Question:
The line $x - 2y + 4z + 4 = 0$, $x + y + z - 8 = 0$ intersects the plane $x - y + 2z + 1 = 0$ at the point
The line $x - 2y + 4z + 4 = 0$, $x + y + z - 8 = 0$ intersects the plane $x - y + 2z + 1 = 0$ at the point
Updated On: Jan 29, 2026
- (-2,5,1)
- (2,-5,1)
- (2,5,-1)
- (2,5,1)
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The Correct Option is D
Solution and Explanation
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:
\[
\boxed{(2,\ 5,\ 1)}
\]
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