Let the plane containing the line of intersection of the planes $P 1: x+(\lambda+4) y+z=1$ and $P 2: 2 x+$ $y+z=2$ pass through the points $(0,1,0)$ and $(1,0,1)$ Then the distance of the point $(2 \lambda, \lambda,-\lambda)$ from the plane $P 2$ is
- $5 \sqrt{6}$
- $2 \sqrt{6}$
- $3 \sqrt{6}$
- $4 \sqrt{6}$
The Correct Option is C
Approach Solution - 1
Equation of plane passing through point of intersection of and
Passing through and
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Also passing
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Approach Solution -2
Step 1: Equation of the plane passing through the line of intersection of \( P_1 \) and \( P_2 \)
The equation of the required plane can be written as:
\[ P = P_1 + kP_2, \] where \( P_1 \) is: \[ x + (\lambda + 4)y + z - 1 = 0 \] and \( P_2 \) is: \[ 2x + y + z - 2 = 0. \] Thus, the equation of the plane is: \[ x + (\lambda + 4)y + z - 1 + k(2x + y + z - 2) = 0. \] Simplifying: \[ (1 + 2k)x + (\lambda + 4 + k)y + (1 + k)z - (1 + 2k) = 0. \]
Step 2: Passing through the points \( (0, 1, 0) \) and \( (1, 0, 1) \)
Substitute \( (0, 1, 0) \) into the plane equation:
\[ 0 + (\lambda + 4 + k)(1) + 0 - (1 + 2k) = 0. \] Simplify: \[ \lambda + 4 + k - 1 - 2k = 0 \quad \Rightarrow \quad \lambda + 3 - k = 0 \quad \text{(Equation 1)}. \]
Substitute \( (1, 0, 1) \) into the plane equation:
\[ (1 + 2k)(1) + (\lambda + 4 + k)(0) + (1 + k)(1) - (1 + 2k) = 0. \] Simplify: \[ 1 + 2k + 1 + k - 1 - 2k = 0 \quad \Rightarrow \quad 1 + k = 0. \] Solve: \[ k = -1. \]
Step 3: Solve for \( \lambda \) Substitute \( k = -1 \) into Equation 1: \[ \lambda + 3 - (-1) = 0 \quad \Rightarrow \quad \lambda + 3 + 1 = 0. \] Therefore: \[ \lambda = -4. \]
Step 4: Find the point \( (2\lambda, \lambda, -\lambda) \) Using \( \lambda = -4 \): \[ (2\lambda, \lambda, -\lambda) = (2(-4), -4, -(-4)) = (-8, -4, 4). \]
Step 5: Distance of \( (-8, -4, 4) \) from the plane \( P_2 \) The distance of a point \( (x_1, y_1, z_1) \) from a plane \( ax + by + cz + d = 0 \) is given by the formula: \[ d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}. \] For \( P_2 : 2x + y + z - 2 = 0 \), substitute \( a = 2 \), \( b = 1 \), \( c = 1 \), and \( d = -2 \): \[ d = \frac{|2(-8) + 1(-4) + 1(4) - 2|}{\sqrt{2^2 + 1^2 + 1^2}}. \] Simplify: \[ d = \frac{|-16 - 4 + 4 - 2|}{\sqrt{4 + 1 + 1}} = \frac{|-18|}{\sqrt{6}} = \frac{18}{\sqrt{6}}. \] Rationalize: \[ d = \frac{18\sqrt{6}}{6} = 3\sqrt{6}. \]
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Concepts Used:
Three Dimensional Geometry
Mathematically, Geometry is one of the most important topics. The concepts of Geometry are derived w.r.t. the planes. So, Geometry is divided into three major categories based on its dimensions which are one-dimensional geometry, two-dimensional geometry, and three-dimensional geometry.
Direction Cosines and Direction Ratios of Line:
Consider a line L that is passing through the three-dimensional plane. Now, x,y and z are the axes of the plane and α,β, and γ are the three angles the line makes with these axes. These are commonly known as the direction angles of the plane. So, appropriately, we can say that cosα, cosβ, and cosγ are the direction cosines of the given line L.
