Q) Let $P_{1},P_{2},$ and $P_{3}$ denote, respectively, the planes defined by

$a_{1}x + b_{1}y + c_{1}z =\alpha_{1}$

$a_{2}x + b_{2}y + c_{2}z = \alpha _{2}$

$a_{3}x + b_{3}y + c_{3}z = \alpha _{3}$

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

(A) do not have any common point of intersection

(B) intersect at a unique point

(C) intersect along a straight line

(D) intersect along a plane

My Approach :- Since, System of equations $Ax=b$ has a solution (either unique or infinitely many) when vector b is in the column space of $A$. So, for given vector $b=(1,1,1)$ , $Ax=b$ gives unique solution it means vector $b$ is in the column space of $A$ and all $3$ vectors of $A$ i.e. $(a_{1},a_{2},a_{3}),(b_{1},b_{2},b_{3}),(c_{1},c_{2},c_{3})$ are linearly independent and So, matrix formed by these $3$ vectors is invertible.
Now, if vector $b$ changed into $(2,3,4)$ then how to sure that this vector will be in the same column space. If it is in the same column space then it will give unique solution because all the vectors of $A$ will still be linearly independent and if it is not in the column space of $A$ then there will be no solution , So these $3$ planes neither cut at a point nor a straight line(infinitely many solution). So, I am confused with option $(A)$ and $(B)$. Please help.


The key thing you pointed out is that $Ax = b$ having a unique solution for some $b$ shows that the columns of $A$ are linearly independent. In particular, the columns of $A$ are three linearly independent vectors in $\mathbb{R}^3$, so they actually form a full basis for $\mathbb{R}^3$. By the definition of a basis, every vector in $\mathbb{R}^3$, such as the vector $(2,3,4)$, can be uniquely written as a linear combination of $A$'s columns. This unique linear combination corresponds to the unique solution to $Ax = (2,3,4)$, so the answer is (B).


The values of $\alpha_i$ are irrelevant, only that the planes intersect at a unique point. The coefficient matrix is thus of full rank, and for any given $\alpha_i$ said matrix can be inverted and left-multiplied with the $\alpha_i$ vector to yield a unique solution for $x,y,z$. Thus B is the correct answer.

  • $\begingroup$ Thanks @Parcly Taxel $\endgroup$ – ankit Feb 24 at 8:10
  • $\begingroup$ @ankit Really? Why don't you upvote and accept my answer? $\endgroup$ – Parcly Taxel Feb 24 at 8:11
  • $\begingroup$ I have already upvoted all the answers. I have less than 15 reputation that's why it is not showing. I have selected that answer because that includes the concept of basis and spanning a vector space which I forgot this concept while solving this problem. I really appreciate your and other's efforts while answering the answer of my query :) $\endgroup$ – ankit Feb 24 at 8:16
  • $\begingroup$ @ankit Go ask another question. $\endgroup$ – Parcly Taxel Feb 24 at 8:17
  • $\begingroup$ math.stackexchange.com/questions/3124615/… $\endgroup$ – ankit Feb 24 at 9:09

Since $Ax=b$ has unique solution for a particular $b$, we know that $A$ is invertible; i.e. $A^{-1}$ exists. Therefore, no matter what other $b$ you pick, there will still be the solution $x=A^{-1}b$ to the equation. Can you continue?

  • 1
    $\begingroup$ yes, Thank you !. $\endgroup$ – ankit Feb 24 at 8:09
  • $\begingroup$ Can the downvoter please explain? $\endgroup$ – YiFan Feb 24 at 8:42

Changing any of the $\alpha$ constants is equivalent to a simple translation of the corresponding plane in space. Translating any plane just moves the point of intersection around. As you are never changing the orientation of the planes, they will only ever intersect at a point.


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