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Question: The perimeter of triangle with vertices at (1,0,0).(0,1,0) and (0,0,1) is:- a.)3 b.) 2 c.) \(2\sqr...

The perimeter of triangle with vertices at (1,0,0).(0,1,0) and (0,0,1) is:-
a.)3 b.) 2 c.) 222\sqrt 2 d.) 323\sqrt 2

Explanation

Solution

Hint:- The given question can be done using the distance formula in the three coordinate plane (x,y,z).\left( {x,y,z} \right).

Complete step-by-step answer:
In this problem, we are asked for the perimeter of the triangle. Perimeter means sum of all the sides of a figure i.e closed figure.
So, the perimeter of the triangle means the sum of all the sides of the triangle. So at first, we would find all the sides of the triangle.

By using distance formula, we would find the sides, AB, BC, and AC;
So, AB =(x2−x1)2+(y1−y2)2+(z1−z2)2= \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_1} - {y_2}} \right)}^2} + {{\left( {{z_1} - {z_2}} \right)}^2}}
=(0−1)2+(1−0)2+(0−0)2= \sqrt {{{\left( {0 - 1} \right)}^2} + {{\left( {1 - 0} \right)}^2} + {{\left( {0 - 0} \right)}^2}}
=(−1)2+(1)2+02= \sqrt {{{\left( { - 1} \right)}^2} + {{\left( 1 \right)}^2} + {0^2}}
=1+1+0=2= \sqrt {1 + 1 + 0} = \sqrt 2
Similarly, BC =(x3−x2)2−(y3−y2)2+(z3−z2)2= \sqrt {{{\left( {{x_3} - {x_2}} \right)}^2} - {{\left( {{y_3} - {y_2}} \right)}^2} + {{\left( {{z_3} - {z_2}} \right)}^2}}
=(0−0)2+(0+1)2+(1−0)2= \sqrt {{{\left( {0 - 0} \right)}^2} + {{\left( {0 + 1} \right)}^2} + {{\left( {1 - 0} \right)}^2}}
=02+(−1)2+(1)2= \sqrt {{0^2} + {{\left( { - 1} \right)}^2} + {{\left( 1 \right)}^2}}
=2\sqrt 2
Similarly, AC = =(x3−x1)2+(y3−y1)2+(z3−z1)2= \sqrt {{{\left( {{x_3} - {x_1}} \right)}^2} + {{\left( {{y_3} - {y_1}} \right)}^2} + {{\left( {{z_3} - {z_1}} \right)}^2}}
=(0−1)2+(0−0)2+(1−0)2= \sqrt {{{\left( {0 - 1} \right)}^2} + {{\left( {0 - 0} \right)}^2} + {{\left( {1 - 0} \right)}^2}}
=(−1)2+(0)2+(1)2= \sqrt {{{\left( { - 1} \right)}^2} + {{\left( 0 \right)}^2} + {{\left( 1 \right)}^2}}
1+0+1    = 2\sqrt {1 + 0 + 1} \;\; = \,\sqrt 2
Hence, perimeter of triangle is given by:-
AB + BC + AC
=2+2+2= \sqrt 2 + \sqrt 2 + \sqrt 2
323\sqrt 2

Note:- The distance formula for the three coordinate plane with points A(x1,y1,z1){\text{A}}\left( {{x_1},{y_1},{z_1}} \right) and B (x2,y2,z2)\left( {{x_2},{y_2},{z_2}} \right) are:-
AB =(x2−x1)2+(y2−y1)2+(z2−z1)2= \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}}