Animated Solution for Mathematics - Three Dimensional Geometry: Let γ∈R be such that the lines L1:1x+11=2y+21=3z+29 and L2:3x+16=2y+11=γz+4 intersect. Let R1 be the point of intersection of L1 and L2. Let O=(0,0,0), and n^ denote a unit normal vector to the plane containing both the lines L1 and L2. Match each entry in List-I to the correct entries in List-II.
List-I
(P)
γ equals
(Q)
A possible choice for n^ is
(R)
OR1 equals
(S)
A possible value of OR1⋅n^ is
List-II
(1)
−i^−j^+k^
(2)
23
(3)
1
(4)
61i^−62j^+61k^
(5)
32
Select Matching Pairs:
* Multiple Allowed
PMatches
QMatches
RMatches
SMatches
Visualized Solution
Intersecting Lines in 3D
We are given two lines L1 and L2 in 3D space.
They intersect at a specific point R1.
Our goal is to find the unknown γ, the intersection point, and the normal vector to their plane.
Parametric Coordinates
Let L1=α and L2=β.
Point on L1: P1=(α−11,2α−21,3α−29)
Point on L2: P2=(3β−16,2β−11,γβ−4)
Equating x and y Coordinates
Since P1 and P2 represent the same point R1:
x-coordinate: α−11=3β−16
y-coordinate: 2α−21=2β−11
Solving for α and β
From y-eq: 2α−2β=10⟹α−β=5
Substitute α=β+5 into x-eq:
(β+5)−11=3β−16
β−6=3β−16⟹2β=10⟹β=5
α=5+5=10
Equating z Coordinates
Now equate the z-coordinates:
3α−29=γβ−4
Substitute α=10 and β=5:
3(10)−29=γ(5)−4
Solving for γ
30−29=5γ−4
1=5γ−4
5γ=5⟹γ=1
Position Vector OR1
Substitute α=10 into P1:
R1=(10−11,2(10)−21,3(10)−29)
R1=(−1,−1,1)
The position vector of R1 from origin O(0,0,0) is:
OR1=−i^−j^+k^
Direction Vectors of Lines
Direction vector of L1: d1=i^+2j^+3k^
Direction vector of L2 (with γ=1): d2=3i^+2j^+k^
Normal Vector via Cross Product
The normal vector n′ is perpendicular to both d1 and d2.
n′=d1×d2
n′=i^13j^22k^31
Calculating the Cross Product
i^(2(1)−3(2))−j^(1(1)−3(3))+k^(1(2)−2(3))
=i^(2−6)−j^(1−9)+k^(2−6)
n′=−4i^+8j^−4k^
Unit Normal Vector n^
Magnitude ∣n′∣=(−4)2+82+(−4)2=16+64+16=96=46
n^=±∣n′∣n′=±46−4i^+8j^−4k^
n^=±61(−i^+2j^−k^)
The Unit Normal Vector
Choosing the opposite direction to match options:
n^=61i^−62j^+61k^
Setup Dot Product OR1⋅n^
We need to find OR1⋅n^
OR1=−1i^−1j^+1k^
n^=61i^−62j^+61k^
Calculating Dot Product
OR1⋅n^=(−1)(61)+(−1)(−62)+(1)(61)
=6−1+2+1
=62
Conclusion
Key Takeaway: Intersecting lines allow us to find unknown parameters by equating coordinates.
Key Takeaway: The cross product of direction vectors yields the normal to the plane.
Next Challenge: What if the lines were skew (non-intersecting)? How would you find the shortest distance between them?
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The Sigma Insight: Equation of a Plane
Solution Diagram
Analyzing the Setup
Welcome, future engineers! Today, we are not just solving a problem; we are visualizing the architecture of 3D space. Imagine you are standing in a vast, empty room where two lines, L1 and L2, are suspended in the air.
They are not parallel; they are destined to meet. Our goal is to find the point where they collide, the unknown parameter γ that defines the second line, and the orientation of the plane they inhabit.
The Parametric Dance
To find where these lines meet, we cannot rely on simple 2D algebra. We need a way to 'travel' along the lines by introducing parameters, α and β.
By setting the equations of L1 and L2 equal to these parameters, we transform the lines into a set of coordinates. For L1, any point P1 can be written as:
P1=(α−11,2α−21,3α−29)
Similarly, for L2, any point P2 is defined as:
P2=(3β−16,2β−11,γβ−4)
Think of α and β as sliders on a control panel; as you move them, you move along the lines.
The Moment of Collision
At the exact moment of intersection, point P1 and point P2 must be at the exact same location in space. This is our point of intersection R1. Therefore, their x,y, and z coordinates must be perfectly equal.
We start by equating the x and y coordinates:
α−11=3β−16
2α−21=2β−11
This gives us a beautiful system of two linear equations. Solving this, we find α−β=5 and, through careful substitution, we arrive at β=5 and α=10. We have found the 'time' at which these lines meet!
Unlocking the Unknown
Now, we turn to the z-coordinate. Since the lines intersect, the z-coordinates must also be equal:
3α−29=γβ−4
Substituting our hard-earned values of α=10 and β=5, we get:
3(10)−29=γ(5)−4
A quick computation reveals 1=5γ−4, which simplifies elegantly to γ=1. We have cracked the code!
The Geometry of the Plane
Now that we know the lines, we need the normal vector n^ to the plane containing them. We extract the direction vectors from the denominators of our line equations:
d1=i^+2j^+3k^
d2=3i^+2j^+k^
To find a vector perpendicular to both, we use the cross product n′=d1×d2. Calculating the determinant, we find:
n′=−4i^+8j^−4k^
To make this a unit normal vector n^, we divide by the magnitude ∣n′∣=46, giving us:
n^=±61(−i^+2j^−k^)
The Final Connection
Finally, we calculate the dot product OR1⋅n^. With OR1=−i^−j^+k^ and our unit normal vector, the calculation is straightforward:
(−1)(61)+(−1)(−62)+(1)(61)=62
We have successfully navigated the geometry, solved the system, and arrived at the solution. Remember, in JEE, it is not just about the answer; it is about the elegance of the path you take to get there.
