Let three planes be:P_1 : x - y + z = 1P_2 : | vedclass

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(D) The normal vectors of the planes are $\vec{n_1} = (1, -1, 1)$,$\vec{n_2} = (1, 1, -1)$,and $\vec{n_3} = (1, -3, 3)$.The direction vectors of the l

Vedclass · Accepted Answer

Statement-$1$ is false,Statement-$2$ is true.

Vedclass · Answer

(D) The normal vectors of the planes are $\vec{n_1} = (1, -1, 1)$,$\vec{n_2} = (1, 1, -1)$,and $\vec{n_3} = (1, -3, 3)$.The direction vectors of the lines of intersection are given by the cross products of the normals:$\vec{v_1} = \vec{n_2} 	imes \vec{n_3} = \begin{vmatrix} i & j & k \ 1 & 1 & -1 \ 1 & -3 & 3 \end{vmatrix} = (0, -4, -4) \parallel (0, 1, 1)$.$\vec{v_2} = \vec{n_3} 	imes \vec{n_1} = \begin{vmatrix} i & j & k \ 1 & -3 & 3 \ 1 & -1 & 1 \end{vmatrix} = (0, 2, 2) \parallel (0, 1, 1)$.$\vec{v_3} = \vec{n_1} 	imes \vec{n_2} = \begin{vmatrix} i & j & k \ 1 & -1 & 1 \ 1 & 1 & -1 \end{vmatrix} = (0, 2, 2) \parallel (0, 1, 1)$.Since all direction vectors are parallel to $(0, 1, 1)$,the lines $L_1, L_2, L_3$ are all parallel to each other. Thus,Statement-$1$ is false.For Statement-$2$,we check the determinant of the coefficient matrix:$D = \begin{vmatrix} 1 & -1 & 1 \ 1 & 1 & -1 \ 1 & -3 & 3 \end{vmatrix} = 1(3-3) + 1(3+1) + 1(-3-1) = 0 + 4 - 4 = 0$.Since $D=0$,the planes do not have a unique common point. Checking for consistency: $P_1+P_2 = 2x = 0 \implies x=0$. Substituting $x=0$ into $P_1$ and $P_2$: $-y+z=1$ and $y-z=-1$. These are the same equation. Substituting $x=0$ into $P_3$: $-3y+3z=2 \implies y-z = -2/3$. Since $1 
eq -2/3$,there is no common point. Thus,Statement-$2$ is true.

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