How do you find the volume of the solid tetrahedron bounded by the coordinate planes and the plane x + 3y + z = 6?

Question

Evelyn Agard · Accepted Answer

#V = 12#

The plane equation can be put in a formulation more useful

#Pi->x+3y+z-6 equiv << p -p_0, vec n >> = 0# with

#p = {x,y,z}# a generic plane point #p_0 = {x_0,y_0,y_0}# given plane point #vec n = {1,3,1}# normal plane vector.

The determination of #p_0# is straightforward

#(x-x_0)+3(y-y_0)+(z-z_0) =x+3y+z-(x_0+3y_0+z_0) = x+3y+z-6#

so

#x_0+3y_0+z_0=6#. Fixing #x_0,y_0# we obtain #z_0=6# so #p_0 = {0,0,6}#

The three points limiting the plane area contained in the first quadrant are:

#p_1={6,0,0}# #p_2={0,2,0}# #p_3={0,0,6}#

Those points are obtained fixing to zero two coordinates into the plane equation, and computing the third.

Considering #p_1,p_2,p_3# as the vertices of the piramid whose cusp vertice is the origin of coordinates, knowing the distance between this vertice and the basis plane, we will obtain the pyramid height and consequently we will be able to compute its volume.

The height or distance between the origin and the plane is obtained as follows.

Given a plane # << p -p_0, vec n >> = 0# and a point #q# the distance between #q# and the plane is given by

#norm(q-p_i)# where #p_i# is given by the solution of

# << p_i -p_0, vec n >> = 0# and #p_i=q+lambda vec n#

substituting we have

#<< q+lambda vec n-p_0,vec n >> = << q,vec n >> + lambda << vec n, vec n >> - << p_0, vec n >> = 0# for

#lambda =( << p_0, vec n >> - << q,vec n >> )/<< vec n, vec n >># but here #q = {0,0,0}# then

#lambda = << p_0, vec n >> /<< vec n, vec n >># so #p_i# is #p_i = q+ << p_0, vec n >> /<< vec n, vec n >> vec n# and

#d = norm(p_i-q) = norm(<< p_0, vec n >> /<< vec n, vec n >> vec n) = abs(<< p_0, vec n >>/norm(vec n)) = 6/sqrt(11)#.

The basis area is given by

#A=1/2norm((p_2-p_1) xx (p_3-p_1)) = 6 sqrt(11)#

and finally the pyramid volume is given by

#V = 1/3 cdot d cdot A = 12#

Christopher Agresta · Answer

undefined To find the volume of the solid tetrahedron bounded by the coordinate planes and the plane $x + 3y + z = 6$, you can use the formula for the volume of a tetrahedron:

$$V = \frac{1}{3} \times \text{Area of base} \times \text{Height}$$

The base of the tetrahedron is a triangle formed by the intersection of the coordinate planes and the plane $x + 3y + z = 6$. To find the area of this triangle, you can find the points of intersection of the plane with the coordinate axes by setting two of the variables to zero and solving for the third. Then use these points to find the lengths of the sides of the triangle using distance formula or other geometric methods. Once you have the lengths of the sides, you can use Heron's formula or other methods to find the area of the triangle.

The height of the tetrahedron can be found by dropping a perpendicular from any vertex of the triangle to the plane $x + 3y + z = 6$, then finding the distance from that point to the plane.

After finding the area of the base triangle and the height of the tetrahedron, plug these values into the formula to calculate the volume.

Charles Anctil · Answer

undefined To find the volume of the solid tetrahedron bounded by the coordinate planes and the plane $x + 3y + z = 6$, you can use the formula for the volume of a tetrahedron given by:

$$ V = \frac{1}{6} \times \text{area of base} \times \text{height} $$

First, find the intersection points of the plane $x + 3y + z = 6$ with the coordinate axes. These points will be the vertices of the base of the tetrahedron.

When $x = 0$, $y = 0$, and $z = 0$, the corresponding points are $ (0,0,6) $, $ (0,2,0) $, and $ (6,0,0) $ respectively.

Calculate the area of the base formed by these three points using the formula for the area of a triangle.

Then, find the distance between the plane $x + 3y + z = 6$ and the origin. This distance will be the height of the tetrahedron.

Once you have the area of the base and the height, plug them into the formula for the volume of a tetrahedron to find the volume of the solid.