I am really Confused about this question:

Let $x=(x_1,x_2,x_3), y=(y_1,y_2,y_3)\in \mathbb{R}^3 $ be linearly independent. Let $$\delta_1 =x_2y_3-y_2x_3$$ $$\delta_2 =x_1y_3-y_1x_3$$ $$\delta_3 =x_1y_2-y_1x_2$$ If V is the span of $x,y$ then,

  • $V=\{(u,v,w):\delta_1u-\delta_2v+\delta_3w=0\}$
  • $V=\{(u,v,w):-\delta_1u+\delta_2v+\delta_3w=0\}$
  • $V=\{(u,v,w):\delta_1u+\delta_2v-\delta_3w=0\}$
  • $V=\{(u,v,w):\delta_1u+\delta_2v+\delta_3w=0\}$

What I have understood so far that all deltas are actually co-factors of the determinant.

\begin{vmatrix} u & v & w \\ x_1 &x_2&x_3 \\ y_1&y_2&y_3 \\ \notag \end{vmatrix}

So, since $\{u,v,w\}$ are independent so, 1st option is correct.

Am I correct?

  • 1
    $\begingroup$ You can simply write $(u,v,w)=c_1\cdot (x_1,x_2,x_3)+c_2\cdot (y_1,y_2,y_3)$ and then insert them inside the equations and then see which one is correct. $\endgroup$ – Levent Jul 5 '16 at 18:14

The first option is correct, since the expression $\delta_1u-\delta_2v+\delta_3w$ correctly expresses the determinant

$$\begin{vmatrix} u & v & w \\ x_1 &x_2&x_3 \\ y_1&y_2&y_3 \\ \notag \end{vmatrix}.$$

Since $x$ and $y$ are assumed to be linearly independent, a vector $(u,v,w)$ is in the plane spanned by $x$ and $y$ if and only if the above determinant is zero.

  • $\begingroup$ oh yes! thanks for pointing out my mistake $\endgroup$ – Learner Jul 5 '16 at 18:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.