Let L_1 and L_2 denote the lines overrightarr | vedclass

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(A) The lines are $L_1: \overrightarrow{r} = (1, 0, 0) + \lambda(-1, 2, 2)$ and $L_2: \overrightarrow{r} = \mu(2, -1, 2)$.Let $A$ be a point on $L_1$

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$1, 2, 4$

Vedclass · Answer

(A) The lines are $L_1: \overrightarrow{r} = (1, 0, 0) + \lambda(-1, 2, 2)$ and $L_2: \overrightarrow{r} = \mu(2, -1, 2)$.Let $A$ be a point on $L_1$ and $B$ be a point on $L_2$. $A = (1-\lambda, 2\lambda, 2\lambda)$ and $B = (2\mu, -\mu, 2\mu)$.The vector $\overrightarrow{AB} = (2\mu + \lambda - 1, -\mu - 2\lambda, 2\mu - 2\lambda)$.The direction of the common perpendicular is $\vec{v} = (-1, 2, 2) 	imes (2, -1, 2) = (6, 6, -6)$,which is parallel to $(2, 2, -1)$.Since $AB$ is the shortest distance line,$\overrightarrow{AB}$ must be parallel to $(2, 2, -1)$.Thus,$\frac{2\mu + \lambda - 1}{2} = \frac{-\mu - 2\lambda}{2} = \frac{2\mu - 2\lambda}{-1} = k$.Solving this system gives $\lambda = 1/9$ and $\mu = 2/9$.Then $A = (8/9, 2/9, 2/9)$ and $B = (4/9, -2/9, 4/9)$.The line $L_3$ passes through $A$ and $B$ with direction $(2, 2, -1)$.Equation of $L_3$: $\overrightarrow{r} = A + t(2, 2, -1) = (8/9, 2/9, 2/9) + t(2, 2, -1)$.Option $(1)$ passes through $(2/3, 0, 1/3)$,which is the midpoint of $AB$. Since the direction is $(2, 2, -1)$,it represents $L_3$.Option $(2)$ passes through $B$,so it represents $L_3$.Option $(4)$ passes through $(8/9, 2/9, 2/9)$,which is point $A$,so it represents $L_3$.Thus,options $1, 2, 4$ are correct.

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