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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
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3 Answers

2 votes

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 $(B).$ 


edited by
1 vote
All you're doing is translating the plane. Translations have no impact on intersection. The normal vectors for the planes remain the same, so their orientation is not changed and they still intersect at one point.
0 votes

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

can be written as $\begin{bmatrix} a_{1} & b_{1} & c_{1}\\ a_{2} & b_{2} & c_{2}\\ a_{3} & b_{3} & c_{3} \end{bmatrix} \begin{bmatrix} x\\ y\\ z \end{bmatrix} = \begin{bmatrix} \alpha _{1}\\ \alpha _{2}\\ \alpha _{3} \end{bmatrix}$

Now it says that when $\alpha _1 = \alpha _2 =\alpha _3=1$, then a unique solution exists, that means $\begin{vmatrix} a_{1} & b_{1} & c_{1}\\ a_{2} & b_{2} & c_{2}\\ a_{3} & b_{3} & c_{3} \end{vmatrix} \neq 0$

Now whatever be the value of $\alpha _1, \alpha _2 ,\alpha _3$, the fact that determinant of our basis vectors isn’t zero won’t change hence we’ll still get one unique solution.

Hence the answer is D, because the solution is unique and had it been not unique then i don't think that with the given information it would have been possible to determine how many dimensions collapsed.

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