Let $P_1$, $P_2$ and $P_3$ denote, respectively, the planes defined by \begin{align*} a_1 x + b_1 y + c_1 z &= \alpha_1 \\ a_2 x + b_2 y + c_2 z &= \alpha_2 \\ a_3 x + b_3 y + c_3 z &= \alpha_3. \end{align*}

It is given that $P_1$, $P_2$ and $P_3$ intersect exactly at one point when $\alpha_1 = \alpha_2 = \alpha_3 = 1$. If now $\alpha_1 = 2$, $\alpha_2 = 3$ and $\alpha_3 = 4$ then the planes

  1. do not have any common point intersection
  2. intersect at a unique point
  3. intersect along a straight line
  4. intersect along a plane.

I have done questions like this. But i donot know how to start this one. Help will be appriciated


  • $\begingroup$ Go for elimiantion of z from 1-2 and then 2-3 $\endgroup$ – Archis Welankar Dec 31 '15 at 11:17


1.The system $Au=b$ has a solution $\iff $ $\operatorname{Rank A}=\operatorname{Rank (A|b)}$

2.If $\operatorname{Rank } A=n;$ Then the system $Ax=0$ has a unique solution and $Ax=b$ has at most one solution if it exists.

Now consider the system $Au=b$ where $A=$ \begin{bmatrix} a_1 &b_1 &c_1\\a_2 & b_2 & c_2\\ a_3 & b_3 &c_3\end{bmatrix} $u=$\begin{bmatrix} x \\ y\\z \end{bmatrix} and $b=$ \begin{bmatrix} \alpha_1=1\\ \alpha_2=1 \\ \alpha_3=1\end{bmatrix}

By your hypothesis $\operatorname{Rank A}=\operatorname{Rank (A|b)}$.Moreover the solution is unique implies $\operatorname{Rank A}=\operatorname{Rank (A|b)}=3$ in this case.

Now focus on the problem given.

We have $Au=b$ where $b=$ \begin{bmatrix} \alpha_1=2\\ \alpha_2=3 \\ \alpha_3=4\end{bmatrix}.

so $A|b=$ \begin{bmatrix} a_1 &b_1 &c_1 & 2\\a_2 & b_2 & c_2 &3\\ a_3 & b_3 &c_3 & 4\end{bmatrix}

Now $\operatorname{Rank (A|b)}=3$ in this case because $\operatorname{Rank (A|b)}$ must be $\leq 3$ as $A|b$ is a $ 3\times 4 $ matrix and $\operatorname{Rank (A|b)}\geq 3$ as $\operatorname{Rank (A)=3}$ .Thus $\operatorname{Rank (A|b)}=\operatorname{Rank (A)=3}$ .

Thus the planes intersect at a unique point.

  • $\begingroup$ i didnot get it(the last part) $\endgroup$ – Taylor Ted Dec 31 '15 at 11:52
  • $\begingroup$ I have edited the answer @TaylorTed $\endgroup$ – Learnmore Dec 31 '15 at 13:04
  • $\begingroup$ is rank of new augmented matrix 3 because of rank A=3.So we cannot have all zeroes in all last row, which makes rank of new aug mat =3 $\endgroup$ – Taylor Ted Jan 1 '16 at 5:18
  • $\begingroup$ Yes we will always get 3 linearly independent columns /rows in $A|b$ because of $A$ $\endgroup$ – Learnmore Jan 1 '16 at 6:08

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