Let sigma_1, sigma_2, sigma_3 be planes passi | vedclass

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Question

(C) The equations of the planes passing through the origin are given by the dot product of the normal vectors with the position vector $(x, y, z) = 0$

Vedclass · Accepted Answer

Any three distinct positive values of $a, b$,and $c$.

Vedclass · Answer

(C) The equations of the planes passing through the origin are given by the dot product of the normal vectors with the position vector $(x, y, z) = 0$.$\sigma_1: x + y + z = 0$$\sigma_2: ax + by + cz = 0$$\sigma_3: a^2x + b^2y + c^2z = 0$For the intersection of these three planes to be a single point (the origin),the determinant of the coefficient matrix must be non-zero:$\Delta = \begin{vmatrix} 1 & 1 & 1 \ a & b & c \ a^2 & b^2 & c^2 \end{vmatrix} 
eq 0$This is a standard Vandermonde determinant. We can simplify it by performing column operations:Apply $C_2 ightarrow C_2 - C_1$ and $C_3 ightarrow C_3 - C_1$:$\Delta = \begin{vmatrix} 1 & 0 & 0 \ a & b-a & c-a \ a^2 & b^2-a^2 & c^2-a^2 \end{vmatrix} = (b-a)(c-a) \begin{vmatrix} 1 & 0 & 0 \ a & 1 & 1 \ a^2 & b+a & c+a \end{vmatrix}$Apply $C_3 ightarrow C_3 - C_2$:$\Delta = (b-a)(c-a) \begin{vmatrix} 1 & 0 & 0 \ a & 1 & 0 \ a^2 & b+a & c-b \end{vmatrix} = (b-a)(c-a)(c-b) = -(a-b)(b-c)(c-a)$For $\Delta 
eq 0$,we must have $a 
eq b$,$b 
eq c$,and $c 
eq a$. Thus,$a, b$,and $c$ must be three distinct positive values.

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