Given the vectors $\vec{a}=\vec{i}+2\vec{j}-\vec{k}, \vec{b}=\vec{i}-\vec{j}+\vec{k}, \vec{c}=\lambda\vec{i}+\vec{k},\lambda\in\Bbb R$ in $V^{3}(0)$ find all scalars $\lambda$ for which the vectors $\vec{a}, \vec{b}, \vec{c}$ are coplanar.


In general, for these vectors to be coplanar, does that mean $\alpha\neq 0, \beta\neq 0, \gamma\neq 0$ or is it sufficient that at least one of them is not 0?

Here's what I did:














In the end I got $\beta=2\alpha, \gamma(\lambda-3)=0$

Now if I put $\gamma=0$ then that would mean $\alpha=0$ and that would mean $\beta=0$. So my vectors would be non-coplanar.

So I have to put $\lambda=3$.

Is that the final solution?

  • 1
    $\begingroup$ Your solution is correct: good job. $\endgroup$ – Crostul Nov 7 '16 at 9:54
  • $\begingroup$ @Crostul Wait, don't I have to write down in the end $(\alpha, \beta, \gamma)=(t, 2t, -t), t\in \Bbb R$? If I just put $\lambda=3$ that would mean $\alpha, \beta, \gamma$ can be anything, right? $\endgroup$ – lmc Nov 7 '16 at 10:18
  • $\begingroup$ If you put $\lambda = 3$, then these three vectors are coplanar since $(\alpha, \beta , \gamma) = (1,2,-1)$ is a non-zero linear combination which adds up to the zero vector. $\endgroup$ – Crostul Nov 7 '16 at 10:20
  • $\begingroup$ @Crostul I don't think you understand what I'm saying: if I just say that $\lambda=3$ for these vectors to be coplanar, then I could take for example $\alpha=\beta=\gamma=1$ but $1\times \vec{a}+1\times\vec{b}+1\times\vec{c}\neq 0$ in that case. $\endgroup$ – lmc Nov 7 '16 at 12:09
  • $\begingroup$ Nicely done. Another way to go about this would be to observe that for the three vectors to be coplanar, they must be linearly dependent, so the matrix with those vectors as columns (or rows) must be singular. I.e., solve $\det\begin{bmatrix}\vec a&\vec b&\vec c\end{bmatrix}=0$ for $\lambda$. This particular determinant expands into a simple expression for $\lambda$. $\endgroup$ – amd Nov 7 '16 at 18:53

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