Finding a plane given a line and plane ortogonal

[the7joker7]

Homework Statement

Find a plane containing the line r(t) = <5,5,-1> + t<-6,-7,-7> and orthogonal to the plane -5 x + 5 y -2 z = 1

The Attempt at a Solution

I...honestly have no idea where to even begin. I don't recall ever being taught how to find a plane given a line and the orthogonal plane. I know how to do it with three points and 1 point plus a line and two vectors...but not a line and an orthogonal plane...could someone point me in the right direction?

 

[gordilloedwin]

Here goes my best guess ( I'm sorry am new at this forum )

get the orthogonal plane and use it to express a vector ( -5i , 5j , -2k ) call it 'a'
then using the scalar product rule express the other vector r(t) = 'b' but make the line within 'b' hit a point in the plane. that is easy .. you may already know

since the formula is

cos thetha = (a·b) / |a||b|

if thetha = 0 then both planes will be orthogonal .. however this is my best fast guess, i know there will be someone who can help you better .. my response just quite sucks .. sorry

 

[HallsofIvy]

You should not have to be taught how to do every possible case! You were taught the definitions and that should be enough. If you can find a single point in the plane and its normal vector, you can write down the equation of the plane.

If a line is in a plane then its direction vector is perpendicular to the normal vector of the plane. If two planes perpendicular then their normal vectors are perpendicular. In order that the plane contain the line r(t) = <5,5,-1> + t<-6,-7,-7> and be perpendicular to the plane -5 x + 5 y -2 z = 1, its normal vector must be perpendicular to both <-6, -7, -7> and <-5, 5, -2>. You can find a vector perpendicular to both by taking the cross product of those two vectors. Of course, any point in the line (take t= 0, say) is a point in the plane.