Skip to main content
Mathematics LibreTexts

4.14.E: Exercise for Section 4.10

  1. Last updated
  2. Save as PDF
  • Page ID
    197444

    • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \( \newcommand{\dsum}{\displaystyle\sum\limits} \)

    \( \newcommand{\dint}{\displaystyle\int\limits} \)

    \( \newcommand{\dlim}{\displaystyle\lim\limits} \)

    \( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

    ( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\id}{\mathrm{id}}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\kernel}{\mathrm{null}\,}\)

    \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\)

    \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\)

    \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    \( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

    \( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

    \( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vectorC}[1]{\textbf{#1}} \)

    \( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

    \( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

    \( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \(\newcommand{\longvect}{\overrightarrow}\)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)
    Exercise \(\PageIndex{1}\)

    Let the subspace \( W \) of \( \mathbb{R}^3 \) be spanned by the vectors: \[ \vec{w}_1 = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}, \quad \vec{w}_2 = \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} \nonumber \] Find the orthogonal projection of the vector: \[ \vec{v} = \begin{bmatrix} 2 \\ 3 \\ 4 \end{bmatrix} \nonumber \] onto the subspace \( W \).

    Answer

    \[ \operatorname{proj}_W \vec{v} = \begin{bmatrix} \frac{3}{2} \\ \frac{5}{2} \\ 1 \end{bmatrix} \nonumber \]

    Exercise \(\PageIndex{2}\)

    Let \( W \) be the subspace of \( \mathbb{R}^3 \) given by the solution space of the equation: \[ x_1 + 2x_2 + 3x_3 = 0 \nonumber \] Find a basis for \( W^\perp \), the orthogonal complement of \( W \).

    Answer

    \[ W^\perp = \text{span} \left\{ \begin{bmatrix} 1 \\ -2 \\ -3 \end{bmatrix} \right\} \nonumber \]

    Exercise \(\PageIndex{3}\)

    Let \( W \) be the subspace of \( \mathbb{R}^4 \) spanned by: \[ \vec{w}_1 = \begin{bmatrix} 1 \\ 1 \\ 0 \\ 0 \end{bmatrix}, \quad \vec{w}_2 = \begin{bmatrix} 0 \\ 1 \\ 1 \\ 1 \end{bmatrix} \nonumber \] Find a basis for \(W^\perp\), the orthogonal complement.

    Exercise \(\PageIndex{4}\)

    Let \( W \) be the subspace of \( \mathbb{R}^{4} \) spanned by the vectors: \[ \vec{w} = \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \end{bmatrix} \nonumber \] Find a basis for the orthogonal complement \( W^\perp \).

    Answer

    \[ W^\perp = \text{span} \left\{ \begin{bmatrix} -2 \\ 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} -3 \\ 0 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -4 \\ 0 \\ 0 \\ 1 \end{bmatrix} \right\} \nonumber \]

    Exercise \(\PageIndex{5}\)

    Consider the following scalar equation of a plane. \[2x-3y+z=0\nonumber\] Find the point on the plane which is closest to \(Y=(3,4,1)\) using the steps outlined in Example 4.10.5.

    1. Find a basis \(X\) of the subspace \(W\) of \(\mathbb{R}^3\) defined by the equation \(2x-3y+z=0\).
    2. Orthogonalize the basis \(X\) to get an orthogonal basis \(B\) of \(W\).
    3. Find the projection on \(W\) of the position vector of the point \(Y\).
    Exercise \(\PageIndex{6}\)

    Consider the following scalar equation of a plane. \[x+3y+z=0\nonumber\] Find the point on the plane which is closest to \(Y=(1,2,1)\).

    Exercise \(\PageIndex{7}\)

    Let \(\vec{v}\) be a vector and let \(\vec{n}\) be a normal vector for a plane through the origin. Find the equation of the line through the point determined by \(\vec{v}\) which has direction vector \(\vec{n}\). Show that it intersects the plane at the point determined by \(\vec{v}−proj_{\vec{n}}\vec{v}\).

    Hint

    \[The line:\(\vec{v}+t\vec{n}\). It is in the plane if \(\vec{n}•(\vec{v}+t\vec{n}) = 0\). Determine \(t\). Then substitute in to the equation of the line.

    Exercise \(\PageIndex{8}\)

    As shown in the above problem, one can find the closest point to \(\vec{v}\) in a plane through the origin by finding the intersection of the line through \(\vec{v}\) having direction vector equal to the normal vector to the plane with the plane. If the plane does not pass through the origin, this will still work to find the point on the plane closest to the point determined by \(\vec{v}\). Here is a relation which defines a plane \[2x+y+z=11\nonumber\] and here is a point: \((1, 1, 2)\). Find the point on the plane which is closest to this point. Then determine the distance from the point to the plane by taking the distance between these two points.

    Hint:

    Line: \((x, y,z) = (1, 1, 2) +t(2, 1, 1)\). Now require that it intersect the plane.

    Exercise \(\PageIndex{9}\)

    As shown in the above problem, one can find the closest point to~v in a plane through the origin by finding the intersection of the line through \(\vec{v}\) having direction vector equal to the normal vector to the plane with the plane. If the plane does not pass through the origin, this will still work to find the point on the plane closest to the point determined by \(\vec{v}\). Here is a relation which defines a plane \[2x+y+z=11\nonumber\] and here is a point: \((1, 1, 2)\). Find the point on the plane which is closest to this point. Then determine the distance from the point to the plane by taking the distance between these two points.

    Hint:

    Line: \((x, y,z) = (1, 1, 2) +t(2, 1, 1)\). Now require that it intersect the plane.

    Exercise \(\PageIndex{10}\)

    Consider the plane \[3x+4y-5z=7\nonumber\]

    a. Find the point on this plane that is closest to the point \((2,-1,3)\).
    b. Compute the distance between \((2,-1,3)\) and the plane.

    Exercise \(\PageIndex{11}\)

    Provide a proof for proposition 4.10.2: Let \(W\) be a subspace of \(\mathbb{R}^n\). Then the orthogonal complement \(W^{\perp}\) is also a subspace of \(\mathbb{R}^n\).

    This page titled 4.14.E: Exercise for Section 4.10 is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Doli Bambhania, Fatemeh Yarahmadi, and Bill Wilson.