Skip to main content
Back to 3D Geometry
JEE Main 2020
3D Geometry
3D Geometry
Easy

Question

If the line, x32=y+21=z+43{{x - 3} \over 2} = {{y + 2} \over { - 1}} = {{z + 4} \over 3}\, lies in the planes, lx+myz=9,lx+my-z=9, then l2+m2{l^2} + {m^2} is equal to :

Options

Solution

Key Concepts and Formulas

  1. Line in Symmetric Form: A line passing through a point (x0,y0,z0)(x_0, y_0, z_0) with direction vector d=(a,b,c)\vec{d} = (a, b, c) is represented as xx0a=yy0b=zz0c\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}.
  2. Plane Equation: The general equation of a plane is Ax+By+Cz=DAx+By+Cz=D, where n=(A,B,C)\vec{n} = (A, B, C) is the normal vector to the plane.
  3. Conditions for a Line Lying in a Plane: For a line to lie entirely within a plane, two crucial conditions must be satisfied:
    • Point Condition: Any point on the line must also satisfy the equation of the plane.
    • Direction Perpendicularity Condition: The direction vector of the line (d\vec{d}) must be perpendicular to the normal vector of the plane (n\vec{n}). Mathematically, their dot product must be zero: dn=0\vec{d} \cdot \vec{n} = 0.

Step-by-Step Solution

1. Extract Information from the Given Line and Plane Equations

  • From the line equation: The given line is x32=y+21=z+43{{x - 3} \over 2} = {{y + 2} \over { - 1}} = {{z + 4} \over 3}\, From this, we identify:

    • A point on the line, P0(x0,y0,z0)=(3,2,4)P_0(x_0, y_0, z_0) = (3, -2, -4).
    • The direction vector of the line, d=(2,1,3)\vec{d} = (2, -1, 3).
  • From the plane equation: The given plane is lx+myz=9lx+my-z=9 From this, we identify:

    • The normal vector to the plane, n=(l,m,1)\vec{n} = (l, m, -1).

2. Apply the Point Condition

  • What we're doing: We use the fact that if the line lies in the plane, then the point P0(3,2,4)P_0(3, -2, -4) which lies on the line must also satisfy the equation of the plane lx+myz=9lx + my - z = 9.
  • Why: This condition ensures that the line actually touches the plane at a specific point, which is a prerequisite for it to lie within the plane. It also helps establish a linear relationship between the unknown coefficients ll and mm.

Substituting the coordinates of P0(3,2,4)P_0(3, -2, -4) into the plane equation lx+myz=9lx + my - z = 9, we establish the first linear equation relating ll and mm: 3l2m=4(1)3l - 2m = 4 \quad \ldots(1)

3. Apply the Direction Perpendicularity Condition

  • What we're doing: We use the condition that the direction vector of the line must be perpendicular to the normal vector of the plane.
  • Why: If the line lies in the plane, it is parallel to the plane. Since the normal vector is perpendicular to the plane, it must also be perpendicular to any line lying within or parallel to that plane. This gives us another linear relationship between ll and mm.

The condition for perpendicularity is that the dot product of the two vectors is zero: dn=0\vec{d} \cdot \vec{n} = 0 Substituting the components of d=(2,1,3)\vec{d} = (2, -1, 3) and n=(l,m,1)\vec{n} = (l, m, -1): (2)(l)+(1)(m)+(3)(1)=0(2)(l) + (-1)(m) + (3)(-1) = 0 2lm3=02l - m - 3 = 0 Adding 3 to both sides, we obtain our second linear equation: 2lm=3(2)2l - m = 3 \quad \ldots(2)

4. Solve the System of Linear Equations

We now have a system of two linear equations with two variables ll and mm:

  1. 3l2m=43l - 2m = 4
  2. 2lm=32l - m = 3

We can solve this system using the substitution method. From equation (2), express mm in terms of ll: m=2l3m = 2l - 3 Substitute this expression for mm into equation (1): 3l2(2l3)=43l - 2(2l - 3) = 4 3l4l+6=43l - 4l + 6 = 4 l+6=4-l + 6 = 4 l=46-l = 4 - 6 l=2-l = -2 l=2l = 2 Now substitute the value of l=2l=2 back into the expression for mm: m=2(2)3m = 2(2) - 3 m=43m = 4 - 3 m=1m = 1 So, we have found l=2l=2 and m=1m=1.

5. Calculate the Required Value

The question asks for the value of l2+m2l^2 + m^2. Substitute the values l=2l=2 and m=1m=1: l2+m2=(2)2+(1)2l^2 + m^2 = (2)^2 + (1)^2 l2+m2=4+1l^2 + m^2 = 4 + 1 l2+m2=5l^2 + m^2 = 5

Common Mistakes & Tips

  • Correctly Extracting Information: Ensure precise identification of the point (x0,y0,z0)(x_0, y_0, z_0) and direction vector (a,b,c)(a, b, c) from the line's symmetric form, and the normal vector (A,B,C)(A, B, C) from the plane's equation. Pay close attention to signs.
  • Understanding Conditions: Don't just memorize the two conditions; understand their geometric meaning. Visualizing a line inside a plane helps reinforce why the direction vector of the line must be perpendicular to the normal vector of the plane.
  • Algebraic Accuracy: Be meticulous with arithmetic, especially when solving the system of linear equations. A small calculation error can lead to an incorrect final answer.

Summary

To determine unknown parameters when a line lies within a plane, we apply two fundamental conditions: first, a known point on the line must satisfy the plane's equation; and second, the line's direction vector must be perpendicular to the plane's normal vector. These conditions generate a system of linear equations, which can then be solved for the unknown parameters. In this problem, solving the system for ll and mm yielded l=2l=2 and m=1m=1, leading to l2+m2=5l^2+m^2=5.

The final answer is 5\boxed{5}, which corresponds to option (A).

Practice More 3D Geometry Questions

View All Questions