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pottersheep
Grade 12 Vectors . Find a point P on the line L situated 19√2 units distance from plane W.? Find a point P on the line L situated 19√2 units distance from plane W. W = 5x + 3y - 4z = 18 L = (x-1)/2 = (y+2)/-1 = z/-3 Please help me. How do I do this? Thank you so much
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?
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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)
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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?
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 :)