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pottersheep
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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 = (x1)/2 = (y+2)/1 = z/3
Please help me. How do I do this? Thank you so much
 one year ago
 one year ago
pottersheep Group Title
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 = (x1)/2 = (y+2)/1 = z/3 Please help me. How do I do this? Thank you so much
 one year ago
 one year ago

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amistre64 Group TitleBest ResponseYou've already chosen the best response.1
find the point on the plane that the xyz of the line fits
 one year ago

pottersheep Group TitleBest ResponseYou've already chosen the best response.0
For paramatric equation of the line I got X = 1+2t Y = 2 –t Z = 3t
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
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?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
dw:1371056632838:dw
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
im thinking we need to go another route, find the angle between the normal and the line
 one year ago

pottersheep Group TitleBest ResponseYou've already chosen the best response.0
We don't deal with angles really..we use this formula though http://easycalculation.com/analytical/pointplane.gif
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
i can never remember the formula, so i tend to go a longer route.
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
if we plug in xyz from the line eq, into the plane we get: 5(1+2t)+3(2t)4(3t) = 18 5+10t 63t +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)
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
dw:1371057451397:dw
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
19*sqrt(2) = 19sqrt(2) 10sqrt(7)*sqrt(2) = 10sqrt(14) = t to define the point is what im thinking
 one year ago

pottersheep Group TitleBest ResponseYou've already chosen the best response.0
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?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
by all means :)
 one year ago

pottersheep Group TitleBest ResponseYou've already chosen the best response.0
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, tt, 3t) If I plug it into the equation for finding the distance of a point to a plane then 19/√2 = 5x +3y4z – 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
 one year ago

pottersheep Group TitleBest ResponseYou've already chosen the best response.0
Once again thank you so much for your help, you're many people's hero :)
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
youre welcome :)
 one year ago
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