Determine the parametric equations of each of the following planes:
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g) the plane containing the 2 intersecting lines:
r = (5,4,2) + t(4,-2,1)
r = (1,6,-6) + s(6,-4,4)
My Attempt:
x = 5 + 4t + 6s
y = 4 -2t - 4s
z = 2 + t + 4s
####What did I do wrong in this question? Please point out and correct my mistake!
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h)the plane containing the point (8,3,5) and the line:
r = (1,3,-1) + t(2,2,-5)
My Attempt:
x = 8 + 2t
y = 3 + 2t
z = 5 - 5t
####How do I find the other direction vector for this question?
I can not find values of s and t to make those two lines intersect.
At the intersection, the x, y and z values must be the same but
for x : 5 + 4 t = 1 + 6 s
for y : 4 - 2 t = 6 - 4 s
for z : 2 + 1 t = -6 + 4 s
In the second problem find the equation of a vector through the point perpendicular to the first line ( the line better not go through (8,3,5) )
the plane will then contain those two lines
Oh or more easily in the second problem, a line containing (1,3,-1) where your given vector starts and (8,3,5) lies in the plane so you could use it as the second vector.
h)the plane containing the point (8,3,5) and the line:
r = (1,3,-1) + t(2,2,-5)
first find the vector from (1,3,-1) to (8,3,5)
7 i + 0 j + 6 k
the given vector has components
2 i + 2 j -5 k
the cross (vector ) product is perpendicular to both and therefore perpendicular to the desired plane
I get for the normal vector:
N = -12 i + 47 j + 14 k
now we know the point (8,3,5) lies in the plane so a line from Point P in the plane perpendicular to N lies in the plane so the dot product of that with N is 0
(-12)(x-x1) + 47 (y -y1) +14(z-z1) = 0
-12(x-8) + 47 (y-3) + 14 (z-5) = 0
-12 x + 47 y + 14 z = 115
CHECK my arithmetic!
For question g), to determine the parametric equations of the plane containing the two intersecting lines, the mistake in your attempt lies in the equations you set up.
To find the parametric equations of a plane, we need a point on the plane and two vectors that lie on the plane.
First, let's find a point on the plane. Since the plane contains the two intersecting lines, we can set the two equations equal to each other:
(5,4,2) + t(4,-2,1) = (1,6,-6) + s(6,-4,4)
Comparing corresponding components, we have the following system of equations:
5 + 4t = 1 + 6s
4 - 2t = 6 - 4s
2 + t = -6 + 4s
Solving this system, we find t = -2 and s = 1.
Substituting these values back into either of the initial lines, we obtain the point (1, 0, -1).
Now, let's find two vectors that lie on the plane. One vector can be found by taking the difference of the two points on the intersecting lines:
(4,-2,1) - (6,-4,4) = (-2,2,-3)
Another vector can be found by taking the cross product of the direction vectors of the two lines:
(4,-2,1) × (6,-4,4) = (-6, 4, -4)
So, the parametric equations of the plane containing the two intersecting lines are:
x = 1 - 2t - 6s
y = 0 + 2t + 4s
z = -1 - 3t - 4s
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For question h), to find the other direction vector for the plane containing the point (8,3,5) and the line, you can choose any vector that is not parallel to the line direction vector (2,2,-5).
One way to do this is to take the cross product of the line direction vector and some other arbitrary vector. Let's choose the vector (1,0,0):
(2,2,-5) × (1,0,0) = (0, 5, -2)
So, the vector (0, 5, -2) is perpendicular to the line direction vector and can be used as another direction vector for the plane.
Therefore, the parametric equations of the plane containing the point (8,3,5) and the line are:
x = 8 + 2t + 0s
y = 3 + 2t + 5s
z = 5 - 5t - 2s