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 the $x$-axis is

• A. y+3z+6=0
• B. y+3z-6=0
• C. y-3z+6=0
• D. y-3z-6=0

Answer 1

The Correct Option is C

We are given two planes: Plane 1: \( x + y + z = 1 \) Plane 2: \( 2x + 3y - z + 4 = 0 \Rightarrow 2x + 3y - z = -4 \) Let the equation of a plane through their intersection be: \[ \Pi: (x + y + z - 1) + \lambda(2x + 3y - z + 4) = 0 \] Expand: \[ x + y + z - 1 + \lambda(2x + 3y - z + 4) = 0 \Rightarrow x + y + z - 1 + 2\lambda x + 3\lambda y - \lambda z + 4\lambda = 0 \] Group like terms: \[ (1 + 2\lambda)x + (1 + 3\lambda)y + (1 - \lambda)z + (4\lambda - 1) = 0 \tag{1} \] This is the general form of the plane through the line of intersection. We now require this plane to be parallel to the x-axis. So the direction vector \( \vec{d} = (1, 0, 0) \) must lie on the plane. That means the normal vector of the plane should be orthogonal to \( (1, 0, 0) \) From (1), the normal vector is: \[ \vec{n} = ((1 + 2\lambda),\ (1 + 3\lambda),\ (1 - \lambda)) \] For the plane to be parallel to the x-axis, we must have: \[ \vec{n} \cdot (1, 0, 0) = 0 \Rightarrow 1 + 2\lambda = 0 \Rightarrow \lambda = -\frac{1}{2} \] Substitute \( \lambda = -\frac{1}{2} \) into (1): - Coefficient of \( x \): \( 1 + 2(-\frac{1}{2}) = 0 \) - Coefficient of \( y \): \( 1 + 3(-\frac{1}{2}) = 1 - \frac{3}{2} = -\frac{1}{2} \) - Coefficient of \( z \): \( 1 - (-\frac{1}{2}) = \frac{3}{2} \) - Constant term: \( 4(-\frac{1}{2}) - 1 = -2 - 1 = -3 \) So the equation becomes: \[ 0x - \frac{1}{2}y + \frac{3}{2}z - 3 = 0 \Rightarrow -y + 3z - 6 = 0 \Rightarrow \boxed{y - 3z + 6 = 0} \] Final Answer: \[ \boxed{y - 3z + 6 = 0} \]