Hello, i has been little problems with a calculus test problem of a plane and her vectors:
It is only 6 fast subsections that i trying to resolved without sucess (exept the first...or i just want to think so)
The vectors to work are:
V1 <3,4,4> V2 <-4,-4,3> V3 <3,-2,1> V4 <1,3,-3>
1) Calculate the Triple Scalar Product (TSP) and the volume-- i used the cross product with the vectors 1,2 and 3.
3 4 4 (-4+6)(3)-(-4-9)(4)+(8+12)(4)
TSP -4 -4 3 6+52+80=138
3 -2 1 TSP = 138
V = 138 u
TSP -7 -8 -1 (24-6)i - (21+0) j + (42+0)k
0 -6 -3 N = <18,-21,42>
2) Calculate ecuation of the plane that passing by V1, V2 and V3 --- i calculated the scalar ecuation and the vectorial ecuation
3) Calculate the distance D1 without exit to the plane ---- I used a imaginary intersection point 3 in the origin of the intersections (under the plane) and imaginary stright.
4) The vectorial ecuation and the parametrics of L1 that is passing by V4 and have the NORMAL DIRECTION --- I only remplaced the V4 in the Vectorial ecuation
I really sorry for the big text, but, i tried to resume the max that was possible. Thank you, very much!
It is only 6 fast subsections that i trying to resolved without sucess (exept the first...or i just want to think so)The vectors to work are:
V1 <3,4,4> V2 <-4,-4,3> V3 <3,-2,1> V4 <1,3,-3>
1) Calculate the Triple Scalar Product (TSP) and the volume-- i used the cross product with the vectors 1,2 and 3.
3 4 4 (-4+6)(3)-(-4-9)(4)+(8+12)(4)
TSP -4 -4 3 6+52+80=138
3 -2 1 TSP = 138
V = 138 u
first of all, for the next subsections i need the NORMAL ---- i used the Freedom vectors 2 and 3 for a new resulting vectorTSP -7 -8 -1 (24-6)i - (21+0) j + (42+0)k
0 -6 -3 N = <18,-21,42>
2) Calculate ecuation of the plane that passing by V1, V2 and V3 --- i calculated the scalar ecuation and the vectorial ecuation
N⋅r = N⋅V1 = <18,-21,42> ⋅ <x,y,z> = <18,-21,42> ⋅ <3,4,4> 18x-21y+42z = 138
Vectorial ecuation and Scalar ecuation
Vectorial ecuation and Scalar ecuation
3) Calculate the distance D1 without exit to the plane ---- I used a imaginary intersection point 3 in the origin of the intersections (under the plane) and imaginary stright.
x= 18+3t y=-21+4t 0=42+4t <---- t= 21/2 = 10.5 sec that imaginary stright touching the "floor". So....the intersection point 3 is (49.5,21,84)
√(3-49.5)^2 + (4-21)^2 + (4-84)^2 = 94.08 u
BUT: V1 concurring with D1, right? So, is it possible resolve by proyection: <3,4,4> ⋅ <18,-21,42> / √2646 = 3.26 u is it not the same result √(3-49.5)^2 + (4-21)^2 + (4-84)^2 = 94.08 u
4) The vectorial ecuation and the parametrics of L1 that is passing by V4 and have the NORMAL DIRECTION --- I only remplaced the V4 in the Vectorial ecuation
<18,-21,42> ⋅ <x,y,z> = <18,-21,42> ⋅ <1,3,-3>
x= 18+t y= -21+3t z= 42-3t
5) Calculate D2 ---- Free Vector 4 concurring with D2, i need it. x= 18+t y= -21+3t z= 42-3t
Free Vector 4= V1 - V4 = <3,4,4> - <1,3,-3> = <2,1,1>
Proyection of Free Vector 4 = <2,1,1> ⋅ <19,-21,42> / √2646 = 1.10 u
6) Calculate the distance between the intersection points 1 in the plane and in the xy plane (Intersection point 2) ---- Here i dont know will do.Proyection of Free Vector 4 = <2,1,1> ⋅ <19,-21,42> / √2646 = 1.10 u
I really sorry for the big text, but, i tried to resume the max that was possible. Thank you, very much!