Question

The equation of the plane through the intersection of the planes \(x + y + z = 1\) and \(2x + 3y - z + 4 = 0\) and parallel to X-axis, is

• A. \(y - z + 6 = 0\)
• B. \(3y - z + 6 = 0\)
• C. \(y + 3z + 6 = 0\)
• D. \(3y - 2z + 6 = 0\)

Hint:

Plane parallel to X-axis means its equation has no x term.

Answer 1

The Correct Option is A

To find the equation of the plane through the intersection of the given planes and parallel to the X-axis, we proceed as follows:

The equation of any plane passing through the intersection of two planes \(P_1: x + y + z = 1\) and \(P_2: 2x + 3y - z + 4 = 0\) can be given by the linear combination:

\(\pi_1 + \lambda \pi_2 = 0\)

This translates to:

\( (x + y + z - 1) + \lambda (2x + 3y - z + 4) = 0\)

Simplifying, we get:

\( x + y + z - 1 + \lambda(2x + 3y - z + 4) = 0 \) \( (1 + 2\lambda)x + (1 + 3\lambda)y + (1 - \lambda)z + (4\lambda - 1) = 0 \)

Since the plane is parallel to the X-axis, the coefficient of \(x\) should be zero:

\(1 + 2\lambda = 0\)

Solving for \(\lambda\), we get:

\(\lambda = -\frac{1}{2}\)

Substituting \(\lambda = -\frac{1}{2}\) in the equation, we get:

\(0 \cdot x + \left(1 - \frac{3}{2}\right)y + \left(1 + \frac{1}{2}\right)z + \left(4\left(-\frac{1}{2}\right) - 1\right) = 0\) \(-\frac{1}{2}y + \frac{3}{2}z - 3 = 0\)

Multiplying throughout by 2 for simplicity:

\(-y + 3z - 6 = 0\) or \(y - z + 6 = 0\)

Therefore, the equation of the plane parallel to the X-axis through the intersection is:

\(y - z + 6 = 0\)

Thus, the correct answer is \((y - z + 6 = 0)\).

Other options are incorrect because they don't satisfy the condition of being parallel to the X-axis.