# Vector how to find the equation..

• May 27th 2009, 06:40 AM
NaNa
Vector how to find the equation..
here is the question that I have and I have confuse

find the equation of the straight line which is perpendicular to the plane
2x+3y+4z=5
and which goes through the point (1,1,7)
is the point (5,7,15) on this line?

r=(1,1,7)+t(2,3,4)
that maybe is something!?

(Thinking)
• May 27th 2009, 06:56 AM
artvandalay11
yes that's correct
• May 27th 2009, 06:58 AM
NaNa
so the solution is that: r=(1,1,7)+t(2,3,4)?!
• May 27th 2009, 07:13 AM
artvandalay11
yes because the normal vector of a plane given by the equation ax+by+cz+d=0 is [a,b,c]
• May 27th 2009, 08:04 AM
NaNa
ok thx!
• May 27th 2009, 08:33 AM
HallsofIvy
You still have to answer the question "is the point (5,7,15) on this line?"
• May 27th 2009, 08:35 AM
NaNa
for that i have to say x=1+2t y=1+3t and z=7+4t and then solve with respect to t right!?
• May 27th 2009, 09:14 AM
HallsofIvy
Quote:

Originally Posted by NaNa
for that i have to say x=1+2t y=1+3t and z=7+4t and then solve with respect to t right!?

Yes. x= 1+ 2t= 5, y= 1+ 3t= 7, z= 7+ 4t= 15. Obviously you can solve the first equation, 1+ 2t= 5, then check to see if that value works for the other two equations.
• May 27th 2009, 09:15 AM
NaNa
Yeah i have done it!!!:)thx!!!