# Equation of plane with a point and a perpendicular line

Discussion in 'Undergraduate Math' started by jajcorda, Dec 3, 2008.

1. ### jajcordaGuest

The question is:
An equation of the plane containing the point P (1, âˆ’1, 2) and perpendicular to the line (x, y, z) = (2 âˆ’ 3t, 1 + 5t, 7 + t) is:

So, I know I can find the equation of the plane by getting the cross product of two vectors in the plane (to get the normal)... since the point is already given.
...And I don't really see how I can use projections here...

Since I am given a line that is perpendicular to the plane, is it right to assume that the vector (-3,5,1) is in the plane? I'm really not sure where to start... Could someone help on where to begin the question...

Thanks

jajcorda, Dec 3, 2008

2. ### VirgilGuest

An equation of a line containing the point P(a1,a2,a3) and perpendicular
to the line

f(t) = (x,y z) = (b1 + c1*t, b2 + c2*t, b3 + c3 w2*t)
is
g(t) = (x,y,z) = (a1 + c1*t, a2 + c2*t, a3 + c3 w2*t)

Virgil, Dec 3, 2008

3. ### jajcordaGuest

I actually tried doing this, but didn't know where to go from there... to get the scalar equation of the plane, I would need to find another vector other than the direction vector of that line (d = (-3,5,1)). If I can find another vector, then I can take the cross product and get the normal.

Is it possible to take any arbitrary point in the plane (A) and then get vector AP by subtracting the arbitrary Point A from Point P given in the equation? I'm really not sure what to do...?

Thanks so much.

jajcorda, Dec 3, 2008
4. ### Mike TerryGuest

perpendicular to the line (x, y, z) = (2 âˆ’ 3t, 1 + 5t, 7 + t) is:
product of two vectors in the plane (to get the normal)... since the point
assume that the vector (-3,5,1) is in the plane? I'm really not sure where
to start... Could someone help on where to begin the question...
In effect, you have a point P in the plane, and a normal vector n to the
plane. For a point Q in your plane, the vector PQ will be perpendicular to
the normal, so you can use the dot product to express this.

Regards,
Mike.

Mike Terry, Dec 3, 2008
5. ### Ken PledgerGuest

But the normal has been given to you, free.

Ken Pledger.

Ken Pledger, Dec 3, 2008
6. ### jajcordaGuest

Thanks!

This was my original reasoning for the problem, but I made a really stupid mistake... because what I did was:

I took (-3,5,1) as the normal, and then then took
(-3,5,1) * ((x,y,z)-(1,-1,2)) = 0

As, (-3,5,1) is a normal, and ((x,y,z)-(1,-1,2)) would represent a vector PQ...

To get:
-3x + 5y + z + 6 = 0, or -3x + 5y + z = -6

I took this question from a practice exam, which had this option, which is obviously correct:
(B) 6x - 10y - 2z = 12

I was just having a very stupid moment and not realizing this was the same thing. Thanks a bunch for helping me realize this was the correct process!!!

jajcorda, Dec 4, 2008