Animated Solution for Mathematics - Three Dimensional Geometry: If (a,b,c) is the image of the point (1,2,−3) in the line 2x+1=−2y−3=−1z, then a+b+c is equal to:
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Visualized Solution
P and Line L
Given point:P(1,2,−3)
Given lineL: 2x+1=−2y−3=−1z
Goal: Find the image (a,b,c) and calculate a+b+c.
The Foot of Perpendicular Q
To find the image R, we first need the foot of the perpendicular Q on line L.
General Point Q
Let 2x+1=−2y−3=−1z=λ
General pointQ(2λ−1,−2λ+3,−λ)
Vector PQ
PQ=(Qx−Px,Qy−Py,Qz−Pz)
PQ=(2λ−1−1,−2λ+3−2,−λ−(−3))
PQ=(2λ−2,−2λ+1,−λ+3)
Perpendicularity Condition
Direction vector of L is d=(2,−2,−1)
Since PQ⊥d, their dot product is zero:
PQ⋅d=0
Setting up the Dot Product
Equation:2(2λ−2)+(−2)(−2λ+1)+(−1)(−λ+3)=0
Solving for λ
4λ−4+4λ−2+λ−3=0
Simplify:9λ−9=0
Result:λ=1
Exact Coordinates of Q
Substitute λ=1 into Q(2λ−1,−2λ+3,−λ)
Q(2(1)−1,−2(1)+3,−1)
Result:Q(1,1,−1)
Midpoint Relation
Let the image be R(a,b,c).
Q is the exact midpoint of segment PR.
Midpoint Formula
(2a+1,2b+2,2c−3)=(1,1,−1)
Finding a,b,c
a+1=2⇒a=1
b+2=2⇒b=0
c−3=−2⇒c=1
ImageR(1,0,1)
Final Calculation
We need a+b+c.
1+0+1=2
The correct answer is 2.
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The Sigma Insight: Equation of a Line in Space
Solution Diagram
The Mirror in 3D Space
A Journey of Reflection
Imagine you are standing in a room, holding a point P(1,2,−3) in your hand. Before you lies a mirror, but this mirror is not a flat surface—it is a straight line L defined by the equation:
2x+1=−2y−3=−1z
Your goal is to find the reflection of point P on the other side of this line. This is not just a calculation; it is a beautiful dance of vectors and symmetry.
Phase 1
The Stepping Stone
To find the image R(a,b,c), we cannot simply jump across the line. We need a bridge. That bridge is the foot of the perpendicular, Q.
If we drop a perpendicular from P onto the line L, the point where it hits, Q, is the exact midpoint between P and its image R. Think of Q as the anchor point that holds the symmetry together.
Phase 2
The Parametric Power
Since Q lies on the line L, we can describe its position using a single parameter, λ. By setting the line equation equal to λ, we can express any point on the line as:
Q(2λ−1,−2λ+3,−λ)
This is the power of parametric form—it turns a 3D line into a simple 1D path.
Phase 3
The Perpendicularity Condition
Now, we construct the vector PQ by subtracting the coordinates of P from Q:
PQ=(2λ−2,−2λ+1,−λ+3)
We know that PQ must be perpendicular to the line L. The direction vector of L is d=(2,−2,−1).
Because they are perpendicular, their dot product must be zero: PQ⋅d=0. This gives us the equation:
2(2λ−2)+(−2)(−2λ+1)+(−1)(−λ+3)=0
Solving this, we find 4λ−4+4λ−2+λ−3=0, which simplifies to 9λ−9=0, or λ=1.
Phase 4
The Final Symmetry
With λ=1, we find the exact coordinates of Q by substituting λ back into our general point:
Q(2(1)−1,−2(1)+3,−1)=(1,1,−1)
Now, we use the midpoint formula: 2P+R=Q. This implies:
2a+1=1,2b+2=1,2c−3=−1
Solving these gives a=1,b=0,c=1. The image point is R(1,0,1).
Finally, the sum a+b+c=1+0+1=2. You have successfully navigated the mirror of 3D space!