Q. 54 - Calculus

Question

(a) Show that the three points (1, 0, 0), (0, 1, 0), and (0, 0, 1) are the vertices of an equilateral triangle.

(b) Determine the two values of a so that the four points (1, 0, 0), (0, 1, 0), (0, 0, 1), and (a, a, a) are the vertices of a regular tetrahedron.

Step-by-Step Solution

Answer

(a).  The given three points (1, 0, 0), (0, 1, 0), and (0, 0, 1) are the vertices of an equilateral triangle. 

(b) The two values of a are \(- \frac{1}{3},1\)

Step 1. part (a) of the solution

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.

Step 2. Part (b) of the solution- calculating distances between given points

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

Step 3. Finding the values of a

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