In a triangle BC, if the mid points of sides AB, BC, CA are (3,0,0), (0,4,0),(0,0,5) respectively, then AB2 + BC2 + CA2 =
In a triangle BC, if the mid points of sides AB, BC, CA are (3,0,0), (0,4,0),(0,0,5) respectively, then AB2 + BC2 + CA2 =
50
200
300
400
The Correct Option is D
Solution and Explanation
To solve the problem, we need to find the sum of the squares of the side lengths of triangle ABC given the midpoints of its sides.
1. Define the Vertices and Midpoints:
Let the vertices of triangle ABC be A = (x₁, y₁, z₁), B = (x₂, y₂, z₂), C = (x₃, y₃, z₃).
Let D, E, F be the midpoints of BC, CA, AB respectively, given as D = (3, 0, 0), E = (0, 4, 0), F = (0, 0, 5).
2. Set Up Midpoint Equations:
For D = ((x₂+x₃)/2, (y₂+y₃)/2, (z₂+z₃)/2) = (3, 0, 0):
x₂ + x₃ = 6, y₂ + y₃ = 0, z₂ + z₃ = 0.
For E = ((x₃+x₁)/2, (y₃+y₁)/2, (z₃+z₁)/2) = (0, 4, 0):
x₃ + x₁ = 0, y₃ + y₁ = 8, z₃ + z₁ = 0.
For F = ((x₁+x₂)/2, (y₁+y₂)/2, (z₁+z₂)/2) = (0, 0, 5):
x₁ + x₂ = 0, y₁ + y₂ = 0, z₁ + z₂ = 10.
3. Solve for x-Coordinates:
From x₁ + x₂ = 0, x₁ = -x₂.
From x₃ + x₁ = 0, x₃ = -x₁ = x₂.
From x₂ + x₃ = 6, x₂ + x₂ = 6, so 2x₂ = 6, x₂ = 3.
Thus, x₁ = -3, x₃ = 3.
4. Solve for y-Coordinates:
From y₁ + y₂ = 0, y₁ = -y₂.
From y₃ + y₁ = 8, y₃ = 8 - y₁ = 8 + y₂.
From y₂ + y₃ = 0, y₂ + 8 + y₂ = 0, so 2y₂ = -8, y₂ = -4.
Thus, y₁ = 4, y₃ = 4.
5. Solve for z-Coordinates:
From z₁ + z₂ = 10, z₁ = 10 - z₂.
From z₃ + z₁ = 0, z₃ = -z₁ = z₂ - 10.
From z₂ + z₃ = 0, z₂ + z₂ - 10 = 0, so 2z₂ = 10, z₂ = 5.
Thus, z₁ = 5, z₃ = -5.
6. Determine the Vertices:
The vertices are A = (-3, 4, 5), B = (3, -4, 5), C = (3, 4, -5).
7. Calculate the Side Lengths Squared:
AB² = (3 - (-3))² + (-4 - 4)² + (5 - 5)² = 6² + (-8)² = 36 + 64 = 100.
BC² = (3 - 3)² + (4 - (-4))² + (-5 - 5)² = 0 + 8² + (-10)² = 64 + 100 = 164.
CA² = (-3 - 3)² + (4 - 4)² + (5 - (-5))² = (-6)² + 0 + 10² = 36 + 100 = 136.
8. Sum the Squares:
AB² + BC² + CA² = 100 + 164 + 136 = 400.
Final Answer:
The sum of the squares of the side lengths is 400.
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Concepts Used:
Three Dimensional Geometry
Mathematically, Geometry is one of the most important topics. The concepts of Geometry are derived w.r.t. the planes. So, Geometry is divided into three major categories based on its dimensions which are one-dimensional geometry, two-dimensional geometry, and three-dimensional geometry.
Direction Cosines and Direction Ratios of Line:
Consider a line L that is passing through the three-dimensional plane. Now, x,y and z are the axes of the plane and α,β, and γ are the three angles the line makes with these axes. These are commonly known as the direction angles of the plane. So, appropriately, we can say that cosα, cosβ, and cosγ are the direction cosines of the given line L.
