Question

Find the plane that contains the line
x=−2+3t,y=4+2t, z = 3 − t
and is perpendicular to the plane
x − 2y + z = 5.

Answer 1

Normal vector of the given plane x-2y+z=5 is \(\vec n_1=<1,-2,1>\)

Then normal vector of the required plane, namely \(n_2\), is a vector such that \(\vec n_1\bullet \vec n_2 =0\)

Answer 2

Let \(\vec n_2=<a,b,c>\), then \(\vec n_1\bullet\vec n_2 = a-2b+c=0\)

Answer 3

@Loser66 how did u get a−2b+c ?

Answer 4

the first plane is defined as \(\pi_0: x - 2y + z = 5 \)

and \(\pi_o\)'s normal vector is: \(\vec n_o =<1, -2, 1> \)


the plane you are looking for, \(\pi_1\), will follow both of these ideas:

- it is perpendicular to \(\pi_o\), so it contains \(\pi_o\)'s normal vector

and

- it contains the line \(\vec l = <-2,4,3> + t <3,2,-1>\), so it contains the \(<3,2,-1>\) direction vector

So, IOW, it contains the direction vectors \(<1, -2, 1>\) and \(<3,2,-1>\)

So \(\pi_1\)'s normal vector \(\vec {n}_1\) is the cross product of those 2 vectors, ie:

\(\vec {n}_1 = \det \left( \begin {matrix}
\hat i & \hat j & \hat k \\
3 & 2 & -1\\
1 & -2 & 1
\end{matrix} \right)
\)

\(\equiv <0, 1, 2>\)

using the usual equation for a plane, ie

\((\vec r - \vec r_o) \bullet \vec n = 0\)

\(\implies \vec r \bullet \vec n = \vec r_o \bullet \vec n\)


So.....from there you plug in the common point, \(\vec r_o = <-2, 4, 3>\)


ie

\(<x,y,z> \bullet <0,1,2> = <-2,4,3> \bullet <0,1,2>\)

that's:

\(y + 2 z = 10\)

that's a complete solution but i have no idea whether it's right or not, so it's really a steer on methodology.