anonymous one year ago Calc. III Find an equation of a plane containing the line r = <-5,2,-5> + t<11,2,-1> which is parallel to the plane -1x+3y-5z=31 in which the coefficient of x is -1.

1. jim_thompson5910

anything parallel to -1x+3y-5z=31 will be in the form -1x+3y-5z=D where D is a constant

2. jim_thompson5910

Find an equation of a plane containing the line r = <-5,2,-5> + t<11,2,-1> so we know the point (x,y,z) = (-5,2,-5) is on the line and on this unknown plane

3. Jhannybean

Well, we're also given the $$\vec n$$ of the parallel plane

4. Jhannybean

$\vec n = \langle -1~,~3~,~-5\rangle$

5. anonymous

Okay, so how do we put all the given information together to find the equation of the plane??

6. jim_thompson5910

at this point you just plug (x,y,z) = (-5,2,-5) into -1x+3y-5z=D to find D

7. anonymous

Found out the D = 26

8. anonymous

Scratch that D = 36

9. jim_thompson5910

side note: take the dot product of <11,2,-1> and the normal vector to the plane that Jhannybean wrote you should get a dot product of 0 which shows that the entire line is contained in the parallel plane

10. anonymous

So the final answer will be -1x+3y-5z-36 :) TYYY!!!!

11. jim_thompson5910

-1x+3y-5z-36 = 0 or -1x+3y-5z=36

12. Jhannybean

Oh I see, your $$P_0 = (-5~,~2~,~-5)$$ , cross checking whether your normal vector and position vector = 0, we'll know if theyre orthogonal to one another. Then with $$\vec n = \langle -1~,~3~,~-5\rangle$$ and $$P_0 = (-5~,~ 2~,~-5)$$ we can fidn the equation of a parallel plane by :$a(x-x_0)+b(y-y_0)+c(z-z_0)=0$

13. Jhannybean

Thats how I would find it...

14. Jhannybean

$-1(x-(-5)) +3(y-2)-5(z-(-5))=0$$-(x+5)+3(y-2)-5(z+5)=0$$-x-5 +3y-6-5z-25=0$$-x+3y-5z=25+6+5$$-x+3y-5z=36$

15. Jhannybean

|dw:1444016189424:dw| So you'll have like a vector running in between planes...Haha