Animated Solution for Mathematics - Three Dimensional Geometry: Let L1 and L2 denotes the lines r=i^+λ(−i^+2j^+2k^),λ∈R and r=μ(2i^−j^+2k^),μ∈R respectively. If L3 is a line which is perpendicular to both L1 and L2 and cuts both of them, then which of the following options describe(s) L3?
Select Answer:
* Multiple Correct
Visualized Solution
L1 and L2 in Space
Given lines:
L1:r=i^+λ(−i^+2j^+2k^)
L2:r=μ(2i^−j^+2k^)
The Common Perpendicular L3
L3 is perpendicular to both L1 and L2.
L3 intersects L1 at A and L2 at B.
Extracting Direction Vectors
Direction of L1: b1=−i^+2j^+2k^
Direction of L2: b2=2i^−j^+2k^
Direction of L3
Direction of L3 is parallel to b1×b2
b1×b2=i^−12j^2−1k^22
Evaluating Cross Product
b1×b2=i^(4−(−2))−j^(−2−4)+k^(1−4)
b1×b2=6i^+6j^−3k^
Simplifying Direction Vector
b1×b2=3(2i^+2j^−k^)
Direction of L3 is parallel to 2i^+2j^−k^
General Points A and B
General point on L1: A(1−λ,2λ,2λ)
General point on L2: B(2μ,−μ,2μ)
Vector AB
AB=(2μ−(1−λ))i^+(−μ−2λ)j^+(2μ−2λ)k^
AB=(2μ+λ−1)i^+(−μ−2λ)j^+(2μ−2λ)k^
Proportionality Condition
AB is parallel to L3's direction: 2i^+2j^−k^
Therefore, their direction ratios are proportional:
22μ+λ−1=2−μ−2λ=−12μ−2λ
Solving for λ and μ (Part 1)
From 1st and 2nd parts:
2μ+λ−1=−μ−2λ
3μ+3λ=1⟹μ+λ=31 --- (Eq 1)
Solving for λ and μ (Part 2)
From 2nd and 3rd parts:
2−μ−2λ=−12μ−2λ
μ+2λ=4μ−4λ
3μ=6λ⟹μ=2λ --- (Eq 2)
Finding λ and μ
Substitute μ=2λ into Eq 1:
2λ+λ=31⟹3λ=31
λ=91 and μ=2(91)=92
Coordinates of Points A and B
Substitute λ=91 into A: A(98,92,92)
Substitute μ=92 into B: B(94,−92,94)
Checking Options 2 and 4
Equation of L3 is r=a+t(2i^+2j^−k^)
Option 4 uses point A. Option 4 is Correct.
Option 2 uses point B. Option 2 is Correct.
Checking Option 1
Option 1 uses point (32,0,31)
Midpoint of A and B is M=(298+94,292−92,292+94)=(32,0,31)
Since M lies on L3, Option 1 is Correct.
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The Sigma Insight: Shortest Distance Between Two Skew Lines
Solution Diagram
The Geometry of Connection
Finding the Common Perpendicular
Imagine you are standing in a vast, three-dimensional void. You see two lines, L1 and L2, drifting through space. They are not parallel, and they do not intersect; they are skew lines.
Your mission is to find the unique line L3 that acts as a bridge between them—a line that is perfectly perpendicular to both and touches both. This is the classic problem of the common perpendicular, and it is a beautiful exercise in vector algebra.
Phase 1
The Direction of the Bridge
To define any line in space, we need two things: a point on the line and a direction vector. We are given:
L1:r=i^+λ(−i^+2j^+2k^)L2:r=μ(2i^−j^+2k^)
The direction vectors are b1=−i^+2j^+2k^ and b2=2i^−j^+2k^. Since L3 must be perpendicular to both, its direction vector v must be parallel to the cross product b1×b2.
Calculating this, we get:
b1×b2=i^−12j^2−1k^22=6i^+6j^−3k^
We can simplify this by factoring out a 3, giving us the direction vector v=2i^+2j^−k^. This vector is the compass that guides our line L3.
Phase 2
The Intersection Points
Now, we need to find where this bridge touches our skew lines. Let A be the point on L1 and B be the point on L2.
We represent these as general points using the parameters λ and μ:
A(1−λ,2λ,2λ)B(2μ,−μ,2μ)
The vector AB connects these two points. Because AB lies on L3, it must be parallel to our direction vector v=(2,2,−1). This means the components of AB must be proportional to (2,2,−1).
Phase 3
Solving the System
Constructing AB=(2μ−1+λ)i^+(−μ−2λ)j^+(2μ−2λ)k^, we set up the proportionality:
22μ+λ−1=2−μ−2λ=−12μ−2λ
Solving this system is where the magic happens. From the first two parts, we find μ+λ=1/3. From the second and third parts, we find μ=2λ.
Substituting the second into the first, we get 3λ=1/3, so λ=1/9 and μ=2/9. Plugging these back into our general points, we find:
A=(8/9,2/9,2/9)B=(4/9,−2/9,4/9)
Phase 4
The Final Synthesis
We have the direction (2,2,−1) and two points A and B on the line. Any equation of the form r=a+tv is valid.
Option 4 uses point A, Option 2 uses point B, and Option 1 uses the midpoint M=(2/3,0,1/3). All three are correct!
This problem teaches us that while the path to the solution requires rigorous calculation, the result is a flexible, elegant description of a line in space. Keep practicing, and soon, visualizing these 3D structures will become second nature.
