The Geometry of Infinite Possibilities
Imagine you are standing in a vast, three-dimensional space. You have three flat, infinite sheets of paper—our planes—defined by the equations:
P1:x+2y+3z=3
P2:4x+3y−4z=4
P3:8x+4y−λz=9+μ
Usually, three planes intersect at a single point, like the corner of a room. However, the problem states that this system has infinitely many solutions.
This means our three planes are not meeting at a point; they are all slicing through space to meet along a single, shared line. Think of the binding of an open book, where every page meets at the same spine. That spine is our line of intersection.
The Art of Reduction
We want to find the values of λ and μ that force this geometric alignment. To do this, we need to simplify our world. We have three equations in three variables, but we can reduce this to two equations in two variables by eliminating x.
First, let's look at P1 and P2. To eliminate x, we multiply P1 by 4 to match the coefficient of x in P2:
4(x+2y+3z)=4(3)⇒4x+8y+12z=12
Now, subtract P2 from this new equation:
(4x+8y+12z)−(4x+3y−4z)=12−4
This gives us our first reduced equation, which we will call Equation 4:
The Second Slice
Now, we repeat the process for P2 and P3. We need to eliminate x again. The coefficient of x in P2 is 4, and in P3 it is 8. So, we multiply P2 by 2:
2(4x+3y−4z)=2(4)⇒8x+6y−8z=8
Now, subtract P3 from this modified P2:
(8x+6y−8z)−(8x+4y−λz)=8−(9+μ)
This simplifies beautifully to our second reduced equation, Equation 5:
The Condition for Coincidence
Here is the crux of the problem. We have reduced our 3D system to two lines in a 2D plane: 5y+16z=8 and 2y+(λ−8)z=−1−μ. For the original system to have infinitely many solutions, these two lines must be the same line.
In algebra, two lines are identical if and only if their coefficients are proportional. This is the moment where the geometry meets the algebra:
The Final Calculation
Now, it is just a matter of solving for our unknowns. First, we equate the first two ratios to find λ:
25=λ−816⇒5(λ−8)=32⇒5λ−40=32⇒5λ=72⇒λ=572
Next, we equate the first and third ratios to find μ:
25=−1−μ8⇒5(−1−μ)=16⇒−5−5μ=16⇒−5μ=21⇒μ=−521
We have arrived at our destination. The ordered pair (λ,μ) is (572,−521). By reducing the complexity of the 3D system, we found the hidden proportionality that keeps these planes locked together in an infinite embrace.