Question

Find an equation of a plane containing the line r=<4,1,5> + t<2,10,-5> which is parallel to the plane -5x+2y+2z=-13 in which the coefficient of x=-5

Answer 1

this is a very similar problem
what is the normal vector n?

Answer 2

<2,10,-5>?

Answer 3

no, the equation of the plane is found by solving\[\vec n\cdot(\vec P-\vec P_0)=0\]let\[\vec n=\langle a,b,c\rangle\]and\[(\vec P-\vec P_0)=\langle x-p,y-q,z-r\rangle\]and you see that you will get some form\[ax+by+cz=d\]after the dot product. Hence the coefficients of the plane equation form the normal vector to the plane

Answer 4

so the normal vector is n = <-5,2,2>??

Answer 5

yes
now all you need is to form the general equation of a vector in the plane that contains the point Po=(4,1,5)
can you do that? (it's the same deal as last time)

Answer 6

n dot product of P-Po??

Answer 7

yes, exactly

Answer 8

<-5,2,2> dot product <4-x,1-y,5-z>??

Answer 9

yep

Answer 10

its correct thanks again :)

Answer 11

you did most of the work this time :)
welcome though