1}{\sqrt{2}}(-1)\), thus, constant = 0. Check which point satisfies \( x + \frac{1}{\sqrt{2}}y - \frac{1}{\sqrt{2}}z = 0\) for each listed option.

Key Concepts

Plane Equation

A plane equation is a mathematical expression that characterizes all the points lying on a specific plane in a three-dimensional space. Typically, a plane equation in 3D space is given by:

• \( ax + by + cz = d \)

where \(a, b,\) and \(c\) are the coefficients that determine the "position" and "orientation" of the plane, and \(d\) is a constant. The set of all points \((x, y, z)\) that satisfies this equation forms the plane.

The normal vector to the plane is \((a, b, c)\), which is crucial for understanding the plane's orientation. In this exercise, we're aiming to find a plane that passes through given points and meets specific angle conditions with another plane. By solving the step-by-step conditions, we determine the proper coefficients \(a, b,\) and \(c\) that fit all these criteria.

Direction Vector

In vector geometry, a direction vector indicates the direction of a line in three-dimensional space. This vector is crucial when determining the equation of a line and understanding its geometric properties.

Given two points, \(A = (x_1, y_1, z_1)\) and \(B = (x_2, y_2, z_2)\), the direction vector, \(\mathbf{AB}\), is calculated as:

• \( \mathbf{AB} = B - A = (x_2 - x_1, y_2 - y_1, z_2 - z_1) \)

In the original exercise, the direction vector of the line passing through points \( (0, -1, 0) \) and \( (0, 0, 1) \) was found to be \( (0, 1, 1) \). This vector helps to define the line segment connecting the two points.

The direction vector is instrumental in ensuring that a plane is perpendicular to a line, which is necessary for certain mathematical conditions while solving plane equations.

Normal Vector

A normal vector is a vector that is perpendicular to a surface. In the context of a plane, the normal vector is instrumental in determining the plane's orientation and is often used in the equation of the plane.

For a plane defined by \(ax + by + cz = d\), the normal vector is \((a, b, c)\). This vector points directly orthogonal to the plane's surface and is fundamental in geometry, calculations involving the plane's orientation, and determining angles with other planes.

In the given exercise, one plane had a known normal vector of \((0, 1, -1)\), derived directly from the equation \(y - z + 5 = 0\). Identifying the correct normal vector for a new plane can be challenging, as it must satisfy conditions like being perpendicular to certain direction vectors from a line. This perpendicular relation is confirmed by the dot product equation \(\mathbf{n} \cdot \mathbf{d} = 0\), where \(\mathbf{d}\) is a direction vector.

Angle Between Planes

The angle between two planes is defined by the angle formed by their normal vectors. The mathematical formula to find this angle, \(\theta\), is based on the dot product of the normal vectors \(\mathbf{n_1}\) and \(\mathbf{n_2}\) as follows:

• \(\text{cos} \theta = \frac{|\mathbf{n_1} \cdot \mathbf{n_2}|}{||\mathbf{n_1}|| ||\mathbf{n_2}||}\)

This relation helps us solve problems where we need to know how much one plane is tilted in relation to another.

In our exercise, the given condition was that the plane makes an angle \(\frac{\pi}{4}\) (or 45 degrees) with another plane. To fulfill this condition and find the correct normal vector for the unknown plane, you must solve equations that result from substituting known values into this formula.
This helps ensure that both geometric constraints (like intersecting directions and point belonging) and numeric constraints (angle relationships) are satisfied.