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Let $P_1$, $P_2$ and $P_3$ denote, respectively, the planes defined by

$$\begin{array} {} a_1x +b_1y+c_1z=\alpha _1 \\ a_2x +b_2y+c_2z=\alpha _2 \\ a_3x +b_3y+c_3z=\alpha _3 \end{array}$$

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 $\alpha _1 =2, \: \alpha _2=3, \: \alpha _3 = 4$ then the planes

  1. do not have any common point of intersection
  2. intersect at a unique point
  3. intersect along a straight line
  4. intersect along a plane
in Linear Algebra by Veteran (431k points)
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1 Answer

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Given that $$\begin{array} {} a_1x +b_1y+c_1z=\alpha _1 \\ a_2x +b_2y+c_2z=\alpha _2 \\ a_3x +b_3y+c_3z=\alpha _3 \end{array}$$

Non-homogeneous equations is the form of $AX = B$ and augmented matrix $[A:B]$

Case$1:$ If $\text{rank}(A) \neq \text{rank}([A:B]),$ then no solution. 

Case$2:$ If $\text{rank}(A) = \text{rank}([A:B]),$ then

  1.  If  $\text{rank}(A) = \text{rank}([A:B]) = \text{number of variables(unknown)},$ then unique solution.
  2.  If  $\text{rank}(A) = \text{rank}([A:B]) < \text{number of variables(unknown)},$ then infinite solution.

Now, we can write the above equations into an augmented form

$[A:B] = \begin{bmatrix} a_{1}& b_{1}&c_{1} & :\alpha_{1}\\ a_{2}&b_{2} &c_{2} &:\alpha_{2} \\a_{3} & b_{3}&c_{3} &:\alpha_{3} \end{bmatrix}$

Given that these planes intersect exactly at one point when $\alpha _1 = \alpha _2 =\alpha _3=1.$ It means unique solution. Or we can say that  $\text{rank}(A) = \text{rank}([A:B]) = \text{number of variables(unknown)}.$

$[A:B] = \begin{bmatrix} a_{1}& b_{1}&c_{1} & :1 \\ a_{2}&b_{2} &c_{2} &:1 \\a_{3} & b_{3}&c_{3} &:1 \end{bmatrix}$

Operation$:R_{3} \rightarrow R_{3} - R_{2}, R_{2} \rightarrow R_{2} - R_{1}$ by doing the elementary transformation rank of the matrix doesn't change.

$[A:B] = \begin{bmatrix} a_{1}& b_{1}&c_{1} & :1 \\ a_{2} - a_{1} &b_{2} - b_{1} &c_{2} - c_{1} &: 0 \\a_{3} - a_{2} & b_{3}- b_{2}&c_{3}-c_{2} &:0 \end{bmatrix}$

If $c_{2} - c_{1}\neq 0 $ and $c_{3}-c_{2} \neq 0,$ than $\text{rank}(A) = \text{rank}([A:B]) = \text{number of variables(unknown)}=3.$

If now $\alpha _1 =2, \: \alpha _2=3, \: \alpha _3 = 4$ then 

$[A:B] = \begin{bmatrix} a_{1}& b_{1}&c_{1} & :2 \\ a_{2}&b_{2} &c_{2} &:3 \\a_{3} & b_{3}&c_{3} &:4 \end{bmatrix}$

Operation$:R_{3} \rightarrow R_{3} - R_{2}, R_{2} \rightarrow R_{2} - R_{1}$

$[A:B] = \begin{bmatrix} a_{1}& b_{1}&c_{1} & :2 \\ a_{2} - a_{1} &b_{2} - b_{1} &c_{2} - c_{1} &: 1 \\a_{3} - a_{2} & b_{3}- b_{2}&c_{3}-c_{2} &:1 \end{bmatrix}$

Operation$:R_{3} \rightarrow R_{3} - R_{2}$

$[A:B] = \begin{bmatrix} a_{1}& b_{1}&c_{1} & :2 \\ a_{2} - a_{1} &b_{2} - b_{1} &c_{2} - c_{1} &: 1 \\a_{3} - 2a_{2} + a_{1} & b_{3} - 2b_{2} + b_{1} & c_{3} - 2c_{2} + c_{1} &:0 \end{bmatrix}$

Here, also  $\text{rank}(A) = \text{rank}([A:B]) = \text{number of variables(unknown)}=3.$

Therefore, intersect at a unique point.

So, the correct answer is $(B).$ 

by Veteran (58.9k points)
edited by

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