Two Sides of a Plane - Practice Questions & MCQ

Let ax + by + cz + d = 0 be the plane, then the points (x₁, y₁, z₁) and (x₂, y₂, z₂)  lie on the same side or opposite side according as

$\frac{a x_1+b y_1+c z_1+d}{a x_2+b y_2+c z_2+d}>0$ or $<0$

Let the line segment joining the point P (x₁, y₁, z₁) and Q (x₂, y₂, z₂) divided by a point R (which lies in the plane) internally in the ratio m:n.

Point R lies in the plane ax+by+cz+d=0.

From the section formula

$
R \equiv\left(\frac{m x_2+n x_1}{m+n}, \frac{m y_2+n y_1}{m+n}, \frac{m z_2+n z_1}{m+n}\right)
$

Since, point $R$ lies in the plane
Therefore,
$
\begin{array}{cc} 
& a\left(\frac{m x_2+n x_1}{m+n}\right)+b\left(\frac{m y_2+n y_1}{m+n}\right)+c\left(\frac{m z_2+n z_1}{m+n}\right)+d=0 \\
\Rightarrow & a\left(m x_2+n x_1\right)+b\left(m y_2+n y_1\right)+c\left(m z_2+n z_1\right)+d(m+n)=0 \\
\Rightarrow & m\left(a x_2+b y_2+c z_2+d\right)+n\left(a x_1+b y_1+c z_1+d\right)=0 \\
\Rightarrow & \frac{m}{n}=-\frac{\left(a x_1+b y_1+c z_1+d\right)}{\left(a x_2+b y_2+c z_2+d\right)}
\end{array}
$

Now, if $\mathrm{ax}_1+\mathrm{by}_1+\mathrm{cz}_1+\mathrm{d}$ and $\mathrm{ax}_2+\mathrm{by}_2+\mathrm{cz}_2+\mathrm{d}$ are of same sign then $\mathrm{m} / \mathrm{n}<0$ (external division) and if opposite sign then $\mathrm{m} / \mathrm{n}>0$ (internal division).
Therefore, if
$
\begin{array}{ll}
\frac{a x_1+b y_1+c z_1+d}{a x_2+b y_2+c z_2+d}>0 & \text { (same side) } \\
\frac{a x_1+b y_1+c z_1+d}{a x_2+b y_2+c z_2+d}<0 & \text { (opposite side) }
\end{array}
$