Question: Let \({\bf{p}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{3}}}\\{\bf{1}}\\{\bf{2}}\end{array}} \right)\), \({\bf{n}} = \left( {\begin{array}{*{20}{c}}{\bf{2}}\\{\bf{1}}\\{\bf{5}}\\{ - {\bf{1}}}\end{array}} \right)\), \({{\bf{v}}_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{1}}\\{\bf{1}}\\{\bf{1}}\end{array}} \right)\), \({{\bf{v}}_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{ - {\bf{2}}}\\{\bf{0}}\\{\bf{1}}\\{\bf{3}}\end{array}} \right)\), and \({{\bf{v}}_{\bf{3}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{\bf{4}}\\{\bf{0}}\\{\bf{4}}\end{array}} \right)\), and let H be the hyperplane in\({\mathbb{R}^{\bf{4}}}\) with normal n and passing through p. Which of the points \({{\bf{v}}_{\bf{1}}}\), \({{\bf{v}}_{\bf{2}}}\), and \({{\bf{v}}_{\bf{3}}}\) are on the same side of H as the origin, and which are not?

Consider \({\bf{v}} = \left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right)\).

Let \({H_0}\) be the origin that contains the normal vector, and then it can be written as shown below:

\(\begin{array}{c}\left( {\begin{array}{*{20}{c}}2&1&5&{ - 1}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right) = 0\\2{x_1} + {x_2} + 5{x_3} - {x_4} = 0\end{array}\)

So, it can be written as \(\left( {{H_0}:f} \right) = 2{x_1} + {x_2} + 5{x_3} - {x_4}\).

As hyperplaneis passing through the point p as shown below:

\(\begin{array}{c}d = f\left( p \right)\\ = 2\left( 1 \right) + \left( { - 3} \right) + 5\left( 1 \right) - 2\\ = 2\end{array}\)

So, \(d = 2\).

The dot product of \(n\) and \({v_1}\) is:

\(\begin{array}{c}f\left( {{{\bf{v}}_1}} \right) = n \cdot {{\bf{v}}_1}\\ = 2\left( 0 \right) + \left( 1 \right) + 5\left( 1 \right) - 1\\ = 5\end{array}\)

As the dot product is more significant than d, so \({{\bf{v}}_1}\) is outside the origin.

The dot product of \(n\) and \({{\bf{v}}_2}\) is:

\(\begin{array}{c}f\left( {{{\bf{v}}_2}} \right) = n \cdot {{\bf{v}}_2}\\ = 2\left( { - 2} \right) + \left( 0 \right) + 5\left( 1 \right) - 3\\ = - 2\end{array}\)

The value is lesser than d, so \({{\bf{v}}_2}\) lies the same side of the hyperplane.

The dot product of \(n\) and \({{\bf{v}}_3}\) is:

\(\begin{array}{c}f\left( {{{\bf{v}}_3}} \right) = n \cdot {{\bf{v}}_3}\\ = 2\left( 1 \right) + \left( 4 \right) + 5\left( 0 \right) - 4\\ = 2\end{array}\)

The value is equal to d, so \({{\bf{v}}_3}\) lies on the hyperplane.