Asked by Anonymous
Write a vector an parametric equation for the plane which contains the point P(9, 1, -2) and is parallel to [x, y, z] = (4, 1, 8) + [1, -1, 1) and [x, y, z] = (-5, 0, 10) + t(-6, 2, 5).
Answers
Answered by
Reiny
slightly puzzled by your notation.
Looks like you are using [....] to represent vector directions, but in
[x, y, z] = (4, 1, 8) + [1, -1, 1) , there is confusion and [1,-1,1] lacks a parameter.
So assuming the two lines have direction [1,-1,1] and [-6,2,5]
We need a normal to both of those, which is the cross-product
one such normal is [7,11,4] , (I assume you know how find that normal)
So the equation of the plane is 7x + 11y + 4z = k
with (9,1,-2) on it, so ....
63 + 11 - 8 = k = 66
7x + 11y + 4z = 66
Looks like you are using [....] to represent vector directions, but in
[x, y, z] = (4, 1, 8) + [1, -1, 1) , there is confusion and [1,-1,1] lacks a parameter.
So assuming the two lines have direction [1,-1,1] and [-6,2,5]
We need a normal to both of those, which is the cross-product
one such normal is [7,11,4] , (I assume you know how find that normal)
So the equation of the plane is 7x + 11y + 4z = k
with (9,1,-2) on it, so ....
63 + 11 - 8 = k = 66
7x + 11y + 4z = 66
Answered by
Reiny
If you want a parametric equation for the plane, it is very simply:
[x,y,z] = (9, 1, -2) + s[1, -1, 1] + t[-6, 2, 5]
[x,y,z] = (9, 1, -2) + s[1, -1, 1] + t[-6, 2, 5]
Answered by
Anonymous
Isn't [x,y,z] = (9, 1, -2) + s[1, -1, 1] + t[-6, 2, 5] the vector equation not the parametric equation?
Answered by
Reiny
Surely you are not going to tell me that you don't know how to change the vector equation to parametric form.
here is the x,
x = 9 + s - 6t
how about:
y = ...
z = ...
here is the x,
x = 9 + s - 6t
how about:
y = ...
z = ...
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