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 Entrance Exam
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Answer : Given : line r = i  2j  3k + L(2i 3j +2k) and plane r.(i +3j  2k) = 3.
To find : the distance of (1,2,0) from the point of intersection of line r = i  2j  3k + L(2i 3j +2k) and plane r.(i +3j  2k) = 3.
Let us assume the point A be the intersection of the line and the plane given and B be (1,2,0)
we can represent the line equation in the quadric form as
x= 12t , y= 2 3t , z=3 +2t (where t is a parameter corresponding to direction of the line )
and the plane as
x+3y  2z =3
As we know the intersected point would lie on the plane as well the line , therefore it should satisfy both there equation as follows:
x_{A} = 12t................(1)
y_{A}= 2 3t...............(2)
z_{A}=3 +2t...............(3)
and
x_{A} + 3y_{A}  2z_{A} =3 ...................(4)
Now substituting the eq 1 , 2 , 3 in eq 4 we get
=> 12t + 3(23t) 2(3+2t) = 3
=>12t69t+64t = 3
=> t = 2 /15 ..............(5)
now putting the value of t in eq 1,2,3 we get
the point A is (19/15,24/15,49/15)
the distance between point A and B is
= sqrt ((x_{2}x_{1})^{2} + ((y_{2}y_{1})^{2} + (z_{2}z_{1})^{2} )
= sqrt ( (19/15) 1) ^{2} + ((24/15) +2)^{2} + ((49/15) 0)^{2} )
= 49.53 /15
=3.301 units is required distance Answer
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