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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
 10 months ago
 10 months 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
 10 months ago
 10 months ago

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amistre64Best ResponseYou've already chosen the best response.1
find the point on the plane that the xyz of the line fits
 10 months ago

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

amistre64Best 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?
 10 months ago

amistre64Best ResponseYou've already chosen the best response.1
dw:1371056632838:dw
 10 months ago

amistre64Best 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
 10 months ago

pottersheepBest 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
 10 months ago

amistre64Best ResponseYou've already chosen the best response.1
i can never remember the formula, so i tend to go a longer route.
 10 months ago

amistre64Best 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)
 10 months ago

amistre64Best ResponseYou've already chosen the best response.1
dw:1371057451397:dw
 10 months ago

amistre64Best 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
 10 months ago

amistre64Best 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
 10 months ago

pottersheepBest 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?
 10 months ago

pottersheepBest 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
 10 months ago

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