Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

**jacks****Member**- Registered: 2012-11-21
- Posts: 80

equation of plane passes through

and making angles of with the planeOffline

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,046

hi jacks,

hhmmm, this looks like it could get complicated. Maybe it will simplify as you go along. I won't get a chance to do this until much later when I'll post again.

Here's the way I'll start:

(i) You need to find the line of intersection of the planes, call it L .

(ii) Then pick a line perpendicular to L in each of the planes.

(iii) Introduce the 45 degree constraint to find the missig parameter for the second plane.

So

write x-4y+z-9=0 in the form r = a + lambda b + mu c

write the other plane in the form r = d + eta e

where a,b,c,d and e need to be found.

Get the line of intersection.

Pick the two lines.

Set the angle to 45.

Eliminate any remaining unknowns.

LATER EDIT:

OR

Maybe let (f,g,h) be a point on the line of intersection.

Get the equation of the target plane using the three points you know have.

Get the normal to plane 2.

Put the two normals at 45 to determine plane 2.

I THINK THIS IS PROBABLY QUICKER AND LESS EQUATIONS TO DEAL WITH.

EVEN LATER EDIT.

I think this is even quicker:

let plane 2 be Lx + my + nz = constant

then vectors (L,m,n) and (1,-4,1) are the normals to the planes and must be at 45 to each other.

So use the scalar product, and the known points to find L,m,n.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline