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find the point on the plane that the xyz of the line fits
For paramatric equation of the line I got X = 1+2t Y = -2 –t Z = -3t
the normal of the plane is <5,3,-4> a line with that normal is X = 1+5t Y = -2 +3t Z = 0 -4t would you agree?
im thinking we need to go another route, find the angle between the normal and the line
We don't deal with angles really..we use this formula though http://easycalculation.com/analytical/pointplane.gif
i can never remember the formula, so i tend to go a longer route.
if we plug in xyz from the line eq, into the plane we get: 5(1+2t)+3(-2-t)-4(-3t) = 18 5+10t -6-3t +12t = 18 19t = 19, when t=1 ---------------------- X = 1+2 = 3 Y = -2–1 = -3 Z = -3 so the point (3,-3,-3) is the point on the plane that the line peirces thru and the angle between the normal and line is: arccos(19/10sqrt(7)
if we attach the point in the plane to the line vector .. we get an equivalent setup for the line to play with x = 3 +2t y = -3 -1t z = -3 -3t we can use t as a scalar, using the appropriate "length" of the traingle
19*sqrt(2) = 19sqrt(2) 10sqrt(7)*sqrt(2) = 10sqrt(14) = t to define the point is what im thinking
Thanks to what you did, I figured out what I think is another way, and got t = 11. can I type it out for you?
by all means :)
For paramatric equation of the line I got X = 1+2t Y = -2 –t Z = -3t So any point on the line is (1+2t, -t-t, -3t) If I plug it into the equation for finding the distance of a point to a plane then 19/√2 = 5x +3y-4z – 18/(√50) 19/√2 = 5(1+2t) + 3 (-2 –t) -4(-3) = 18/(√50) 19/√2= (-19 + 19t)/√50 19 √100 = (-19 + 19t) 190 = (-19 + 19t) T = 11
Once again thank you so much for your help, you're many people's hero :)
youre welcome :)