Suggested languages for you:

Americas

Europe

Q13E

Expert-verified
Found in: Page 437

### Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

# 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 $${H_{\bf{1}}} \cap {H_{\bf{2}}}$$.)

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

See the step by step solution

## Step 1: 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 $${H_2}$$.

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

## Step 2: 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:

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

## Step 3: 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, $${H_1} \cap {H_2} = \left\{ {{\bf{x}}:{\bf{x}} = {\bf{p}} + {x_3}{{\bf{v}}_1} + {x_4}{{\bf{v}}_2}} \right\}$$.

## Recommended explanations on Math Textbooks

94% of StudySmarter users get better grades.