Question

The equations \[ a_1x + b_1y + c_1z = d_1,\quad a_2x + b_2y + c_2z = d_2,\quad a_3x + b_3y + c_3z = d_3 \] represent three planes. Given that \( d_1d_2d_3 \ne 0 \) and \[ \frac{a_i}{a_j} \ne \frac{b_i}{b_j} \ne \frac{c_i}{c_j} \quad \text{for } i, j = 1, 2, 3,\ i \ne j, \] Let \[ A = \begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix}, \quad B = \begin{bmatrix} d_1 \\ d_2 \\ d_3 \end{bmatrix}. \] If \( \text{rank}(A) = \text{rank}([A:B]) = 2 \), then the planes are such that

• A. line of intersection of any two is parallel to the third plane
• B. line of intersection of any two planes cuts the third plane
• C. line of intersection of any two planes lies on the third plane
• D. the three planes are adjacent faces of a box

Hint:

If \( \text{rank}(A) = \text{rank}([A:B]) = 2 \), the system is consistent and the solution represents a line.

Answer 1

The Correct Option is C

The given problem involves three planes defined by the equations: 

\[a_1x + b_1y + c_1z = d_1,\quad a_2x + b_2y + c_2z = d_2,\quad a_3x + b_3y + c_3z = d_3\]

where \(d_1d_2d_3 \ne 0\) and \(\frac{a_i}{a_j} \ne \frac{b_i}{b_j} \ne \frac{c_i}{c_j}\) for \(i, j = 1, 2, 3\), \(i \ne j\). We are also given two matrices, A and B:

\[A = \begin{bmatrix}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{bmatrix}, \quad B = \begin{bmatrix}d_1 \\ d_2 \\ d_3\end{bmatrix}\]

The problem states that \(\text{rank}(A) = \text{rank}([A:B]) = 2\).

When \(\text{rank}(A) = 2\), it implies that the three planes represented by the rows of matrix A are not parallel and are not coincident because a rank of less than 3 means that the vectors (normals of the planes) are linearly dependent but not collinear.

Since \(\text{rank}([A:B]) = \text{rank}(A) = 2\), the system of equations is consistent, and the solution set forms a line because the number of free variables will be 1 (3 variables - 2 rank).

The condition \(\frac{a_i}{a_j} \ne \frac{b_i}{b_j} \ne \frac{c_i}{c_j}\) ensures no two planes are parallel, implying the combination of any two planes will intersect in a line.

In summary, given \(\text{rank}(A) = \text{rank}([A:B]) = 2\), the line of intersection of any two planes lies on the third plane. Hence, the correct answer is: line of intersection of any two planes lies on the third plane.