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pottersheep
 3 years ago
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
pottersheep
 3 years ago
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

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amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1find the point on the plane that the xyz of the line fits

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

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1the 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?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1371056632838:dw

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1im thinking we need to go another route, find the angle between the normal and the line

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

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1i can never remember the formula, so i tend to go a longer route.

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1if 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)

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1371057451397:dw

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1if 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

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.119*sqrt(2) = 19sqrt(2) 10sqrt(7)*sqrt(2) = 10sqrt(14) = t to define the point is what im thinking

pottersheep
 3 years ago
Best ResponseYou've already chosen the best response.0Thanks to what you did, I figured out what I think is another way, and got t = 11. can I type it out for you?

pottersheep
 3 years ago
Best ResponseYou've already chosen the best response.0For 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

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