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).
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
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]
Isn't [x,y,z] = (9, 1, -2) + s[1, -1, 1] + t[-6, 2, 5] the vector equation not the parametric equation?
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 = ...
To find a vector and parametric equation for a plane, we need to determine a normal vector to the plane and a point on the plane.
First, we find the normal vector by taking the cross product of the given vectors that are parallel to the plane.
Vector A = (1, -1, 1)
Vector B = (-6, 2, 5)
The normal vector N is given by the cross product of A and B:
N = A x B
Using the determinant method, we can calculate the cross product:
N = (i, j, k)
(1, -1, 1)
(-6, 2, 5)
N = ((-1 * 5) - (1 * 2), (1 * -6) - (1 * 5), (1 * 2) - (-1 * -6))
= (-7, -11, 8)
So, the normal vector to the plane is N = (-7, -11, 8).
Now, we have the normal vector and a point on the plane P(9, 1, -2). We can use this information to write the vector and parametric equations of the plane.
1. Vector Equation:
(x, y, z) = P + tN
Substituting the values:
(x, y, z) = (9, 1, -2) + t(-7, -11, 8)
= (9 - 7t, 1 - 11t, -2 + 8t)
So, the vector equation of the plane is (x, y, z) = (9 - 7t, 1 - 11t, -2 + 8t).
2. Parametric Equations:
x = 9 - 7t
y = 1 - 11t
z = -2 + 8t
So, the parametric equations of the plane are:
x = 9 - 7t
y = 1 - 11t
z = -2 + 8t
These equations represent the plane that 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).