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(A) Let $P$ be a point on $L_1$ given by $(1+\lambda, 2-\lambda, 3+\lambda)$ and $Q$ be a point on $L_2$ given by $(4+\mu, 5+\mu, 6-\mu)$.The vector $

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$21$

Vedclass · Answer

(A) Let $P$ be a point on $L_1$ given by $(1+\lambda, 2-\lambda, 3+\lambda)$ and $Q$ be a point on $L_2$ given by $(4+\mu, 5+\mu, 6-\mu)$.The vector $\vec{PQ} = (4+\mu-(1+\lambda))\hat{i} + (5+\mu-(2-\lambda))\hat{j} + (6-\mu-(3+\lambda))\hat{k} = (3+\mu-\lambda)\hat{i} + (3+\mu+\lambda)\hat{j} + (3-\mu-\lambda)\hat{k}$.The direction vectors of $L_1$ and $L_2$ are $\vec{v_1} = \hat{i}-\hat{j}+\hat{k}$ and $\vec{v_2} = \hat{i}+\hat{j}-\hat{k}$ respectively.Since $PQ$ is the shortest distance line,it is perpendicular to both $L_1$ and $L_2$. Thus,$\vec{PQ} \cdot \vec{v_1} = 0$ and $\vec{PQ} \cdot \vec{v_2} = 0$.$\vec{PQ} \cdot \vec{v_1} = (3+\mu-\lambda) - (3+\mu+\lambda) + (3-\mu-\lambda) = 3 - 3\lambda - \mu = 0 \implies 3\lambda + \mu = 3$.$\vec{PQ} \cdot \vec{v_2} = (3+\mu-\lambda) + (3+\mu+\lambda) - (3-\mu-\lambda) = 3 + 3\mu + \lambda = 0 \implies \lambda + 3\mu = -3$.Solving these equations: $3(3\lambda + \mu) - (\lambda + 3\mu) = 3(3) - (-3) \implies 8\lambda = 12 \implies \lambda = \frac{3}{2}$.Substituting $\lambda = \frac{3}{2}$ into $3\lambda + \mu = 3$,we get $\frac{9}{2} + \mu = 3 \implies \mu = -\frac{3}{2}$.Coordinates of $P = (1+\frac{3}{2}, 2-\frac{3}{2}, 3+\frac{3}{2}) = (\frac{5}{2}, \frac{1}{2}, \frac{9}{2})$.Coordinates of $Q = (4-\frac{3}{2}, 5-\frac{3}{2}, 6+\frac{3}{2}) = (\frac{5}{2}, \frac{7}{2}, \frac{15}{2})$.The midpoint $(\alpha, \beta, \gamma) = (\frac{5/2+5/2}{2}, \frac{1/2+7/2}{2}, \frac{9/2+15/2}{2}) = (\frac{5}{2}, 2, 6)$.Thus,$2(\alpha+\beta+\gamma) = 2(\frac{5}{2} + 2 + 6) = 5 + 4 + 12 = 21$.

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