Question

Question #fd431

Answer 1

`3x-y+z-20=0`

Explanation:

If you don't immediately know what to do, ALWAYS draw a diagram.

Here, I have the plan given, and the one intercepting it. I've marked a normal vector to the plane given, `uln`, as well as the point P (which I marked as X so sorry about that).

If the two planes are perpendicular, then the normal vector to one plane will be a vector parallel to the other plane. Also, the normal vector to each plane will be perpendicular to each other, and dot product to zero.

Let `uln` be the normal vector to the given plane.

`uln=((1),(1),(-2))`

To find a vector perpendicular to this one:

Let `ulm` be a vector normal to the desired plane.

Let `ulm=((m_1),(m_2),(m_3))`

`uln*ulm=0`

`m_1+m_2-2m_3=0`

There are no unique solutions to this, so let's find a perpendicular vector.

Let `m_3=1`

`m_1+m_2=2`

Let `m_2=-1 => m_1=3`

`:. ulm=((3),(-1),(1))`

To check this is normal to the desired plane,

`ulm*uln=((3),(-1),(1))*((1),(1),(-2))`
`=3-1-2`
`=0`

Since we now have a vector normal to our desired product, we can use the scalar product equation of a plane:

`ulr*ulm=ulp*ulm`
`ulr*((3),(-1),(1))=((5),(-1),(4))*((3),(-1),(1))`

`ulr*((3),(-1),(1))=15+1+4`

`ulr*((3),(-1),(1))=20`

In Cartesian form:

`3x-y+z-20=0`