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  • Please help! Equation of plane 2x-2y+z+6=0 in normal form becomes

Equation of plane 2x-2y+z+6=0 in normal form becomes

 

 

  • Option 1)

    \frac{-2}{3}x+\frac{2}{3}y+\frac{z}{3}=2

  • Option 2)

    \frac{-2}{3}x+\frac{2}{3}y-\frac{z}{3}=2

  • Option 3)

    \frac{2}{3}x-\frac{2}{3}y+\frac{z}{3}=2

  • Option 4)

    none of these

 

Answers (1)

best_answer

As we have learned

Conversion of equation in normal form (vector form ) -

The equation ax+by+cz+D= 0

is converted in normal by

(i)    ax+by+cz= -D

(ii)    Making RHS position

    -ax-by-cz= D

(iii)    Dividing by \sqrt{a^{2}+b^{2}+c^{2}}

(iv)    \frac{-ax-by-cz}{\sqrt{a^{2}+b^{2}+c^{2}}}= \frac{D}{\sqrt{a^{2}+b^{2}+c^{2}}}

We get lx+my+nz= d

-

 

 2x-2y+z+6=0\Rightarrow-2x+2y-z=6

dividing throughout by  \sqrt{(-2)^{2}+(2)^{2}+(-1)^{2}}=3

we get  \frac{-2}{3}x+\frac{2}{3}y-\frac{z}{3}=2


Option 1)

\frac{-2}{3}x+\frac{2}{3}y+\frac{z}{3}=2

Option 2)

\frac{-2}{3}x+\frac{2}{3}y-\frac{z}{3}=2

Option 3)

\frac{2}{3}x-\frac{2}{3}y+\frac{z}{3}=2

Option 4)

none of these

Posted by

Himanshu

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