Question-56

Find the smallest subspace of \(\displaystyle \mathbb{R}^{3}\) containing the vectors \(\displaystyle ( 2,-3,-3)\) and \(\displaystyle ( 0,3,2)\).

• \(\displaystyle S=\{c( 2,-3,-3) ;c\in \mathbb{R}\}\)
• \(\displaystyle S=\{( x,y,z) :3x-4y+6z=0;x,y,z\in \mathbb{R}\}\)
• \(\displaystyle S=\{( x,y,z) :x-4y+z=0;x,y,z\in \mathbb{R}\}\)
• \(\displaystyle S=\{c( 0,3,2) ;c\in \mathbb{R}\}\)

Answer

• \(\displaystyle S=\{c( 2,-3,-3) ;c\in \mathbb{R}\}\)
• \(\displaystyle S=\{( x,y,z) :3x-4y+6z=0;x,y,z\in \mathbb{R}\}\)
• \(\displaystyle S=\{( x,y,z) :x-4y+z=0;x,y,z\in \mathbb{R}\}\)
• \(\displaystyle S=\{c( 0,3,2) ;c\in \mathbb{R}\}\)

Solution

The smallest subspace that contains these two vectors is their span. Since the two vectors are linearly independent, their span will be a two-dimensional subspace of \(\displaystyle \mathbb{R}^{3}\), which is a plane passing through the origin:

\[ \begin{equation*} S=\{( x,y,z) :ax+by+cz=0\} \end{equation*} \]

We have:

\[ \begin{equation*} \begin{aligned} 2a-3b-3c & =0\\ 3b+2c & =0 \end{aligned} \end{equation*} \]

From the options, we see that \(\displaystyle c=6\) is one possibility. This gives us:

\[ \begin{equation*} \begin{aligned} 2a-3b-3c & =0\\ 3b+2c & =0 \end{aligned} \Longrightarrow a=3,b=-4 \end{equation*} \]

Therefore, the required subspace is:

\[ \begin{equation*} S=\{( x,y,z) :3x-4y+6z=0\} \end{equation*} \]