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  • Need explanation for: Equation of plane 3x-6y+2z-14=0 in normal form is

Equation of plane 3x-6y+2z-14=0 in normal form is

  • Option 1)

    \frac{3}{7}x-\frac{6}{7}y+\frac{2}{7}z=2

  • Option 2)

    3x-6y+2z=14

  • Option 3)

    3x+6y-2z=14

  • Option 4)

    3x-6y-2z=14

 

Answers (1)

best_answer

As we have learned

Normal form (cartesian form ) -

lx+my+nz=d

where d is the distance from origin.  

- wherein

\vec{r}= x\hat{i}+y\hat{j}+z\hat{k}

\hat{n}= l\hat{i}+m\hat{j}+n\hat{k}

putting in \vec{r}\cdot \hat{n}= d

We get -  lx+my+nz= d

 

 3x-6y+2z=14

it will be comparable with \vec{r}\cdot\hat{n}=d after dividing by |3\hat{i}-6\hat{j}+2\hat{k}| i.e 7

\frac{3}{7}x-\frac{6}{7}y+\frac{2}{7}z=2


Option 1)

\frac{3}{7}x-\frac{6}{7}y+\frac{2}{7}z=2

Option 2)

3x-6y+2z=14

Option 3)

3x+6y-2z=14

Option 4)

3x-6y-2z=14

Posted by

Himanshu

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