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Let $u_1=(1,-1,a)$, $u_2=(a,0,1)$, $u_3=(1,1,a)$.

(a) Show that there are no values of a such that span{$u_1,u_2,u_3$} is a line in $\mathbb{R}^3$ that passes through the origin.

I figured out that $a = \pm1$ for it to result in a linear line. But i'm stuck after.

For $a=1$, i found that there does exist the possibility it could be the zero vector when $z=s$ where $s$ is the parameter.

am i doing something wrong?

(b) Find all values of $a$ such that span{$u_1,u_2,u_3$} is a plane in $\mathbb{R}^3$ containing the origin.

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  • $\begingroup$ I don't have much time now, but for a), you need the matrix you construct from your coordinate vectors to have rank 1, since you need all your vectors to be linearly dependent, even if you choose two from the three at random. The rank of a matrix however is the same as the rank of the linear map it represents, which is the same as the dimension of its range. Try making that matrix as such that it maps into an 1-dimensional space. $\endgroup$ – Bence Racskó Mar 15 '15 at 13:33
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No matter what $a$ is, the line through the origin and $u_2$ will always have $y=0$ everywhere. So neither $u_1$ nor $u_3$ can be on it, again independently of what $a$ is.

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  • $\begingroup$ i don't really understand what you mean by this. $\endgroup$ – Danxe Mar 15 '15 at 14:39
  • $\begingroup$ @Danxe: If the three vectors span a line through the origin, then that line can only be the line that passes through the origin in the direction of $u_2$. The second component of every point is always $0$, but neither $u_1$ nor $u_3$ has $y=0$; therefore they are not on the line through the origin and $u_2$. $\endgroup$ – Henning Makholm Mar 15 '15 at 14:46
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HINT: $$ \left|\begin{array}[ccc]{} 1 &-1&a\\ a&0&1 \\ 1&1&a \end{array}\right| =2(a^2-1)=0 \Leftrightarrow a=\pm1 $$

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