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Given planes P1: cy+bz=x P2 : az+cx =y P3: bx+ay=z P1, P2 and P3 pass through one line, if (A) a 2 +b 2 +c 2 =ab+bc+ca (B) a 2 +b 2 +c 2 +2abc=1 (C)a 2 +b 2 +c 2 =1 (D) a 2 +b 2 +c 2 +2ab+2bc+2ca+2abc=1

Given planes 
P1: cy+bz=x
P2 : az+cx =y
P3: bx+ay=z
 
P1, P2 and P3 pass through one line, if
(A) a2+b2+c2=ab+bc+ca
(B) a2+b2+c2+2abc=1
(C)a2+b2+c2=1
(D) a2+b2+c2+2ab+2bc+2ca+2abc=1

Grade:11

2 Answers

Yash Jain
55 Points
6 years ago
The answer to your question is B) a2+b2+c2+2abc=1. 
I’ve got it using a bit long method so try to solve it with any better and easier one. I used the concept of family of Planes.What I did is as follows:
The equation of family of planes from P1 and P2 is given as,
-→ -x+cy+bz + L(cx-y+az) = 0
-→ (-1+Lc)x + (c-L)y + (b+La)z = 0   ---(i)
Since P3 also passes through the line of intersection, i.e.P1, P2 and P3 pass through one line,Thus the coeff. of P3 and that of eq (i) are proportional. So,
(-1+Lc)/b = (c-L)/a = (b+La)/-1 = k (say)   ---(ii)
Now, lets start eliminating the assumed constants.
From eq (ii) we get L = (bk+1)/c {= c-ka = -(1+b)/a}--(iii)
From (iii), conclude that k = -(bc+a)/(ab+c)
Putting this value of k in the equation (bk+1)/c = c-ka, we’ll get to our answer which is...

a2+b2+c2+2abc=1
I am sure you’ll be able to find more convenient method for the same. Best wishes.

 
Vivekanandan
13 Points
2 years ago
Try it it with coplanarity conditions
 
So
Determinant of |-1 c b|
                            |c -1 a|  =0
                            |b a -1|
So.  -1(1-a2) -c(-c-ab) +b(ac+b)
          = -1+a2 + c2+abc +abc+b2 = 0
           = a2+b2+c2+2abc = 1
 
 
 
 
Hence proved
 
        
 
 
 
 
 
 

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