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# help p6 question driving me crazy watch

1. Hi i did this question from a past paper and got it right when doing it. I have since thrown the paper away, and now, reviewing the papers the night before the exam, i CANT get the answer. please put me out of my misery;

the lines have equations r = (5i + j +5k) + s(i - j -2k)

and r = (i +11j + 2k) + t(-4i - 14j +2k)

find the equation of the plane containg both lines in the form ax + by + cz = d

thanks!
2. You want r.n = a.n

n is found by the cross product of the direction vectors of the lines.

a is any point on the plane, either of the two position vectors given should do.

r is of course xi+yj+zk.
3. (Original post by ZJuwelH)
You want r.n = a.n

n is found by the cross product of the direction vectors of the lines.

a is any point on the plane, either of the two position vectors given should do.

r is of course xi+yj+zk.
thats what i was doing, but just cant get the rite answer
4. (Original post by Leeroy)
thats what i was doing, but just cant get the rite answer
What is the right answer? I get two different ones by using different vectors for a.

EDIT: a is (5i+j+5k).
5. i get 5x - y + 3z = 39

edit: that is...

(-4i -14j + 2k) x (i -j -2k)

you cross the direction vectors
6. This is how I always remember the cross product:

Think of it as finding determinants (watch out for the middle one though!)
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7. show me how
8. i'm sure you know how to do the cross product,

just cross the 2 direction vectors to find the normal to the two lines

the r.n = k

r is (x y z), dot that with the normal, 'n'. k is a constant

so when you 'dot' it out, you get 5x -y +3z = k

put in a know point ie (x y z) to get k
9. (Original post by Leeroy)

the lines have equations r = (5i + j +5k) + s(i - j -2k)

and r = (i +11j + 2k) + t(-4i - 14j +2k)

find the equation of the plane containg both lines in the form ax + by + cz = d

Okay...

Cross product the two vectors in the plane to get a normal... Use a determinant to do this like:
¦i 1 -4 ¦
¦j -1 -14 ¦
¦k -2 2 ¦

expand in terms of the i j and k

Dot product one of the 'a' portions, i.e the 5i + j +5k with this normal

to get r . n = d

where d is the result of the dot product.

hence you know that (ai + bj + ck) . n = d

then put in cartesian.

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