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January 31, 2015

January 31, 2015

Posted by **Alex** on Monday, April 7, 2008 at 9:49pm.

Consider the lines r = (1,-1,1) + t(3,2,1) and r = (-2,-3,0) + u(1,2,3).

a) Find their point of intersection.

b) Find a vector equation for the line perpendicular to both of the given lines that passes through their point of intersection.

For a I got the equations

3t-1u+3 = 0

2t-2u+2 = 0

t-3u+1 = 0

Then I put 2 and 3 together and got u = -4.

In the back of the book the answers says (-2, -3, 0). So am I already wrong?

2) Show that the lines r = (4,7,-1) + t(4,8,-4) and r = (1,5,4) + u(-1,2,3) intersect at right angles and find the point of intersection.

Can someone please walk me through these?

- Vectors -
**Reiny**, Monday, April 7, 2008 at 10:11pmWhen I solved equations 2 and 3 I got u=0 and t=-1, giving me r = (-2,-3,0) in both r = ... equations.

check your arithmetic

for the second, if they are perpendicular, then the dot product of their directions vectors should be zero

(4,8,-4)∙(-1,2,3) = -4 + 16 - 12 = 0

so they are perpendicular.

For the intersection point do exactly the same thing you did for the first one without making arithmetic errors.

BTW, your equations for the first one were correct.

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