Question

find the line that passes through the point ( 2, 0, 1) and is perpendicular to plane 2x -4y-5z + 8 =0

Answer 1

strip the normal and apply it to the stated point

Answer 2

the normal IS perp to the plane .. therefore ....

Answer 3

the normal is (2 , -4 , -5 ) ? how do I find the line ?

Answer 4

add some multiple, t, of the normal vector to the state point

Answer 5

L = P + n

Answer 6

the parametric line would be to define xyz in their own right

Answer 7

x = px + (nx)t
y = py + (ny)t
z = pz + (nz)t

Answer 8

this can also be expressed as the matrix product:
\[
\begin{pmatrix}
px&nx\\
py&ny\\
pz&nz\\
\end{pmatrix}
\begin{pmatrix}
1\\t
\end{pmatrix}
=
\begin{pmatrix}
x\\y\\z
\end{pmatrix}

\]

Answer 9

the nx is the normal vectors ? what is px then?

Answer 10

point (x,y,z) px is my own shorthand for the x component of the given point ... nx has a similar concept

Answer 11

Sorry, ok px is the given point? (2,0,1)

Answer 12

the given point is:( 2, 0, 1)

px = 2
py = 0
pz = 1

Answer 13

Eq. of any line through (x1,y1,z1) with direction ratios a,b,c is
\[\frac{ x-x1 }{a }=\frac{ y-y1 }{ b }=\frac{ z-z1 }{c }\]
Here (x1,y1,z1)=(2,0,1)
a,b,c are proportional to 2,-4,-5 respectively.