Value of x for which (x,x+1,x+2) and (1-x,x+2,x) lies on opposite side of plane  x+y+z-2=0 is

  • Option 1)

    x\in (\frac{1}{3},1)

  • Option 2)

    x\in (\frac{-1}{3},1)

  • Option 3)

    x\in (-1,\frac{-1}{3})

  • Option 4)

    x\in (-1,\frac{1}{3})

 

Answers (1)
H Himanshu

As we have learned

Two sides of plane -

Let the plane be   ax+by+cz+d= 0

and two points be A(x_{1},y_{1},z_{1}),B(x_{2},y_{2},z_{2}) then

If they are on same side:

\left ( ax_{1}+by_{1}+cz_{1}+d \right )\left ( ax_{2}+by_{2}+cz_{2}+d \right )> 0

If they are on opposite side:

\left ( ax_{1}+by_{1}+cz_{1}+d \right )\left ( ax_{2}+by_{2}+cz_{2}+d \right )< 0

 

 

-

 

  On putting points in L.H.S of equation of plane one by one, we get expression 

(x)+(x+1)+(x+2)-2=3x+1

and (1-x)+(x+2)+(x-2)=x+1

For points in opposite side of plane,

(3x+1)(x+1)< 0

\Rightarrow x\in (-1,\frac{-1}{3})


Option 1)

x\in (\frac{1}{3},1)

Option 2)

x\in (\frac{-1}{3},1)

Option 3)

x\in (-1,\frac{-1}{3})

Option 4)

x\in (-1,\frac{1}{3})

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