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Given information:
|vector AB| = 3*|vector AC|
angle betweem AB and AC is 30 deg
The coodinate points are given in the paint picture
Is it possible to locate the point C?
the inner product would be
9(squrt(3)/2) = 3*(x-4) + 3(y-5) + 3(z-6)
x + y + z = 3*squrt(3)/2 + 11
its the x component of BC times the magnitude of AB
AC = (x-1)i + (y-2)j + (z-3)k
Inner product of AB and BC = |AB||BC|Cos(∏/6) = (√27)(√27/3)(√3/2) =
=9√3/2 = 3(x-4) + 3(y-5) + 3(z-6) ... x + y + z = 1/3(9√3/2 + 45)
Eq 1: AC = (x-1)i + (y-2)j + (z-3)k
Eq 2: x + y + z = 1/3(9√3/2 + 45)
Eq 3:
Sorry I am unfamilliar with coplanar points... this is a problem i made up so if you could tell me about the formula you are referring to about coplanar points I can look it up in my calc book.
Would the two other equation be
vector Projection of BC onto AB
and (Look at picture to see point Vector PC)
PC = BC - (vector Projection of BC onto AB)
Ok to find the magnitude of a vector using components you square root the square of its components
so if the components of AB are x,y and z then I believe that the components of BC would be 9x,9y, and 9z because there must have been a factor of 9 for each component to pull a 3 out infront of |BC|
Bc= < 13/3,16/3,19/3>
So I am guessing I could have solved this another way by finding three equations that delt with scalars rather than three equations that delt with vectors. Is that correct?
The equation x+y+z=(3/2)√3 +15 describes the surface whose coordinates of point (x,y,z) add up to (3/2)√3 +15
(x-4)2 + (y-5)2 + (z-6)2=3 defines all points whose distance from (4,5,6) to (x,y,z) is squrt(3).
The intersection of these two would be some type of circle I am guessing.. is any of this correct?