P3 vectors Watch

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idiopathic
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#1
Report Thread starter 13 years ago
#1
Find the point of intersection of the line through the points (2, 0, 1) and (-1, 3, 4) and the line through the points (-1, 3, 0) and (4, -2, 5). Calculate the acute angle between the 2 lines.


for the first part I keep getting (8, -6, 7) which is wrong I think.
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dvs
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#2
Report 13 years ago
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r1 = (2, 0, 1) + t(1, -1, -1)
r2 = (-1, 3, 0) + s(-1, 1, -1)

Equate coefficients of i, j and k:
2+t = -1-s
-t = 3+s
1-t = -s

Solve these smultaneously to get:
s = -2
t = -1

So the point of intersection is:
(2, 0, 1) - (1, -1, -1) = (1, 1, 2)
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RichE
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#3
Report 13 years ago
#3
(Original post by endeavour)
Find the point of intersection of the line through the points (2, 0, 1) and (-1, 3, 4) and the line through the points (-1, 3, 0) and (4, -2, 5). Calculate the acute angle between the 2 lines.


for the first part I keep getting (8, -6, 7) which is wrong I think.

Lines are given parametrically by

(2,0,1) + s(-3,3,3)
(-1,3,0)+t(5,-5,5)

t = 2/5 and s = 1/3 giving point of intersection (1,1,2)

then to find the angle between them this is

[email protected] = (-1,1,1).(1,-1,1)/rt(3)rt(3) = -1/3 (this would give the larger angle)

but we're looking for the acute angle so we need to solve [email protected] = 1/3
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Fermat
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#4
Report 13 years ago
#4
Intersection point is (1,1,2)
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