Asked by Anonymous
Determine the parametric equations of each of the following planes:
--------------------------
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!
--------------------------
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?
--------------------------
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!
--------------------------
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?
Answers
Answered by
Damon
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
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
Answered by
Damon
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.
Answered by
Damon
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!
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!