Question

12.5The vertices of a tetrahedron are located at the points (1,2,-1), (3,1,- 2) , (-1,-2,0) and (2,1,5) . Find the altitude of tetrahedron measured from the vertex at (2,1,5)

Answer 1

so the distance of a point from a plane

Answer 2

point of (2,1,5)

plane of: (1,2,-1), (3,1, -2) , (-1,-2,0)
-1-2 1 -1-2 1 -1-2 1
-------------------------
<0,0,0> <2,-1,-1> <-2,-4,1>

cross the 2 nonzero vectors for a normal


x 2 -2 x = -1-2 = -3
y -1 -2 -y = 2-2 = 0
z -1 1 z = -4-2 = -6

our normal to the plane is: <3,0,2>, and anchor it to our point as a line eq:

x= 2+3t
y= 1
z= 5+2t

when this line intersects the plane; we can define the point that is at the perp under the point

Answer 3

define our plane by the normal vector and any point in it; and make sure our normal isnt messed up by forgetting how to divide by -3 lol

<-3,0,-6>
----------- = <1,0,2> so change x to x = 2+t
-3

Answer 4

1(x-1)+0(y-2)+2(z+1)=0

2+t -1 +10+4t +2 = 0

t = 3/5

Answer 5

insert that into the line equation to determine the points to distance

x= 10+ 3/5 = 13/5
y= 1 = 5/5
z= 25+6/5 = 31/5

those look lovely :)

Answer 6

2,1,5 = 10/5, 5/5, 25/5


(13/5, 5/5, 31/5)
-(10/5, 5/5, 25/5)
-----------------
3/5 , 0 , 6/5

finish out the distance process and we get

sqrt(9+36)
----------
5

Answer 7

of course dbl chk the work to make sure I didnt do any basic mathing errors :)

Answer 8

Thanks man, i,m right on it.

Answer 9

my t value is off ....

Answer 10

i totally ignored the +10 :)

1(2+t-1)+0()+2(5+2t+1)=0

(1+12)2 +t(1+4) = 0

t = -13/5

Answer 11

x= 10/5 - 13/5 = -3/5
y= 5/5 = 5/5
z= 25/5 -26/5 = -1/5

Answer 12

(-3/5, 5/5, -1/5)
-(10/5, 5/5, 25/5)
-----------------
-13/5 , 0 , 26/5

sqrt(159+676)
------------- ; maybe :)
5

Answer 13

my fingers hate me :) 13^2 = 169 not 159

Answer 14

13/sqrt(5)

i recall it being something similar to the t value