Animated Solution for Mathematics - Three Dimensional Geometry: Let L1 and L2 be the following straight line. L1:1x−1=−1y=3z−1 and L2:−3x−1=−1y=1z−1. Suppose the straight line L:lx−α=my−1=−2z−γ lies in the plane containing L1 and L2, and passes through the point of intersection of L1 and L2. If the line L bisects the acute angle between the lines L1 and L2, then which of the following statements is/are TRUE?
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Visualized Solution
The Intersecting Lines
Given lines L1 and L2 intersect at a point P.
Line L is the acute angle bisector of L1 and L2.
Finding the Intersection Point
L1:1x−1=−1y=3z−1
L2:−3x−1=−1y=1z−1
By observation, x=1,y=0,z=1 makes all terms zero.
Intersection point is P(1,0,1).
Direction Vectors
Direction of L1: b1=i^−j^+3k^
Direction of L2: b2=−3i^−j^+k^
Magnitudes of Direction Vectors
∣b1∣=12+(−1)2+32=11
∣b2∣=(−3)2+(−1)2+12=11
Since ∣b1∣=∣b2∣, the bisectors are along b1+b2 and b1−b2.
Acute or Obtuse?
If b1⋅b2>0, the angle between them is acute.
In this case, the sum b1+b2 gives the acute angle bisector.
If b1⋅b2<0, the sum gives the obtuse angle bisector.
Calculating the Dot Product
b1⋅b2=(1)(−3)+(−1)(−1)+(3)(1)
b1⋅b2=−3+1+3=1
Since 1>0, the angle is acute.
Direction of Acute Bisector
Acute bisector direction d=b1+b2
d=(1−3)i^+(−1−1)j^+(3+1)k^
d=−2i^−2j^+4k^
Simplifying the Direction
Divide by −2 to simplify the direction ratios.
Simplified direction dL=i^+j^−2k^
Direction ratios of L are (1,1,−2).
Matching Direction Ratios
Given line L:lx−α=my−1=−2z−γ
Its direction ratios are (l,m,−2).
We found the direction ratios to be proportional to (1,1,−2).
Comparing the z-components, l=1 and m=1.
Point on Line L
Line L passes through the intersection point P(1,0,1).
Substitute x=1,y=0,z=1 into the equation of L.
11−α=10−1=−21−γ
Solving for α
11−α=−1
1−α=−1
α=1+1=2
Solving for γ
−21−γ=−1
1−γ=2
γ=1−2=−1
Evaluating the Options
We found: α=2,γ=−1,l=1,m=1
Option A: α−γ=2−(−1)=3 (True)
Option B: l+m=1+1=2 (True)
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The Sigma Insight: Angle Between Two Lines
Solution Diagram
Analyzing the Setup
Welcome, future engineer! Today, we are exploring the elegant geometry of 3D space. We aim to find the line L that bisects the acute angle between two given lines, L1 and L2.
The lines are defined as:
L1:1x−1=−1y=3z−1
L2:−3x−1=−1y=1z−1
Phase 1
Finding the Intersection Point
First, we locate the heart of the system: the intersection point P. By observation, if we set x=1, y=0, and z=1, every term in both equations becomes zero.
Thus, our point of intersection is P(1,0,1). This point serves as our anchor for the line L.
Phase 2
The Rhombus Property
Next, we extract the direction vectors for the lines. For L1, we have b1=i^−j^+3k^, and for L2, we have b2=−3i^−j^+k^.
We calculate their magnitudes:
∣b1∣=12+(−1)2+32=11
∣b2∣=(−3)2+(−1)2+12=11
Because ∣b1∣=∣b2∣, the parallelogram formed by these vectors is a rhombus. In a rhombus, the diagonal acts as the angle bisector, allowing us to simply add or subtract the vectors.
Phase 3
The Acute Test
To determine if b1+b2 represents the acute bisector, we calculate the dot product:
b1⋅b2=(1)(−3)+(−1)(−1)+(3)(1)=−3+1+3=1
Since the dot product is positive (1>0), the angle between the vectors is acute. Therefore, the sum d=b1+b2 provides the direction of the acute bisector.
Adding the vectors, we get:
d=(1−3)i^+(−1−1)j^+(3+1)k^=−2i^−2j^+4k^
To simplify, we divide by −2, yielding the direction ratios (1,1,−2).
Phase 4
Constructing the Line
We are given the line L:lx−α=my−1=−2z−γ. Comparing the direction ratios, we identify l=1 and m=1.
Since the line passes through P(1,0,1), we substitute these values into the equation:
11−α=10−1=−21−γ
Solving this system:
1. 11−α=−1⇒α=2
2. −21−γ=−1⇒1−γ=2⇒γ=−1
We have successfully navigated the problem. We find that α−γ=2−(−1)=3 and l+m=1+1=2.
