Given vectors $(1, 3, 5), (-2 , -6, -10)$ and $(2, 6 , 10)$ determine whether the linear span of the above is a plane in $\mathbb R^3$.

The vectors are linearly dependent nd hence do not form a basis and it is known that the set of linearly dependent vectors in $R^2$ are collinear.

So based on the above can it be said that linearly dependent vectors in $R^3$ will form a plane.

  • $\begingroup$ You are right. The span a plane. $\endgroup$ – Kavi Rama Murthy Apr 18 at 7:42
  • $\begingroup$ Thanks for the confirmation $\endgroup$ – tendo Apr 18 at 7:49
  • $\begingroup$ Your final sentence is incorrect. For example, the vectors $(1,0,0)$, $(2,0,0)$ and $(3,0,0)$ are linearly dependent, but their span is a line, not a plane. You need another condition to ensure that the span is a plane. $\endgroup$ – amd Apr 18 at 18:18
  • $\begingroup$ @amd But these vectors would only cover one dimension i.e. x axis? $\endgroup$ – tendo Apr 18 at 18:40
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    $\begingroup$ Exactly. It can’t be said that linearly-dependent vectors in $\mathbb R^3$ will form a plane, contrary to your last sentence. They might, but you need other conditions besides linear dependence. $\endgroup$ – amd Apr 18 at 18:47

Your intuition is correct. It can be more rigorously stated, considering that the equation of a plane in $\mathbb R^3$ is :

$$Ax + By + Cz = 0$$

But since your given vectors are linearly dependent, you will arrive at such an equation, thus satisfying the known algebraic expression (form) of a plane.

  • $\begingroup$ Thanks for the explanation $\endgroup$ – tendo Apr 18 at 7:48
  • $\begingroup$ @tendo No problem, welcome to the forum ! In order to "close" question you may use the tick button rewarding the answer you may find the most suitable and the vote buttons reward users ! I also edited your answer a bit, more about MathJax here. $\endgroup$ – Rebellos Apr 18 at 7:50
  • $\begingroup$ Ok sure thank you $\endgroup$ – tendo Apr 18 at 8:31

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