Animated Solution for Mathematics - Three Dimensional Geometry: Let α,β,γ,δ be real numbers such that α2+β2+γ2=0 and α+γ=1. Suppose the point (3,2,−1) is the mirror image of the point (1,0,−1) with respect to the plane αx+βy+γz=δ. Then which of the following statements is/are TRUE?
Select Answer:
* Multiple Correct
Visualized Solution
The Mirror Image Setup
Let point P=(1,0,−1)
Let its mirror image be Q=(3,2,−1)
The mirror is the plane: αx+βy+γz=δ
Properties of Mirror Image
The line segment PQ is perpendicular to the plane.
The midpointM of PQ lies exactly on the plane.
Normal Vector to the Plane
The normal vector n of the plane is parallel to the vector PQ.
Direction ratios of the normal are (α,β,γ).
Calculating Vector PQ
PQ=(x2−x1)i^+(y2−y1)j^+(z2−z1)k^
PQ=(3−1)i^+(2−0)j^+(−1−(−1))k^
Components of PQ
PQ=2i^+2j^+0k^
Direction ratios of PQ are (2,2,0).
Proportionality of Direction Ratios
Since n∥PQ, their direction ratios are proportional.
(α,β,γ)=k(2,2,0)
α=2k, β=2k, γ=0
Using the Given Condition
We are given: α+γ=1
Substitute α=2k and γ=0:
2k+0=1
Finding the Constant k
2k=1
k=21
Therefore: α=2(21)=1, β=2(21)=1, γ=0
The Midpoint M
Midpoint M=(2x1+x2,2y1+y2,2z1+z2)
M=(21+3,20+2,2−1+(−1))
Coordinates of M
M=(24,22,2−2)
M=(2,1,−1)
Finding δ
Point M(2,1,−1) lies on the plane αx+βy+γz=δ.
Substitute x=2,y=1,z=−1 and α=1,β=1,γ=0.
(1)(2)+(1)(1)+(0)(−1)=δ
Calculating δ
2+1+0=δ
δ=3
The plane equation is x+y=3.
Checking the Options
We found: α=1,β=1,γ=0,δ=3
(A) α+β=1+1=2 (True)
(B) δ−γ=3−0=3 (True)
(C) δ+β=3+1=4 (True)
(D) α+β+γ=2=3 (False)
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The Sigma Insight: Equation of a Plane
Solution Diagram
The Geometry of Reflection
Imagine standing before a mirror in three-dimensional space. You are looking at a point P(1,0,−1), and its reflection, Q(3,2,−1), appears to be behind the glass.
The plane, defined by αx+βy+γz=δ, acts as our mirror. To solve this, we must embrace two fundamental truths about mirror images in 3D space.
First, the line segment connecting the object P and its image Q is perfectly perpendicular to the plane. Second, the midpoint M of this segment must lie exactly on the plane. These two insights are the keys that unlock the entire problem.
The Normal Vector
Since the line segment PQ is perpendicular to the plane, the vector PQ must be parallel to the normal vector of the plane. Let us calculate this vector by subtracting the coordinates of P from Q:
PQ=(3−1)i^+(2−0)j^+(−1−(−1))k^=2i^+2j^+0k^
This vector, (2,2,0), gives us the direction ratios of the normal to our plane. In the general equation of a plane αx+βy+γz=δ, the coefficients (α,β,γ) represent the direction ratios of the normal.
Because our vector PQ is the normal, these coefficients must be proportional to (2,2,0). We can write this as (α,β,γ)=k(2,2,0), where k is a non-zero constant.
The Proportionality Constraint
Now, we have a system of equations: α=2k, β=2k, and γ=0. The problem provides us with a crucial constraint: α+γ=1.
Substituting our expressions for α and γ into this condition, we get 2k+0=1, which immediately tells us that k=21.
With k in hand, we can find the exact values for our coefficients:
α=2(21)=1,β=2(21)=1,γ=0
We have successfully determined the orientation of our mirror!
The Midpoint Anchor
We are almost there. We know the orientation of the plane, but we still need the constant δ. This is where the second geometric property comes into play: the midpoint M of the segment PQ must lie on the plane.
Using the midpoint formula:
M=(21+3,20+2,2−1−1)=(2,1,−1)
Since M lies on the plane 1x+1y+0z=δ, we substitute the coordinates of M into the equation:
(1)(2)+(1)(1)+(0)(−1)=δ
This simplifies beautifully to 2+1+0=δ, giving us δ=3. The equation of our plane is simply x+y=3.
Conclusion and Verification
With α=1, β=1, γ=0, and δ=3, we can now evaluate the given statements.
Option (A) α+β=1+1=2 is true. Option (B) δ−γ=3−0=3 is true. Option (C) δ+β=3+1=4 is true.
Finally, Option (D) α+β+γ=1+1+0=2, which is not equal to δ=3, so it is false. By visualizing the geometry and systematically applying the properties of mirror images, we have navigated this problem with precision and confidence.
