Question

Question: Let \({{\bf{p}}_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{2}}\\{ - {\bf{3}}}\\{\bf{1}}\\{\bf{2}}\end{array}} \right)\), \({{\bf{p}}_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{\bf{2}}\\{ - {\bf{1}}}\\{\bf{3}}\end{array}} \right)\), \({{\bf{n}}_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{\bf{2}}\\{\bf{4}}\\{\bf{2}}\end{array}} \right)\), and \({{\bf{n}}_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{\bf{2}}\\{\bf{3}}\\{\bf{1}}\\{\bf{5}}\end{array}} \right)\), let \({H_{\bf{1}}}\) be the hyperplane in \({\mathbb{R}^{\bf{4}}}\) through \({{\bf{p}}_{\bf{1}}}\) with normal \({{\bf{n}}_{\bf{1}}}\), and let \({H_{\bf{2}}}\) be the hyperplane through \({{\bf{p}}_{\bf{2}}}\) with normal \({{\bf{n}}_{\bf{2}}}\). Give an explicit description of \({H_{\bf{1}}} \cap {H_{\bf{2}}}\). (Hint: Find a point p in \({H_{\bf{1}}} \cap {H_{\bf{2}}}\) and two linearly independent vectors \({{\bf{v}}_{\bf{1}}}\) and \({{\bf{v}}_{\bf{2}}}\) that span a subspace parallel to the 2-dimensional flat \({H_{\bf{1}}} \cap {H_{\bf{2}}}\).)

Short answer

\({H_1} \cap {H_2} = \left\{ {{\bf{x}}:{\bf{x}} = {\bf{p}} + {x_3}{{\bf{v}}_1} + {x_4}{{\bf{v}}_2}} \right\}\)

Step-by-step solution

Find the values of \({d_{\bf{1}}}\) and \({d_{\bf{2}}}\)

Find \({d_1}\) for \({H_1}\).

\(\begin{array}{c}{d_1} = {{\bf{n}}_1} \cdot {{\bf{p}}_1}\\ = 1\left( 2 \right) + 2\left( { - 3} \right) + 4\left( 1 \right) + 2\left( 2 \right)\\ = 4\end{array}\)

Find \({d_2}\) for \({H_2}\).

\(\begin{array}{c}{d_2} = {{\bf{n}}_2} \cdot {{\bf{p}}_2}\\ = 2\left( 2 \right) + 3\left( 2 \right) + 1\left( { - 1} \right) + 5\left( 3 \right)\\ = 22\end{array}\)

Write the augmented matrix

The system of equations is:

\(\left( {\begin{array}{*{20}{c}}1&2&4&2\end{array}} \right){\bf{x}} = 4\)

\(\left( {\begin{array}{*{20}{c}}2&3&1&5\end{array}} \right){\bf{x}} = 22\)

The augmented matrix can be written as shown below:

\(\left( {\begin{array}{*{20}{c}}1&2&4&2&4\\2&3&1&5&{22}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&0&{ - 10}&4&{32}\\0&1&7&{ - 1}&{ - 14}\end{array}} \right)\)

Write the general solution

The general solution is:

\(\begin{array}{c}x = \left( {\begin{array}{*{20}{c}}{32}\\{ - 14}\\0\\0\end{array}} \right) + {x_3}\left( {\begin{array}{*{20}{c}}{10}\\{ - 7}\\1\\0\end{array}} \right) + {x_4}\left( {\begin{array}{*{20}{c}}{ - 4}\\1\\0\\1\end{array}} \right)\\ = p + {x_3}{{\bf{v}}_1} + {x_4}{{\bf{v}}_2}\end{array}\)

So,

\(p = \left( {\begin{array}{*{20}{c}}{32}\\{ - 14}\\0\\0\end{array}} \right)\), \({{\bf{v}}_1} = \left( {\begin{array}{*{20}{c}}{10}\\{ - 7}\\1\\0\end{array}} \right)\) and \({{\bf{v}}_2} = \left( {\begin{array}{*{20}{c}}{ - 4}\\1\\0\\1\end{array}} \right)\)

Therefore, \({H_1} \cap {H_2} = \left\{ {{\bf{x}}:{\bf{x}} = {\bf{p}} + {x_3}{{\bf{v}}_1} + {x_4}{{\bf{v}}_2}} \right\}\).