Three lines L_1: overrightarrow{r} = lambda h | vedclass

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(D) Let $P = (\lambda, 0, 0)$ be a point on $L_1$,$Q = (0, \mu, 1)$ be a point on $L_2$,and $R = (1, 1, v)$ be a point on $L_3$. Since $P, Q,$ and $R$

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$3, 4$

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(D) Let $P = (\lambda, 0, 0)$ be a point on $L_1$,$Q = (0, \mu, 1)$ be a point on $L_2$,and $R = (1, 1, v)$ be a point on $L_3$. Since $P, Q,$ and $R$ are collinear,the vectors $\vec{PQ}$ and $\vec{QR}$ must be proportional. $\vec{PQ} = (0 - \lambda, \mu - 0, 1 - 0) = (-\lambda, \mu, 1)$. $\vec{QR} = (1 - 0, 1 - \mu, v - 1) = (1, 1 - \mu, v - 1)$. For collinearity,$\frac{-\lambda}{1} = \frac{\mu}{1 - \mu} = \frac{1}{v - 1}$. This implies $\lambda = -\frac{\mu}{1 - \mu} = \frac{\mu}{\mu - 1}$ and $v - 1 = \frac{1 - \mu}{\mu}$,so $v = 1 + \frac{1 - \mu}{\mu} = \frac{1}{\mu}$. These values of $\lambda$ and $v$ exist for all $\mu \in R$ except $\mu = 0$ and $\mu = 1$. If $\mu = 0$,$Q = (0, 0, 1) = \hat{k}$. If $\mu = 1$,$Q = (0, 1, 1) = \hat{j} + \hat{k}$. Thus,$Q$ can be any point on $L_2$ except $\hat{k}$ and $\hat{j} + \hat{k}$. Checking the given options: $(1)$ $\hat{k} + \hat{j}$ (corresponds to $\mu = 1$,excluded) $(2)$ $\hat{k}$ (corresponds to $\mu = 0$,excluded) $(3)$ $\hat{k} + \frac{1}{2}\hat{j}$ (corresponds to $\mu = 0.5$,valid) $(4)$ $\hat{k} - \frac{1}{2}\hat{j}$ (corresponds to $\mu = -0.5$,valid) Therefore,points $3$ and $4$ are valid.

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