Sunday

April 26, 2015

April 26, 2015

Posted by **sania** on Wednesday, December 12, 2012 at 11:13am.

- maths --plse help me.. -
**Steve**, Wednesday, December 12, 2012 at 11:38amL1:

x = 0 + 4t

y = -1 + 6t

z = -1 + 2t

L2:

x = 3 - 7s

y = 9 - 5s

z = 4 + 0s

L1 intersects L2 if

4t = 3-7s

-1+6t = 9-5s

-1+2t = 4

(t,s) = (5/2, -1) is the solution, so

(x,y,z) = (10,14,4)

- maths --plse help me.. -
**Reiny**, Wednesday, December 12, 2012 at 11:46amdirection vector of 1st line = (4,6,2) or reduced to (2,3,1)

parametric equation of 1st:

x = 0 + 2t

y = -1 + 3t

z = -1 + t

direction vector of 2nd line = (7, 5, 0)

parametric equation of 2nd:

x = 3 + 7k

y = 9 + 5k

z = 4

solving for x's

0+2t = 3 + 7k

2t = 3+ 7k (#1)

solving for y's

3t-1 = 5k+9

3t - 5k = 10 (#2)

using the z's :

t-1 = 4

t = 5

sub that into #1:

10 = 3 + 7k

k = 1

so t=5 and k = 1

check for consistency in #2 (the equation not used to solve)

3t - 5k = 10

LS = 15 - 5 = 10 = RS

so the point of intersection is: (Using the 1st, could use either one)

x = 0+2t = 10

y = -1+3t = 14

z = -1 + t = 4

They intersect at (10,14,4)

- maths --plse help me.. -
**MathMate**, Wednesday, December 12, 2012 at 12:17pmL(ab): (0,-1,-1)+(4u,6u,2u)

L(cd): (3,9,4)+(-7v,-5v,0)

If they intersect at P, then corresponding x,y,z values match, hence

2u-1 = 0 => u=2.5

where

P=(0+4*2.5, -1+6*2.5, -1+2*2.5)=(10,14,4)

For line L(cd),

we have P(3-7v, 9-5v, 4+0v)

=>

3-7v=10 => v=-1

or

9-5v=14 => v=-1

or

4+0v=4 => v=-1 (works).

Therefore they intersect at P, since P(10,14,4) lies on both lines.