Animated Solution for Mathematics - Three Dimensional Geometry: Three lines L1:r=λi^,λ∈R, L2:r=k^+μj^,μ∈R and L3:r=i^+j^+νk^,ν∈R are given. For which point(s) Q on L2 can we find a point P on L1 and a point R on L3 so that P,Q and R are collinear ?
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Visualized Solution
Visualizing the 3D Lines
L1:r=λi^ (x-axis)
L2:r=k^+μj^ (parallel to y-axis)
L3:r=i^+j^+νk^ (parallel to z-axis)
Placing Points P,Q,R
Let P=(λ,0,0) on L1
Let Q=(0,μ,1) on L2
Let R=(1,1,ν) on L3
Condition for Collinearity
For P,Q,R to be collinear:
PQ must be parallel to QR
PQ=k⋅QR for some scalar k
Calculating Vectors PQ and QR
PQ=(0−λ)i^+(μ−0)j^+(1−0)k^=−λi^+μj^+k^
QR=(1−0)i^+(1−μ)j^+(ν−1)k^=i^+(1−μ)j^+(ν−1)k^
Equating Direction Ratios
Since PQ∥QR, their components are proportional:
xQRxPQ=yQRyPQ=zQRzPQ
1−λ=1−μμ=ν−11
Solving for λ
From the first two terms:
1−λ=1−μμ
λ=1−μ−μ
Solving for ν
From the last two terms:
1−μμ=ν−11
ν−1=μ1−μ⟹ν=μ1
Analyzing Constraints on μ
For λ=1−μ−μ to exist, 1−μ=0⟹μ=1
For ν=μ1 to exist, μ=0
Therefore, μ∈R∖{0,1}
Identifying Invalid Points for Q
Recall Q=(0,μ,1)=μj^+k^
If μ=0, Q=k^ (Not possible)
If μ=1, Q=j^+k^ (Not possible)
Checking the Given Options
Option 1: k^+j^⟹μ=1 (Reject)
Option 2: k^⟹μ=0 (Reject)
Option 3: k^+21j^⟹μ=21 (Valid)
Option 4: k^−21j^⟹μ=−21 (Valid)
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The Sigma Insight: Equation of a Line in Space
Solution Diagram
Analyzing the Setup
To begin our journey into 3D space, we first translate the given lines into coordinate form. Any point P on L1 is defined as (λ,0,0).
A point Q on L2 is defined as (0,μ,1). Finally, a point R on L3 is defined as $(1, 1,
u)$.
These coordinates serve as the variables we must manipulate to achieve collinearity. If points P, Q, and R are collinear, the vector PQ must be parallel to the vector QR.
The Vector Bridge
We calculate the vectors as follows:
PQ=Q−P=−λi^+μj^+k^
QR=R−Q=i^+(1−μ)j^+(u−1)k^
For these two vectors to be parallel, their direction ratios must be proportional. This leads to the following symmetric equation:
1−λ=1−μμ=u−11
Solving the System
We now solve the three-part equality. By equating the first two parts, we find:
λ=1−μ−μ
By equating the last two parts, we obtain:
u−1=μ1−μ⟹u=μ1
Respecting the Constraints
We must remain vigilant regarding the physical reality of our variables. If μ=1, the expression for λ becomes undefined. If μ=0, the expression for $
u$ becomes undefined.
These are not merely algebraic quirks; they represent physical boundaries where point Q cannot exist. Consequently, we must reject any options where μ=1 (such as k^+j^) or μ=0 (such as k^).
The remaining valid options, k^+21j^ and k^−21j^, correspond to μ=21 and μ=−21, respectively. Both are mathematically sound solutions to the problem.
