Question

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

• A. \(\frac{1}{11}\)
• B. \(\frac{5}{11}\)
• C. 6
• D. 7

Answer 1

The Correct Option is D

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: 

1. First, identify the equations of the two perpendicular planes:
   • Plane 1: \( 2x + ky - 5z = 1 \)
   • Plane 2: \( 3kx - ky + z = 5 \)
2. 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: \(k^2 - 6k + 5 = 0\).
3. 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 \(k = 1\).
4. Substituting \(k = 1\) into the plane equations, we now have:
   • Plane 1: \(2x + y - 5z = 1\)
   • Plane 2: \(3x - y + z = 5\)
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: \( (2 + 3\lambda)x + (1 - \lambda)y + (-5 + \lambda)z = 1 + 5\lambda\).
6. 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 \(\lambda = \frac{1}{2}\).
7. 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 \(\frac{7}{2}x + \frac{1}{2}y - \frac{9}{2}z = \frac{11}{2}\).
8. 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.

Answer 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.