1.

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 ?

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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