Vikas TU
Last Activity: 7 Years ago
Condition of plane with three focuses
x-x1 y-y1 z-z1 = 0
x2-x1 y2-y1 z2-z1
x3-x1 y3-y1 z3-z1
x-4 y-(- 1) z - 5 = 0
(- 1) – 4 2 - (- 1) 0 - 5
1 – 4 1 - (- 1) 4 - 5
x - 4 y - (- 1) z - 5 = 0
- 5 3 -5
- 3 2 -1
(x - 4 )( 3 · (- 1) - (- 5) · 2 ) - (y - (- 1) )( (- 5) · (- 1) - (- 5) · (- 3) ) + (z - 5 )( (- 5) · 2 - 3 · (- 3) ) = 0
7 (x- 4 )+ 10 (y- (- 1) )+ (- 1) (z- 5 ) = 0
7 x + 10 y - z - 13 =0
a-b={ax-bx;ay-by;az-bz}={(- 1)- 1;2-1;0-4}={-2;1;- 4}
Lets a x,y,z lies on hold opposite to AB and in the plane ,then
x-4,y+1,z-5 will likewise lie in the plane and opposite to AB where x,y,z is area of winged creature whenever.
speck result of these two point ought to be zero
=>8x+4y-3z=0
since sooner or later it ends up plainly collinear with AB
=> triangle framed be A,B and that point will be zero
=>
2x-y+4z=29
additionally it fulfills the plane so
7x+10y-z=13
solvin every one of the three eq.
x=15/7,y=3/7,z=44/7
remove =
Time = remove/speed
=/10
= 1/√14
