Q.Find the vector equation of a plane which is at a distance of $7$ units from the origin having $3,-4,5$ as direction ratios of a normal to it.

Equation of a Plane

A plane is fixed by a point on it and a direction perpendicular to it (its normal $\vec n$). Every standard form below is really that one idea written differently.

Point + normal form

If the plane passes through $\vec a$ with normal $\vec n$, then for any point $\vec r$ on it, $\vec r-\vec a$ lies in the plane, so it is perpendicular to $\vec n$:

In Cartesian form with $\vec n=(A,B,C)$: $A(x-x_1)+B(y-y_1)+C(z-z_1)=0$, i.e. $Ax+By+Cz=d$. The coefficients of $x,y,z$ are the normal's direction ratios.

Normal (perpendicular) form

If $\hat n$ is the unit normal and the plane is at distance $p$ from the origin: $\vec r\cdot\hat n=p$, i.e. $lx+my+nz=p$ with $l^2+m^2+n^2=1$.

Intercept form

A plane cutting the axes at $a,b,c$: $\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1$.

Through three points / a line of intersection

• Three points $A,B,C$: take $\vec n=\vec{AB}\times\vec{AC}$, then use point+normal.
• Family through the line of intersection of $P_1=0$ and $P_2=0$: every such plane is $P_1+\lambda P_2=0$; fix $\lambda$ from the extra condition (a point, a distance, or a perpendicularity).