If the plane P passes through the intersection of two mutually perpendicular planes 2x + ky – 5z = 1 and 3kx – ky + z = 5, k < 3 and intercepts a unit length on positive x-axis, then the intercept made by the plane P on the y-axis is
If the plane P passes through the intersection of two mutually perpendicular planes 2x + ky – 5z = 1 and 3kx – ky + z = 5, k < 3 and intercepts a unit length on positive x-axis, then the intercept made by the plane P on the y-axis is
\(\frac{1}{11}\)
\(\frac{5}{11}\)
- 6
- 7
The Correct Option is D
Approach Solution - 1
To solve this problem, we need to determine the intercept made by the plane P on the y-axis, given that it passes through the intersection of two mutually perpendicular planes and makes a unit length intercept on the positive x-axis.
Step-by-Step Solution:
- First, identify the equations of the two perpendicular planes:
- Plane 1: \( 2x + ky - 5z = 1 \)
- Plane 2: \( 3kx - ky + z = 5 \)
- Two planes are perpendicular if their normal vectors are orthogonal. The normal vector of Plane 1 is \((2, k, -5)\) and for Plane 2 is \((3k, -k, 1)\). Their dot product should be zero:
- \(2 \cdot 3k + k(-k) + (-5) \cdot 1 = 0\)
- This simplifies to \(6k - k^2 - 5 = 0\).
- Rearranging gives: \(k^2 - 6k + 5 = 0\).
- Solve for \(k\) from the quadratic equation:
- Using the quadratic formula: \(k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -6\), \(c = 5\).
- \(k = \frac{6 \pm \sqrt{36 - 20}}{2} = \frac{6 \pm \sqrt{16}}{2} = \frac{6 \pm 4}{2}\).
- The solutions are \(k = 5\) and \(k = 1\).
- Since \(k < 3\), we select \(k = 1\).
- Substituting \(k = 1\) into the plane equations, we now have:
- Plane 1: \(2x + y - 5z = 1\)
- Plane 2: \(3x - y + z = 5\)
- The equation of Plane \(P\), which passes through the intersection of these planes, can be written as:
- \(P: (2x + y - 5z) + \lambda(3x - y + z) = 0\)
- This simplifies to: \( (2 + 3\lambda)x + (1 - \lambda)y + (-5 + \lambda)z = 1 + 5\lambda\).
- Since the plane makes a unit intercept on the x-axis, substitute \(y = 0\), \(z = 0\), and \(x = 1\):
- \((2 + 3\lambda)(1) = 1 + 5\lambda\)
- Solving: \(2 + 3\lambda = 1 + 5\lambda\)
- This gives \(\lambda = \frac{1}{2}\).
- Substituting \(\lambda = \frac{1}{2}\) in the plane equation, the equation becomes:
- \((2 + \frac{3}{2})x + (1 - \frac{1}{2})y + (-5 + \frac{1}{2})z = 1 + 5 \times \frac{1}{2}\)
- This simplifies to \(\frac{7}{2}x + \frac{1}{2}y - \frac{9}{2}z = \frac{11}{2}\).
- Rewriting the equation in x, y, z intercept form:
- \(\frac{x}{\frac{11}{7}} + \frac{y}{11} - \frac{z}{\frac{11}{9}} = 1\)
- The y-intercept is \(11\).
Hence, the y-intercept of the plane \(P\) is 11.
Approach Solution -2
The correct answer is (D):
P1: 2x+ky-5z = 1
P2 : 3kx – ky + z = 5
∵ P1⊥P2 ⇒ 6k-k2+5 = 0
⇒ k = 1,5
∵ k<3
∴ k = 1
P1 : 2x+y-5z = 1
P2 : (2x+y-5z-1)+λ(3x-y+z-5) = 0
Positive x-axis intercept = 1
\(⇒ \frac{1+5λ}{2+3λ} = 1\)
⇒ λ = \(\frac{1}{2}\)
∵ P : 7x+y-4z = 7
y intercept = 7.
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Concepts Used:
Angle between a Line and a Plane
A line is one example of a one-dimensional figure, which has length but no width. A line is made up of a set of points that is stretched in opposite directions infinitely.
Similarly, when an infinite number of points expanded infinitely in either direction to form a flat surface, it is known as a plane. A set of lines when arranged close by to each other a plane is obtained. A plane is one example of a two-dimensional geometric figure that can be measured in terms of length and width.
The line which is adjacent to the plane is the complement of the angle between and the normal of the plane is called the angle between a line and a plain.