Question

The normal form of \[2x - 2y + z = 5\] is
A) \[12x - 4y + 3z = 39\]
B) $\dfrac{{ - 6}}{7}x + \dfrac{2}{7}y + \dfrac{3}{7}z = 1$
C) $\dfrac{{12}}{{13}}x - \dfrac{{ - 4}}{{13}}y + \dfrac{3}{{13}}z = 3$
D) $\dfrac{2}{3}x - \dfrac{2}{3}y + \dfrac{1}{3}z = \dfrac{5}{3}$

Answer 1

Hint: A plane in space is defined by three points which don’t all lie on the same line) or by a point and normal vector to the plane.
The general equation or standard equation of a straight line is:
ax+by+cz-d=0
Where a and b are constants and either a≠0 or b≠0 or c≠0.
Thus to convert from general form to the normal form, divide the general form to equation by $ \pm \sqrt {{A^2} + {B^2} + {C^2}} $ taking the sign of the square root opposite to the sign of D, where D is not 0.
Algorithm to Transform the General Equation to Normal Form
Step I: Transfer the constant term to the right hand side and make it positive.
Step II: Divide both sides by \[\sqrt {(Coefficient{\text{ }}of{\text{ }}x{)^2} + {{\left( {Coefficient{\text{ }}of{\text{ }}y} \right)}^2} + {{\left( {Coefficient{\text{ }}of{\text{ z}}} \right)}^2}} \]
The obtained equation will be in the normal form.

Complete step-by-step answer:
The given equation is \[2x - 2y + z = 5\]
So A, coefficient of x = 2
B, coefficient of y = -2
C, coefficient of z = 1
So, Now determine
$
   \pm \sqrt {{A^2} + {B^2} + {C^2}} \\
   = - \sqrt {{2^2} + {{( - 2)}^2} + {1^2}} \\
   = = \sqrt 9 = 3 \\
 $
Hence dividing the complete equation by 3 we get:-
\[\dfrac{2}{3}x - \dfrac{2}{3}y + \dfrac{1}{3}z = \dfrac{5}{3}\]
Which is the normal form of the given equation \[2x - 2y + z = 5\]
So, option (D) is the correct answer.

Note: This is called the scalar equation of plane. Often this will be written as,
\[ax + by + cz = d\]
This form is often how we are given equations of planes. Notice that if we are given the equation of a plane in this form we can quickly get a normal vector for the plane.
A normal vector is, \[\overrightarrow n = \left\langle a \right.\left. {,b,c} \right\rangle \]