# 1) Obtain the equation of the plane which passes through the intersection of the two planes x+y-z-1=0 and 2x+3y+z-6=0 and which is parallel to the straight line (x/2)= (y/3)= (z/6).

kkbisht
90 Points
4 years ago
We know that equation of a plane $\pi$=0 passing through the intersection of two planes $\pi$1=0 and $\pi$2=0 is given by
$\pi$1 + k $\pi$2=0   therefore we have x+y-z-1 +k(2x+3y+z-6)=0 or (1+2k)x +(1+3k)y +(k-1)z -1-6k=0 ….......(1)
ihere direction ratios of normal to the plane  are 1+2k, 1+3k, k-1 Now the plane is parallel to the line x/2 = y/3 = z/6 whose direction ratios are  2,3 6 .Now since the plane is parallel to the line that  means the normal to the plane is perpendicular to the line and therefore product of the direction ratios must be zero ( as cos 90=0)
so (1+2k).2 +(1+3k).3 +(k-1)6=0 => 2+3-6+4k+9k+6k=0 => 19k=1=> k=1/19
Now substitute this value of k=1/19 in equation (1)
(1+ 2.1/19)x + (1+3.1/19)y + (1/19-1)z -1-6.1/19=0
which simplifies to      21x+22y-18z-25=0

kkbisht