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Q.

If the point x1+tx2−x1,y1+ty2−y1,z1+z2−z1 devides the line segment joining x1,y1,z1) and x2,y2,z2 internally then

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a

t<0

b

0

c

t>1

d

t=1

answer is B.

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Detailed Solution

x1+tx2−x1,y1+ty2−y1,z1+z2−z1=tx2+1-tx1,ty2+1-ty,tz2+1-tz1                                                                =tx2+1-tx1t+1-t,ty2+1-ty1t+1-t,tz2+1-tz1t+1-t ⇒the ratio of division is t:1-t Since the division is internal, t1-t>0 ⇒tt-1<0 ⇒0
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