Line L_1 passes through the points hat{i}+hat | vedclass

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(C) Line $L_1$ passes through $A(1, 1, 0)$ and $B(-1, 0, 1)$. The direction vector of $L_1$ is $\vec{v_1} = B - A = (-1-1)\hat{i} + (0-1)\hat{j} + (1-

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$z$

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(C) Line $L_1$ passes through $A(1, 1, 0)$ and $B(-1, 0, 1)$. The direction vector of $L_1$ is $\vec{v_1} = B - A = (-1-1)\hat{i} + (0-1)\hat{j} + (1-0)\hat{k} = -2\hat{i} - \hat{j} + \hat{k}$.The equation of $L_1$ is $\vec{r} = (1, 1, 0) + s(-2, -1, 1) = (1-2s, 1-s, s)$.Line $L_2$ passes through $C(0, 1, 2)$ and is parallel to $\vec{v_2} = (1, 1, 1)$.The equation of $L_2$ is $\vec{r} = (0, 1, 2) + t(1, 1, 1) = (t, 1+t, 2+t)$.At the intersection point,$(1-2s, 1-s, s) = (t, 1+t, 2+t)$.Equating coordinates: $1-2s = t$,$1-s = 1+t$,$s = 2+t$.From $1-s = 1+t$,we get $s = -t$.Substitute $s = -t$ into $s = 2+t$: $-t = 2+t \implies 2t = -2 \implies t = -1$.Then $s = 1$.Intersection point: $x = 1-2(1) = -1$,$y = 1-1 = 0$,$z = 1$.Check with $L_2$: $x = -1$,$y = 1+(-1) = 0$,$z = 2+(-1) = 1$. Matches.We need $(y-x) = 0 - (-1) = 1$.Since $z = 1$,$(y-x) = z$.

Line $L_1$ passes through the points $\hat{i}+\hat{j}$ and $\hat{k}-\hat{i}$. Line $L_2$ passes through the point $\hat{j}+2\hat{k}$ and is parallel to the vector $\hat{i}+\hat{j}+\hat{k}$. If $x\hat{i}+y\hat{j}+z\hat{k}$ is the point of intersection of the lines $L_1$ and $L_2$,then $(y-x)=$

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