Question

Find the equation of the plane with the given description:
Contains the lines r1(t)=<2,1,0>+ and r2(t)=<2,1,0>+<3t,t,8t> help? :( @ganeshie8 @perl

Answer 1

You need a `point` and `normal vector` to write teh equation of plane

Answer 2

are you telling me the steps or? because thats all the problem gave me ._.

Answer 3

I'll give a hint : cross product of direction vectors of given lines

Answer 4

ok. hold on...

Answer 5

-_- is it 13x-1y-5z=d?

Answer 6

http://www.wolframalpha.com/input/?i=cross+product+%281%2C2%2C3%29%2C+%283%2C1%2C8%29

Answer 7

normal vector is <13, 1, -5>
use any points on the plane and see jf you can write out the equation

Answer 8

i checked the answer and it is however, it equaled
13x-y-5z=27 to find that 27 id have to find a random point? how?

Answer 9

you should get

13x\(\color{red}{+}\)y-5z=d

Answer 10

oh yeah, +y sorry

Answer 11

you can find the value of "d" by plugging in any point (you're given two lines, think a bit.. )

Answer 12

aren't you given two ready made points as well
?

r1(t)= `<2,1,0>`+<t,2t,3t> and r2(t)= `<2,1,0>`+<3t,t,8t>

Answer 13

yeah ._.

Answer 14

you may use that point

Answer 15

13x+1y-5z=d

plugin the point (2, 1, 0) and find out the value of "d"

Answer 16

ohhh! gosh youre better at explaining and teaching than my UCLA professors.