IIT JEE , AIEEE Forum - Mathematics , Vectors - NCERT 3-D Miscellaneous Question

anpm_dev

Q-18. Find the distance of the point (-1,-5,-10) from the point of intersection of the line r= 2i-j+2k +

(3i+4j+2k) and the plane r.(i-j+k)=5

Q-19. find the vector eqn of the line passing thu (1,2,3) and parallel to planes r.(i-j+2k)=5 and r.(3i+j+k)=6

Q-20 find the vector eqn of the line passing thru pt. (1,2,-4) and perpendicular to the lines: (x-8)/3 = (y+19)/-16 = (z-10)/7 and (x-15)/3 = (y-29)/8 = (z-5)/-5

Pls give the soln. i'm in urgent need---> Definite Rating

layman

18.Let the Intersection point be P(2+3y,-1+4y,2+2y)

It lies on plane so,

2+3y+1-4y+2+2y=5

so y=0

so P(2,-1,2)

so distance=root(169)=13

layman

19. let the drs be (a,b,c)

since it is parallel to the plane x-y+2z=5 it is perpendicular to its normal.

so applying the condition a.a1+b.b1+c.c1=0,

a-b+2c=0

similarly for the next plane,

3a+b+c=0

since drs are just proportional quantities elimination them,

a/-3 = b/5 = c/4

a=-3 and b=5 and c=4

so the equation of line is,

x-1/-3 = y-2/5 = z-3/4

layman

Similar to the 19th is the 20th find the dr's and then find the equation of line.

Ans:

x-1/2 = y-2/3 = z+4/6

avi_1214545

avi_1214545

avi_1214545

avi_1214545

rate if useful

layman

I don't bother rates. I just solve for my interests.

paddy.dude

good work guys ............

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