1[11] didnt get u at all
A line passing through A(2,2,4) and parallel to line
L ≡(x-2)=(y-1)=(z-2)/3
is incident of the
mirror plane x+2y+2z-5=0.
The equation of the reflected ray is
a) (x-1)=(y-1)/3=(1-z) b) (x-1)=(y-1)/-3=(z-1)
c) x=(2-y)=z d) x=(y+2)/3=z
I got the point of incidence and image of the point and found the equation....(it was d i think)
but the point is regardless of whether the answer was d or not....
the "i" component of the direction vectors of the incident and reflected rays remains same.
how?????
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12 Answers
1i meant that direction of normal has to be direction of change of direction (i dont think its a typing mistake)
so please tell me where i am doing wrong .......
1dcs of line are(1,1,3) and for the plane the normal,s dcs are(1,2,2)...
did u notice that "i" component is same therefore there wont be any direction change in
that dirction.(as in case of normal incidence all components are same so there is no chane of d.cs)
1first maTCH THE DIREC. COSINES OF 1.THE LINE AND 2.THE PLANES NORMAL.
OK.
THEN THE MATCHING PART WONT CHANGE THE DIRECTION.
1aisa kyon bhai plane agar x axis ke || hoga toh normal toh x axis ke _|_ hoga
1haan bhai isiliye toh har post mein normal ke baaare mein hi likhe hain....
62when a light ray falls on a mirror..
does it component in the horizontal direction change?
only the vertical component becomes negative..
doesnt it?
1yes.....
but i was trying to point that the direction of normal to the mirror...
will be the change in the two vectors.....
i know i messed up some small part of it..... that is why i am asking.....
[2][17]
wont it be like a collision b/w wall and ball..
11YES PHILIP UR THINKING IS PERFECT
U CAN ALSO FIND THESE TYPE OF QUESTION UN OPTICS (LIKE LIGHT RAYS)
1aree bhai mani tum toh meri samasya hal karne ke bajai or confuse kar rahe ho......
baat ye hai ki answer usse kaise laye ??