First slide

Introduction to 3-D Geometry

Question

The mid-points of the sides of a triangle are (5,7 ,11), (0, 8, 5) and (2,3, - 1). Then, the vertices are

Easy

A

(7, 2, 5), (3, 12, 17), (- 3, 4, - 7)
B

(7, 2, 5), (3, 12, 17), (3, 4,7)
C

(7 , 2,5), (- 3, 12, 17), (- 3, - 4, - 7)
D

None of the above

Solution

Let the vertices of a triangle are $A (x_{1}, y_{1}, z_{1}), B (x_{2}, y_{2}, z_{2}) and C (x_{3}, y_{3}, z_{3}) .$
Since, D, E and F are the mid-points of AC ,BC and AB
$\begin{array}{ll} ∴ & (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}, \frac{z_{1} + z_{2}}{2}) = (0, 8, 5) \\ \Rightarrow & x_{1} + x_{2} = 0, y_{1} + y_{2} = 16, z_{1} + z_{2} = 10 - - - - - (i) \\ (\frac{x_{2} + x_{3}}{2}, \frac{y_{2} + y_{3}}{2}, \frac{z_{2} + z_{3}}{2}) = (2, 3, - 1) \\ \Rightarrow & x_{2} + x_{3} = 4, y_{2} + y_{3} = 6, z_{2} + z_{3} = - 2 - - - - (ii) \\ and & (\frac{x_{1} + x_{3}}{2}, \frac{y_{1} + y_{3}}{2}, \frac{z_{1} + z_{3}}{2}) = (5, 7, 11) \\ \Rightarrow & x_{1} + x_{3} = 10, y_{1} + y_{3} = 14, z_{1} + z_{3} = 22 - - - - (iii) \end{array}$
On adding Eqs. (i), (ii) and (iir), we get
$\begin{aligned} 2 (x_{1} + x_{2} + x_{3}) = 14, 2 (y_{1} + y_{2} + y_{3}) = 36, 2 (z_{1} + z_{2} + z_{3}) = 30 \\ \Rightarrow x_{1} + x_{2} + x_{3} = 7, y_{1} + y_{2} + y_{3} = 18, z_{1} + z_{2} + z_{3} = 15 \dots (iv) \end{aligned}$
On solving Eqs. (i), (ii), (iil) and (iv)' we get
$\begin{aligned} x_{3} = 7, x_{1} = 3, x_{2} = - 3 \\ y_{3} = 2, y_{1} = 12, y_{2} = 4 \\ and z_{3} = 5, z_{1} = 17, z_{2} = - 7 \end{aligned}$
Hence, vertices of a triangle are (7, 2,5), (3, 2, 17), (-3, 4, -7)

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