InterviewSolution
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                                    Find the equation of family of planes through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x – y + z = 0 and parallel to r.(i + j + k) = 0 ? | 
                            
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Answer»  The equation of the family of planes through the line of intersection of planes x + y + z = 1 and 2x + 3y + 4z = 5 is, (x + y + z – 1) + k( 2x + 3y + 4z – 5) = 0 ……(1) (2k + 1)x + (3k + 1)y + (4k + 1)z = 5k + 1 It is perpendicular to the plane x – y + z = 0 (2k + 1)(1) + (3k + 1)( – 1) + (4k + 1)(1) = 5k + 1 2k + 1 – 3k – 1 + 4k + 1 = 5k + 1 K = \(-\cfrac13\) Substituting k = in eq.(1) , We get, x – z + 2 = 0 as the equation of the required plane And its vector equation is \(\vec r.(\hat i-\hat k)+2=0\) The equation of the family of a plane parallel to \(\vec r.(\hat i+\hat j+\hat k)=0\) is \(\vec r.(\hat i+\hat j+\hat k)=0\)…… (1) If it passes through (a, b, c) then (\(a\hat i+b\hat j+c\hat k\) )(\(\hat i+\hat j+\hat k\) ) = d a + b + c = d Substituting a + b + c = d in eq.(1), we get, \(\vec r.(\hat i+\hat j+\hat k)\) = a + b + c x + y + z = a + b + c as the equation of the required plane.  | 
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