
#1
Aug2611, 07:43 AM

P: 4

1. The problem statement, all variables and given/known data
The plane P1 contains the points A,B,C, which have position vectors a=(0,0,0), b=(1,1,8) and c=(0,1,5) respectively. Plane P2 passes through A and is orthogonal to the line BC, whilst plane P3 passes through B an is orthogonal to the line AC. Find the coordinates of r, the point of intersection of the three planes 3. The attempt at a solution I have found the equation of plane P1 using the vector product of vectors AB and AC, but I can't find the equation of the line perpendicular to BC which passes through A to find plane P2. Can anyone help? Thanks 



#2
Aug2611, 09:22 AM

Sci Advisor
HW Helper
Thanks
P: 26,167

hi cluivee! welcome to pf!
does the point of intersection lie in the same plane as ABC ? 



#3
Aug2611, 11:05 AM

P: 4

It must do surely, otherwise it wouldn't intersect all 3 planes?




#4
Aug2611, 12:04 PM

Sci Advisor
HW Helper
Thanks
P: 26,167

Find the point of intersection of three planesso doesn't that make it fairly easy to find? 



#5
Aug2611, 01:25 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,886

1. Find the equation of plane p1.
2. Find the equations of plane p2 and p3. 3. Find the equation of the line of intersection of any two of those three planes. 4. Find the point at which that line intersects the third plane. 



#6
Sep2211, 10:53 AM

P: 1

Okay, I can't guarantee if I have understood the question perfectly, but here's what I came up with:
Plane 1 normal vector = CA x BA = (0,1,5) x (1,1,8) = (3,5,1) So the equation can be written as 3x+5yz+k=0. Since it passes point B, 3(1)+5(1)(8)+k=0, so k=0. (3+58+k=0) The equation I got is therefore 3x+5yz=0. Plane 2 This plane is perpendicular to the line passing points B and C as you mentioned. Therefore, the line joining B and C is a normal to this plane. BC = C  B = (0,1,5)  (1,1,8) = (1,0,3) = normal vector of B The equation can be written as x3z+k=0, or x+3z+k=0 if you'd like. Since it also passes point A, (0)+3(0)+k=0, so k=0. The equation I got is therefore x+3z=0. Plane 3 This plane is perpendicular to the line passing A and C as you mentioned. Therefore, the vector AC is a normal to this plane. AC = C  A = (0,1,5)  (0,0,0) = (0,1,5) = normal vector of C The equation can be written as y+5z+k=0. Since it also passes point B, (1)+5(8)+k=0, so k= 41 (1+40+k=0) The equation I got is therefore y+5z41=0 So let's solve for the variables now, given point r = (x,y,z). (A) 3x+5yz=0 (B) x+3z=0 (C) y+5z41=0 I will eliminate 'z' as it appears nonzero in all three equations. 3(A)+(B) (9x+15y3z) + (x+3z) = 10x+15y=0 (1) and 5(A)+(C) (15x+25y5z) + (y+5z41) = 15x+26y41=0 (2) Solve for 'x' and 'y' given these two equations. (1) 10x+15y=0 (2) 15x+26y41=0 Eliminate x. 3(1)  2(2) (30x+45y)  (30x+52y82) = 0 7y+82=0 7y=82 y = 82/7 Substitute to either equation. (1) 10x+15(82/7)=0 10x+1230/7=0 10x=1230/7 x = 123/7 Substitute x and y into plane (A). 3(123/7) + 5(82/7) z = 0 369/7 + 410/7  z = 0 41/7  z = 0 z = 41/7 Point 'r' is therefore r(123/7,82/7,41/7). 


Register to reply 
Related Discussions  
Find the intersection of three planes (a line)?  Precalculus Mathematics Homework  2  
Help!Given a common intersection point, create 3 different planes  Calculus & Beyond Homework  3  
plane through point and intersection of 2 other planes  Calculus & Beyond Homework  9  
Intersection of two planes (without a given point)  Precalculus Mathematics Homework  2  
Why must every homogeneous system of planes have at least 1 intersection point  General Math  1 