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

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(D) The normal vectors to the planes are $\vec{n}_1 = (1, -1, 1)$,$\vec{n}_2 = (1, 1, -1)$,and $\vec{n}_3 = (1, -3, 3)$.The direction vector of line $

Vedclass · Accepted Answer

Statement-$1$ is False,Statement-$2$ is True

Vedclass · Answer

(D) The normal vectors to the planes are $\vec{n}_1 = (1, -1, 1)$,$\vec{n}_2 = (1, 1, -1)$,and $\vec{n}_3 = (1, -3, 3)$.The direction vector of line $L_1$ (intersection of $P_2$ and $P_3$) is $\vec{v}_1 = \vec{n}_2 	imes \vec{n}_3 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & -1 \ 1 & -3 & 3 \end{vmatrix} = (0, -4, -4)$,which is parallel to $(0, 1, 1)$.The direction vector of line $L_2$ (intersection of $P_3$ and $P_1$) is $\vec{v}_2 = \vec{n}_3 	imes \vec{n}_1 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -3 & 3 \ 1 & -1 & 1 \end{vmatrix} = (0, 2, 2)$,which is parallel to $(0, 1, 1)$.The direction vector of line $L_3$ (intersection of $P_1$ and $P_2$) is $\vec{v}_3 = \vec{n}_1 	imes \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & 1 \ 1 & 1 & -1 \end{vmatrix} = (0, 2, 2)$,which is parallel to $(0, 1, 1)$.Since all lines $L_1, L_2, L_3$ have the same direction vector $(0, 1, 1)$,they are all parallel to each other. Thus,$STATEMENT-1$ is False.For $STATEMENT-2$,we check if the system has a common point by solving the equations. Adding $P_1$ and $P_2$ gives $2x = 0$,so $x = 0$. Substituting $x=0$ into $P_1$ and $P_2$ gives $-y+z=1$ and $y-z=-1$,which are identical. Substituting $x=0$ into $P_3$ gives $-3y+3z=2$,or $-y+z=2/3$. Since $1 
eq 2/3$,the system has no common point. Thus,$STATEMENT-2$ is True.

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