Let M and N be the feet of the perpendiculars | vedclass

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(A) The line $L_1$ can be written as $x=t, y=t, z=1$. Any point $M$ on $L_1$ is $(t, t, 1)$. The vector $\vec{PM} = (t-a, t-a, 1-a)$. Since $PM \perp

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(A) The line $L_1$ can be written as $x=t, y=t, z=1$. Any point $M$ on $L_1$ is $(t, t, 1)$. The vector $\vec{PM} = (t-a, t-a, 1-a)$. Since $PM \perp L_1$,the dot product of $\vec{PM}$ and the direction vector of $L_1$,$\vec{v_1} = (1, 1, 0)$,is $0$: $(t-a)(1) + (t-a)(1) + (1-a)(0) = 0 \implies 2(t-a) = 0 \implies t=a$. Thus,$M = (a, a, 1)$ and $\vec{PM} = (0, 0, 1-a)$.Similarly,the line $L_2$ can be written as $x=s, y=-s, z=-1$. Any point $N$ on $L_2$ is $(s, -s, -1)$. The vector $\vec{PN} = (s-a, -s-a, -1-a)$. Since $PN \perp L_2$,the dot product of $\vec{PN}$ and the direction vector of $L_2$,$\vec{v_2} = (1, -1, 0)$,is $0$: $(s-a)(1) + (-s-a)(-1) + (-1-a)(0) = 0 \implies s-a + s+a = 0 \implies 2s = 0 \implies s=0$. Thus,$N = (0, 0, -1)$ and $\vec{PN} = (-a, -a, -1-a)$.Given $\angle MPN = 90^{\circ}$,the dot product $\vec{PM} \cdot \vec{PN} = 0$: $(0)(-a) + (0)(-a) + (1-a)(-1-a) = 0 \implies -(1-a)(1+a) = 0 \implies -(1-a^2) = 0 \implies a^2-1 = 0 \implies a^2 = 1$.

Let $M$ and $N$ be the feet of the perpendiculars drawn from the point $P(a, a, a)$ to the lines $L_1: x-y=0, z=1$ and $L_2: x+y=0, z=-1$ respectively. If $\angle MPN=90^{\circ}$,then $a^2=$

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