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Question

(i) A force of magnitude 5 units acting parallel to 2i2j+k displaces the point of application from (1,2,3) to (5,3,7). Find the work done by the force.
(ii) If A(1,4,3) is one end of a diameter AB of the sphere x2+y2+z23x2y+2z15=0, then find the coordinates of B

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Solution

(i) Let r be the unit vector acting along the force F
Given, r=2i+2j+k
|r|=22+(2)2+12=4+4+1=9=3
^r=r|r|2i2j+k3
Force F=5^r=53(2i2j+3k)
Also, Let OA=i+2j+3k and OB=5i+3j+7k
Let AB=OBOA=(5i+3j+7k)(i+2j+3k)
d=4i+j+4k
work done =Fd=53(2i+2j+k)
(4i+j+4k)
=53[2(4)+2+1(4)]
=53[82+4]=503 units.

(ii) The equation of the given sphere is
x2+y2+z23x2y+2z15=0
2u= Co-efficient of x=3u=32
2v= Co-efficient of y=2v=22=1
2w= Co-efficient of z=2w=22=1
d=15
Centre is (u,v,w)=(32,1,1)
Given one end of the diameter A is (1,4,3) and the centre (mid point of the diameter AB) is (32,1,1)
Let the other end be B(x,y,z)
(x12,y42,32)=(32,1,1) (using mid point formula)
Equating the co-ordinates of x,y and z, we get
x12=32x=3+1x=4
y+42=1y+4=2y=24=2
32=1z3=2z=2+3=5
Other end of the diameter AB is (4,2,1)

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