Wikipedia
equation of a line in space : The graph of the equation is a straight line, and every straight line can be .... in n-dimensional Euclidean space (for example, a plane in 3-space). ..... More from Wikipedia
Space Waves
Space waves travel in (more or less) straight lines. But they depend on line-of-sight conditions. So, they are limited in their propagation by the curvature of the Earth, except in very unusual circumstance..
Space Waves
Space waves travel in (more or less) straight lines. But they depend on line-of-sight conditions. So, they are limited in their propagation by the curvature of the Earth, except in very unusual circumstances. Thus, they propagate very much like electromagnetic waves in..
EINSTEIN'S RELATIVITY: Everything in the universe is traveling through space-time at the speed of light - the maximum speed possible. If you are sitting still in space, then you are traveling through time at the maximum speed. But if you begin traveling through space, then your progress through time slows down. Time Dilation and other relativistic phenomena await you in this interesting series, so hurry up and slow down!
Equation of a straight line in any dimensional space Forget y = mx + b
Question : Suppose i have P(0,0,0) and Q(0,1,2). How would i find and equation of a line that contains this two points?
Answer : The vector PQ points in the direction of the line. PQ = (0-0)i + (1-0)j + (2-0)k Parametric equations for the line are x = 0 + 0s y = 0 + 1s z = 0 + 2s or more simply x = 0 y = s z = 2s.. More from Yahoo Answers
Answer : The vector PQ points in the direction of the line. PQ = (0-0)i + (1-0)j + (2-0)k Parametric equations for the line are x = 0 + 0s y = 0 + 1s z = 0 + 2s or more simply x = 0 y = s z = 2s.. More from Yahoo Answers
Question : Given the line L, defined by the following parametric equations:
x = 2t
y = 1 - t
z = 2 + t
And given the point P (0, 2, 1), what is the equation of the line passing through P that is perpendicular to L?
Answer : Let X be a point on the line. Then X has coordinates (2t, 1-t, 2+t). Since we want P to be perpendicular to L, the distance from P to L must be a minimum. So the problem is equivalent to finding the shortest distance between point P and line L, which is the same as minimizing PX. Now, PX = <2t, -1-t, 1+t>, so PX^2 = 4t^2 + 2(1+t)^2 = 6t^2 + 4t + 2 PX^2 is minimized when 12t + 4 = 0, or t = -1/3. Thus PX is minimzed when t = -1/3 or X = (-2/3, 4/3, 5/3). Now the line in question contains both P and X, and has a direction vector <2/3, 2/3, -2/3> that can be scaled to <1, 1, -1>. So the equation of the line in question is x = t, y = 2+ t, z = 1-t... More from Yahoo Answers
Answer : Let X be a point on the line. Then X has coordinates (2t, 1-t, 2+t). Since we want P to be perpendicular to L, the distance from P to L must be a minimum. So the problem is equivalent to finding the shortest distance between point P and line L, which is the same as minimizing PX. Now, PX = <2t, -1-t, 1+t>, so PX^2 = 4t^2 + 2(1+t)^2 = 6t^2 + 4t + 2 PX^2 is minimized when 12t + 4 = 0, or t = -1/3. Thus PX is minimzed when t = -1/3 or X = (-2/3, 4/3, 5/3). Now the line in question contains both P and X, and has a direction vector <2/3, 2/3, -2/3> that can be scaled to <1, 1, -1>. So the equation of the line in question is x = t, y = 2+ t, z = 1-t... More from Yahoo Answers
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