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 find the sum of squares of the sides of triangle \( \triangle ABC \), given the midpoints of sides AB, BC, and CA, let's denote the vertices of the triangle \( A, B, \) and \( C \) as points \( A(x_1, y_1, z_1) \), \( B(x_2, y_2, z_2) \), and \( C(x_3, y_3, z_3) \) in 3D space.
The midpoints of sides provide coordinates as follows:
- Midpoint of AB is \( \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right) = (3, 0, 0) \)
- Midpoint of BC is \( \left( \frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2}, \frac{z_2 + z_3}{2} \right) = (0, 4, 0) \)
- Midpoint of CA is \( \left( \frac{x_3 + x_1}{2}, \frac{y_3 + y_1}{2}, \frac{z_3 + z_1}{2} \right) = (0, 0, 5) \)
From the midpoints, we can find:
- \( x_1 + x_2 = 6, \; y_1 + y_2 = 0, \; z_1 + z_2 = 0 \)
- \( x_2 + x_3 = 0, \; y_2 + y_3 = 8, \; z_2 + z_3 = 0 \)
- \( x_3 + x_1 = 0, \; y_3 + y_1 = 0, \; z_3 + z_1 = 10 \)
To find the length of sides, we solve these equations. From them, we can deduce:
- \( x_1 = 3, \; x_2 = 3, \; x_3 = -3 \)
- \( y_1 = 0, \; y_2 = 0, \; y_3 = 8 \)
- \( z_1 = 5, \; z_2 = -5, \; z_3 = 5 \)
Now the vertices are:
- \( A(3, 0, 5) \)
- \( B(3, 0, -5) \)
- \( C(-3, 8, 5) \)
Using the distance formula, we find the square of the length of each side:
- For \( AB \): \[ AB^2 = (3 - 3)^2 + (0 - 0)^2 + (5 - (-5))^2 = 0 + 0 + 100 = 100 \]
- For \( BC \): \[ BC^2 = (3 - (-3))^2 + (0 - 8)^2 + (-5 - 5)^2 = 36 + 64 + 100 = 200 \]
- For \( CA \): \[ CA^2 = (-3 - 3)^2 + (8 - 0)^2 + (5 - 5)^2 = 36 + 64 + 0 = 100 \]
Therefore, the sum of the squares of the sides is:
AB^2 + BC^2 + CA^2 = 100 + 200 + 100 = 400Hence, the correct answer 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.
