Exercise 1.1.5

Question

The linear combinations of $\boldsymbol{v}=(1,1,0)$ and $\boldsymbol{w}=(0,1,1)$ fill a plane in $\mathbf{R}^{3}$.
\begin{enumerate}[label=(\alph*)]
\item Find a vector $z$ that is perpendicular to $v$ and $w$. Then $z$ is perpendicular to every vector $c \boldsymbol{v}+d \boldsymbol{w}$ on the plane : $(c \boldsymbol{v}+d \boldsymbol{w})^{\mathrm{T}} z=c \boldsymbol{v}^{\mathrm{T}} \boldsymbol{z}+d \boldsymbol{w}^{\mathrm{T}} z=0+0$.
\item Find a vector $\boldsymbol{u}$ that is not on the plane. Check that $\boldsymbol{u}^{\mathrm{T}} \boldsymbol{z} 
eq 0$.
\end{enumerate}

Neil Z. · Accepted Answer

\begin{enumerate}[label=(\alph*)]
\item Let $z=\begin{bmatrix}z_1\z_2\z_3\end{bmatrix}$, and for every linear combination of $v$ and $w$, i.e. $cv+dw$, we let $(cv+dw)^Tz = 0$, so we have

\begin{align*}
(cv+dw)^Tz &= cv^Tz + dw^Tz\
&= c(z_1 + z_2) + d(z_2 + z_3)\
&= cz_1 + (c+d)z_2 + dz_3\
&= 0
\end{align*}

Solve this for any $c$ and $d$, we see that $z=\begin{bmatrix}-1\1\-1\end{bmatrix}$ solves the equation.

\item We see that $cv+dw = \begin{bmatrix}c\c+d\d\end{bmatrix}$, it's easy to see that $u=\begin{bmatrix}1\3\1\end{bmatrix}$ is not on theplane, and we have $u^Tz = \begin{bmatrix}1 &3&1\end{bmatrix}\begin{bmatrix}-1\1\-1\end{bmatrix} = 1 
e 0$.
\end{enumerate}

Exercise 1.1.5

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Exercise 1.1.5

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