 # A line L through the origin in RR^3 can be represented by parametric equations of the form x = at, y = bt, and z = ct. sjeikdom0 2021-01-30 Answered

A line L through the origin in ${\mathbb{R}}^{3}$ can be represented by parametric equations of the form x = at, y = bt, and z = ct. Use these equations to show that L is a subspase of $R{R}^{3}$  by showing that if   are points on L and k is any real number, then   are also points on L.

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Let ${v}_{1}=\left({x}_{1},{y}_{1},{z}_{1}\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{v}_{2}=\left({x}_{2},{y}_{2},{z}_{2}\right)$ be elements of L.Then
$\left\{\begin{array}{c}{x}_{1}=a{t}_{1}\\ {y}_{1}=b{t}_{1}\\ {z}_{1}=c{t}_{1}\end{array}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\left\{\begin{array}{c}{x}_{2}=a{t}_{2}\\ {y}_{2}=b{t}_{2}\\ {z}_{2}=c{t}_{2}\end{array}$
for some real numbers ${t}_{1}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{t}_{2}$.
let $w={v}_{1}+{v}_{2}=\left({x}_{1}+{x}_{2},{y}_{1}+{y}_{2},{z}_{1}+{z}_{2}\right)$.Then
$\left\{\begin{array}{c}{x}_{1}+{x}_{2}=a{t}_{1}+a{t}_{2}=a\left({t}_{1}+{t}_{2}\right)\\ {y}_{1}+{y}_{2}=b{t}_{1}+b{t}_{2}=b\left({t}_{1}+{t}_{2}\right)\\ {z}_{1}+{z}_{2}=c{t}_{1}+c{t}_{2}=c\left({t}_{1}+{t}_{2}\right)\end{array}$
Hence the components of w are of the form $x=at,y=bt,z=ctf\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}t={t}_{1}+{t}_{2}$.
Now let k be a real number. And consider $w=k{v}_{1}=\left(k{x}_{1},k{y}_{1},k{z}_{1}\right)$.Then
$\left\{\begin{array}{c}k{x}_{1}=ka{t}_{1}=a\left(k{t}_{1}\right)\\ k{y}_{1}=kb{t}_{1}=b\left(k{t}_{1}\right)\\ k{z}_{1}=kc{t}_{1}=c\left(k{t}_{1}\right)\end{array}$
Hence the components of w are of the form $x=at,y=bt,z=ctf\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}t=k{t}_{1}$.
Then, by the subspace theorem, L is a subspace pf ${\mathbb{R}}^{3}$.