Problem 81 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The plane through the intersection of the planes \(x+y+z=1\) and \(2 x+3 y-z+4=0\) and parallel to \(y\)-axis also passes through the point: \(\quad\) [Jan 09 2019I] (a) \((-3,0,-1)\) (b) \((-3,1,1)\) (c) \((3,3,-1)\) (d) \((3,2,1)\)

Step-by-Step Solution

Answer

The point is (-3, 0, -1) which satisfies all conditions for the plane's equation.

Step 1: Understand the Intersection Line Equation

We start by finding the line of intersection of the planes given by \(x+y+z=1\) (Equation 1) and \(2x+3y-z+4=0\) (Equation 2). This line is the set of points satisfying both plane equations.

Step 2: Parameterize the Line of Intersection

To find the line of intersection, express one variable in terms of a parameter. Let \(z = t\). Substitute \(z = t\) into both equations:From Equation 1: \( x + y + t = 1 \Rightarrow x + y = 1 - t \). From Equation 2: \( 2x + 3y - t + 4 = 0 \Rightarrow 2x + 3y = t - 4 \). Solve these equations to express \(x\) and \(y\) in terms of \(t\).Using Equation 1: \( y = 1 - t - x \) and substitute in Equation 2:\(2x + 3(1 - t - x) = t - 4\)\(2x + 3 - 3t - 3x = t - 4\)Simplifying, \(-x = t - 7\) or \( x = 7 - t \).Substitute \( x = 7 - t \) back into \( y = 1 - t - x \):\( y = 1 - t - (7 - t) = 1 - 7 = -6\).This gives us the parametric form of the intersection line: \((7-t, -6, t)\).

Step 3: Identify Plane Parallel to y-axis Condition

For the plane to be parallel to the \(y\)-axis, its normal vector must not have a \(j\) component (the \(y\)-component). The direction vector of the intersection line should lie in any plane that is parallel to the \(y\)-axis. Hence, any parameterized version \((7-\lambda, -6, \lambda)\) for the line can be used to generate the required plane.

Step 4: Write the Equation of the Plane

The normal vector of the required plane can be derived to have no \(j\)-component. Using a point \((x_0, y_0, z_0)\) on the plane like \((7, -6, 0)\), the equation of the plane can be written as:\( a(x - 7) + b(y + 6) + c(z - 0) = 0 \).For the plane to be parallel to \(y\)-axis, \(b=0\), hence: \( a(x - 7) + c(z) = 0 \) simplifies the condition the plane must satisfy. Choose the point to verify.

Step 5: Check Each Option to Find the Correct Point

Plug each given option into the equation \(a(x - 7) + c(z) = 0\) and check if the equation holds true.Let's test each:Option (a) \((-3,0,-1)\):\(a(-3 - 7) + c(-1) = -10a - c = 0\)Option (b) \((-3,1,1)\):\(-10a + c eq 0\)Option (c) \((3,3,-1)\):\(3 - 7 = -4, -4a - c eq 0\)Option (d) \((3,2,1)\):\(3 - 7 + c(1) = 0\)\(-4a + c eq 0\)Option (a) satisfies the condition - \((-3,0,-1)\) is where the planes intersect.

Key Concepts

Plane Equation

A plane in vector algebra is a flat two-dimensional surface that extends infinitely far. To describe a plane mathematically, we use a plane equation. A common form of this equation is:\[ ax + by + cz = d \]Here, \(a\), \(b\), and \(c\) are the components of the normal vector to the plane, and \(d\) is a constant. This normal vector is crucial because it is perpendicular to every line that lies within the plane.
For example, the plane equations \(x+y+z=1\) and \(2x+3y-z+4=0\) each represent different planes in three-dimensional space. The first plane has a normal vector of \([1,1,1]\), while the second plane has a normal vector of \([2,3,-1]\).
These vectors indicate the planes' orientations in space, and their intersection (if any) results in a line or a single point.

Line of Intersection

When two planes intersect, they do so along a line, assuming the planes are not parallel. The line of intersection consists of all points that satisfy both plane equations simultaneously.
To find this line analytically, one can set a variable (usually the one with the simplest coefficients) to a parameter, say, \(t\), and solve for the other variables in terms of this parameter. This is demonstrated in our step-by-step solution, where we set \(z = t\), and substitute into the two plane equations. Doing so yields a system of equations that is simpler to solve.

• First equation: \[ x + y + t = 1 \rightarrow x + y = 1 - t \]
• Second equation: \[ 2x + 3y = t - 4 \]

These modified equations allow us to express \(x\) and \(y\) in terms of \(t\), ultimately revealing the parametric form of the line of intersection.

Direction Vector

In vector algebra, the direction vector is a fundamental concept used to describe the orientation of lines in space. It is a vector parallel to a given line.
For a line resulting from the intersection of two planes, the direction vector is derived by finding a vector parallel to both normal vectors of the intersecting planes, but not perpendicular to the line itself.In our example, using the parametric form \((7-t, -6, t)\), it's clear that the direction vector \([-1, 0, 1]\) directs the line. This vector comes from the change in each coordinate when \(t\) increases by 1.

• The first coordinate decreases by 1.
• The second coordinate remains constant.
• The third coordinate increases by 1.

Understanding these changes helps predict points along the line's path and visualizes its traverse through space.

Parametric Form

Describing a line in parametric form involves expressing each coordinate of a line as a function of a parameter. This method provides insights into how each coordinate "walks" along the line.
The parametric equations derived for the line of intersection in our scenario are:

• \(x = 7 - t\)
• \(y = -6\)
• \(z = t\)

These equations effectively give a formula for every point on the line, using \(t\) as a movable pointer. By adjusting \(t\), we can find any specified point by plugging the value of \(t\) into these expressions.
Thus, the parametric form simplifies studying geometric properties of lines within the context of vector algebra and aids in solving complex spatial problems.