Question

Hello !! Here's my question :
find the equation of a plane that contains the line x-1=(y+1)/2=z and it's perpenticular to the plane 3x+y-5z+1=0

Answer 1

find the equation of a plane means we need to find its normal (vector perpendicular to the plane's surface) its vector will be \( \vec{N} \cdot \vec{P} = b \)

If a plane contains the line x-1=(y+1)/2=z , its normal vector N will be perpendicular to that line (N will be perpendicular to all lines that are in the plane)
the line is in symmetric form (see http://tutorial.math.lamar.edu/Classes/CalcIII/EqnsOfLines.aspx)

in vector form, its equation is \( \vec{P}= (1,-1,0)+ t (1,2,1)\)
its direction is (1,2,1). The planes normal N will be perpendicular to this direction.

Also, we want the plane to be
perpenticular to the plane 3x+y-5z+1=0

that means its normal vector N will be perpendicular to this planes normal = (3,1,-5)

to find a vector perpendicular to the vectors (1,2,1) and (3,1,-5), find the cross product.
this will be the normal you want
then use
N dot (1,-1,0)= b
to solve for b

Answer 2

thanks a lot !!