IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry - from 3-D

ankitagg

from 3-D

ankitagg

Q... the equation of plane throgh point (1,0,-1) ,(3,2,2) and parallel to the line

x-1/1=1-y/2=z-2/3 is..........

Q....the locus of a point which moves such that sum of squares of its distances

from planes x-y+z=0,x+2y+z=0 and x-z=0 is 16 is.......

(please i want solutions quickly)

Conjurer

1) Let the plane be ax + by + cz +d =0

Putting (1,0,-1), we get a -c +d =0
Putting (3,2,2), we get 3a +2b +2c +d =0

Also 2i^ + 2j^ + 3k^ cross ai^ + bj^ + ck^ =0 (Line joining the two points)

And i^ -2j^ + 3k^ cross ai^ + bj^ + ck^ =0 (Direction ratios of the given line) <-- EDIT

Four equations , four unknowns , can be found?

2) Let the point be h,k,l

Square of the distance is

(h-k+l/root3)^2 + (h+2k +l/root6)^2 + (h-l/root2)^2 = 16

Simplify and replace h,k,l by x,y,z respectively.

ankitagg

but in this line [i^ -2j^ + 3k^ cross ai^ + bj^ + ck^ =0 ],acc. to u (1,-2,3) are the dc's but their sum i.e l^2+m^2+n^2 is not equal to one.

sunip_the_mini_mastermind

Q1.        the eqn. of a plane passing thru a pt is given by

              a(x-1)+b(y-0)+c(z+1) = 0.. .
              now since the plane passes thru (3,2,2)
              so,    2a + 2 b +3c.=0 -1)
              also the plane is parallel to the line having dir.ratios...(1,-2,3)
              so the normal to the plane will be perpendicular to the line...
              so.   a -2b +3c =0   -2)
solve the 2 eqns by cross multiplication and get the values of a,b,c


              rate if satisfied..............

Conjurer

They are direction ratios man.

@below: This comment not for you.

sunip_the_mini_mastermind

i cudnot get u....

Ask & Discuss Questions with Community & Experts