Question
Thu August 23, 2012 By: Nevin

# sir/madam

Expert Reply
Wed August 29, 2012
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= 1-2t , 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:
xA = 1-2t................(1)
yA= -2 -3t...............(2)
zA=-3 +2t...............(3)

and

xA + 3yA - 2zA =3  ...................(4)

Now substituting the eq 1 , 2 , 3 in eq 4 we get
=> 1-2t + 3(-2-3t) -2(-3+2t) = 3
=>1-2t-6-9t+6-4t = 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 ((x2-x1)2 + ((y2-y1)2 + (z2-z1)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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