Question

If the volume of tetrahedron formed by planes whose equation are y+z=0,z+x=0,x+y=0 and x+y+z=1 is t cubic unit then 3t is

Answer 1

Find the vertices, which are the planes intersections, i.e..

x+y=0; y+z=0; z+x=0→A(0,0,0)

x+y=0; y+z=0; x+y+z−1=0→B(1,−1,1)

x+y=0; z+x=0; x+y+z−1=0→C(−1,1,1)

y+z=0; z+x=0; x+y+z−1=0→D(1,1,−1)

AB=(1,−1,1); AC=(−1,1,1); AD=(1,1,−1)

Now the volume is given by 
V=1/6 [AB AC AD]

Thus, volume is 2/3

Answer 2

Given:

Equations of plane :

y + z = 0

z + x = 0

x + y = 0

x + y + z = 1

Volume of tetrahedron = t

To find :

Value of 3t

Solution :

We have four planes,

In 3 dimensional figure,

three faces touching each other to make a point of intersection.

So, equation of planes are :

1. y + z = 0
2. z + x = 0
3. x + y = 0
4. x + y + z = 1

On intersecting planes 1, 2 and 3 , the vertices are:

y + z = 0 ,

z + x = 0,

x + y = 0,

So, the vertices will be (x,y,z) as :

P = (0,0,0)

On intersecting planes 1, 2 and 4 , the vertices are:

y + z = 0 ,

z + x = 0,

x + y + z = 1,

So, the vertices will be (x,y,z) as :

Q = (1,1,-1)

On intersecting planes 2,3 and 4 , the vertices are:

z + x = 0,

x + y = 0,

x + y + z = 1,

So, the vertices will be (x,y,z) as :

R = (-1,1,1)

On intersecting planes 1,3 and 4 , the vertices are:

y + z = 0,

x + y = 0,

x + y + z = 1,

So, the vertices will be (x,y,z) as :

S = (1,-1,1)

So,

PQ = Q - P = (1,1,-1)

PR = R - P = (-1,1,1,)

PS = S - P = (1,-1,1)

So, Volume of tetrahedron :

$\bf V = \frac{1}{6} [PQ \: \: \: PR \: \: PS] \:$

$\bf = \frac{1}{6} \left| \begin{array}{cc} 1 & 1 & - 1 \\ -1 & 1&1\\ 1 & - 1&1 \\ \end{array} \right| </strong></p><p></p><p></p><p><strong>[tex] \bf \: V= \frac{4}{6} = \frac{2}{3}$

So Value of t :

$= \frac{2}{3}$

Then value of 3t :

$\bf \: 3 \times \frac{2}{3} = 2 \: \: {unit}^{3}$