91影视

Given vertices of the triangle are (1, 0, 0), (0, 1, 0), and (0, 0, 1)

Let A=(1, 0, 0), B=(0, 1, 0), and C=(0, 0, 1)

Distance AB:

A=(1, 0, 0), B=(0, 1, 0)

AB = \(\sqrt {{{(0 - 1)}^2} + {{\left( {1 - 0} \right)}^2} + {{\left( { 0 -0} \right)}^2}} \)

= \(\sqrt 2 \) units

Distance BC:

B=(0, 1, 0) ,C=(0, 0, 1)

BC = \(\sqrt {{{(0 - 0)}^2} + {{\left( {0 - 1} \right)}^2} + {{\left( { 1-0} \right)}^2}} \)

=\(\sqrt 2 \) units

Distance AC:

A= (1, 0, 0),C=(0, 0, 1)

AC = \(\sqrt {{{(0 - 1)}^2} + {{\left( {0 - 0} \right)}^2} + {{\left( { 1 -0} \right)}^2}} \)

=\(\sqrt 2 \) units

From the distances, we observe that

AB=BC=AC =\(\sqrt 2 \) units

Thus, the given triangle is an equilateral triangle.

Given vertices of the triangle are (1, 0, 0), (0, 1, 0), (0, 0, 1), and (a,a,a)

Let A=(1, 0, 0), B=(0, 1, 0), C=(0, 0, 1) , and D= (a,a,a)

Distance AB:

A=(1, 0, 0), B=(0, 1, 0)

AB = \(\sqrt {{{(0 - 1)}^2} + {{\left( {1 - 0} \right)}^2} + {{\left( { 0 -0} \right)}^2}} \)

= \(\sqrt 2 \) units

Distance BC:

B=(0, 1, 0) ,C=(0, 0, 1)

BC = \(\sqrt {{{(0 - 0)}^2} + {{\left( {0 - 1} \right)}^2} + {{\left( { 1-0} \right)}^2}} \)

=\(\sqrt 2 \) units

Distance AC:

A= (1, 0, 0),C=(0, 0, 1)

AC = \(\sqrt {{{(0 - 1)}^2} + {{\left( {0 - 0} \right)}^2} + {{\left( { 1 -0} \right)}^2}} \)

=\(\sqrt 2 \) units

From the above step,

AB=BC=AC=\(\sqrt 2 \) units

D = (a,a,a)

To form a regular tetrahedron, the distance between point D and the other vertices should be equal since D is also a vertex of a regular tetrahedron.

Considering vertices A (1,0,0) and D(a,a,a)

Distance AD :

AD= \(\sqrt {{{(a - 1)}^2} + {{\left( {a- 0} \right)}^2} + {{\left( { a-0} \right)}^2}} \)

= \(\sqrt {3{a^2} - 2a + 1} \)

Thus, from step 2

\(\sqrt {3{a^2} - 2a + 1} \) =\(\sqrt 2 \)

\( 3(a^2) - 2a + 1 \) =\( 2 \)

\( 3(a^2) - 3a+a -1 \) =\( 0 \)

\((a-1)(3a+1)\)=0

\((a-1)\)=0 , \((3a+1)\) =0

\(a=-1\), \(a = - \frac{1}{3}\)

Thus either \(a=-1\) or \(a = - \frac{1}{3}\)