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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
Verified

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.

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 \[2x - 2y + z = 5\]

\[ax + by + cz = d\]

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, \[\overrightarrow n = \left\langle a \right.\left. {,b,c} \right\rangle \]