Question #fd431

1 Answer
Feb 26, 2018

3x-y+z-20=0

Explanation:

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

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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