IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

joyfrancis

q) Find the vector equation of the plane passing through the intersection of the planes r.(i+j+k) = 6 and r.(2i+3j+4k) = -5 and the point (1,1,1).

elastiboysai

Hi joyfrancis,

r.[(i+j+k)+p(2i+3j+4k)]=6-5p

now r=i+j+k satisfies dis

so plug in

1+2p+1+3p+1+4p=6-5p

and u get

p=-3/14

now plug bac in

joyfrancis

didn't get :(

priyesh

equation of plane passing thru two planes is given as P1 + kP2 = 0

where P1=0 & P2 = 0

P1 is given as x + y + z -6 = 0

P2 is given as 2x + 3y + 4z + 5 =0

=> equation of plane passing thru line of intesrsection of planes P1 & p2 is

2x + 3y + 4z + 5 + kx + ky + kz - 6k = 0

this passes thru (1,1,1)

put 1,1,1 in above equation find k & hence equation of plane

raulrag009

punnima

let pi1 nd pi2 b d two planes with eqns
vector n1= i+j+k= 6 and
vector n2= 2i+3j+4k= -5
(since d posn vector of a point on d line of intrscn of d two given planes must satisfy both the above eqns,4 all real values of x ,we have)
r.(n1+ x n2)=d1+ x d2
on subs 4 n1 nd n2,
we get
r.[(1+2x)i +(1+3x)j + (1+4x)k ]= 6-5x ---------------------------------------(1)
taking r =ai +bj+ck
we get
(a+b+c -6)+x(2a+3b+4c+5)=0
since point is given as 1,1,1
subs it for a,b,c
we get,
(1+1+1-6)+x(2+3+4+5)=0
=> x=3/14
subs 4 x in eqn 1
nd d reqd plane is got

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