Question

The line $x - 2y + 4z + 4 = 0$, $x + y + z - 8 = 0$ intersects the plane $x - y + 2z + 1 = 0$ at the point

• A. (-2,5,1)
• B. (2,-5,1)
• C. (2,5,-1)
• D. (2,5,1)

Answer 1

The Correct Option is D

We are given a line defined by the intersection of two planes: \[ \begin{cases} x - 2y + 4z + 4 = 0 \quad \text{(1)} \\ x + y + z - 8 = 0 \quad \text{(2)} \end{cases} \] and a plane: \[ x - y + 2z + 1 = 0 \quad \text{(3)} \] We need to find the point where this line intersects the plane (3). Step 1: Parametrize the line To do this, we solve (1) and (2) simultaneously and express in terms of a parameter. Let’s solve (1) and (2) for \(x\), \(y\), \(z\) in terms of a parameter (say \(z = t\)): From equation (2): \[ x + y + z = 8 \Rightarrow x + y = 8 - z = 8 - t \tag{A} \] From equation (1): \[ x - 2y + 4z = -4 \Rightarrow x - 2y + 4t = -4 \Rightarrow x - 2y = -4 - 4t \tag{B} \] Now, solve equations (A) and (B) simultaneously: Add (A) and (B): \[ (x + y) + (x - 2y) = (8 - t) + (-4 - 4t) \Rightarrow 2x - y = 4 - 5t \tag{C} \] Now, multiply (A) by 2: \[ 2x + 2y = 16 - 2t \tag{D} \] Now subtract (C) from (D): \[ (2x + 2y) - (2x - y) = (16 - 2t) - (4 - 5t) \Rightarrow 3y = 12 + 3t \Rightarrow y = 4 + t \] Substitute in (A): \[ x + (4 + t) = 8 - t \Rightarrow x = 4 - 2t \] So, the parametric form of the line is: \[ x = 4 - 2t,\quad y = 4 + t,\quad z = t \] Step 2: Plug into the plane equation Substitute into equation (3): \[ x - y + 2z + 1 = 0 \Rightarrow (4 - 2t) - (4 + t) + 2t + 1 = 0 \Rightarrow 4 - 2t - 4 - t + 2t + 1 = 0 \Rightarrow (-t + 1) = 0 \Rightarrow t = 1 \] Step 3: Find the coordinates of the point Using \(t = 1\): \[ x = 4 - 2(1) = 2,\quad y = 4 + 1 = 5,\quad z = 1 \] Final Answer: \[ \boxed{(2,\ 5,\ 1)} \]