Mathematics | VITEEE 2014 | Discussion Page

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If the points (1,2,3) and (2,-1,0) lie on the opposite sides of the plane 2x+3y-2z=k. then

A) k < 1

B) k > 2

D) k < 1 or k > 2

E) 1 < k < 2

Option D

The points (1,2,3) and (2,-1,0) lie on the opposite sides of the plane 2x+3y-2z-k=0

so, (2+6-6-k)(4-3-k)<0

$\Rightarrow$
$(k-1)(k-2)<0$

$\therefore$ $1<k<2$

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