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natashajane
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[SOLVED] A plane and a point and another point
According to the Flat Earth CLub, Earth is a plane described by the equation
x+y+z = 18
Also according to the Flat Earth Club, Earth will be destroyed on the day this assignment is due by an explosion that will spontaneously occur at the so-called "armageddon point" A = (1,1,1). It so happens that the school (considered as a point) will be the first place on Earth that is destroyed by this explosion (how sad)
a) Calculate the coordinates of the school.
b)What is the distance between A and the school. What is the distance between A and Earth?
a. Form the line that passes through point A with a direction vector that is parallel to the normal vector of the plane.
L: (x,y,z) = (i + j + k) + (i + j + k)t
= (1 + t)i + (1+t)j + (1+t)k
Now substitute into the equation of the plane to determine the point of intersection P.
x = t + 1
y = t +1
z = t + 1
(t+1) + (t+1) + (t +1 ) = 18
3t + 3 = 18
3t = 15
t = 5
The point P is then (6, 6, 6)
b. The distance (D) between A and the Earth/Monash Universtiy (P) which are 3-D points, is calculated by:
D = √[(px-ax)2 + (py-ay)2 + (pz-az)2]
Where P = (px, py, pz) and A = (ax, ay, az)
Therefore: P = (6,7,8) and A = (1,1,1) and
D= √[(6-1)2 + (6-1)2 + (6-1)2
= √75
= 8.66
Homework Statement
According to the Flat Earth CLub, Earth is a plane described by the equation
x+y+z = 18
Also according to the Flat Earth Club, Earth will be destroyed on the day this assignment is due by an explosion that will spontaneously occur at the so-called "armageddon point" A = (1,1,1). It so happens that the school (considered as a point) will be the first place on Earth that is destroyed by this explosion (how sad)
Homework Equations
a) Calculate the coordinates of the school.
b)What is the distance between A and the school. What is the distance between A and Earth?
The Attempt at a Solution
a. Form the line that passes through point A with a direction vector that is parallel to the normal vector of the plane.
L: (x,y,z) = (i + j + k) + (i + j + k)t
= (1 + t)i + (1+t)j + (1+t)k
Now substitute into the equation of the plane to determine the point of intersection P.
x = t + 1
y = t +1
z = t + 1
(t+1) + (t+1) + (t +1 ) = 18
3t + 3 = 18
3t = 15
t = 5
The point P is then (6, 6, 6)
b. The distance (D) between A and the Earth/Monash Universtiy (P) which are 3-D points, is calculated by:
D = √[(px-ax)2 + (py-ay)2 + (pz-az)2]
Where P = (px, py, pz) and A = (ax, ay, az)
Therefore: P = (6,7,8) and A = (1,1,1) and
D= √[(6-1)2 + (6-1)2 + (6-1)2
= √75
= 8.66