Let lambda be an integer. If the shortest dis | vedclass

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(D) The given lines are $L_1: \frac{x-\lambda}{1} = \frac{y-1/2}{1/2} = \frac{z-0}{-1/2}$ and $L_2: \frac{x-0}{1} = \frac{y+2\lambda}{1} = \frac{z-\la

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(D) The given lines are $L_1: \frac{x-\lambda}{1} = \frac{y-1/2}{1/2} = \frac{z-0}{-1/2}$ and $L_2: \frac{x-0}{1} = \frac{y+2\lambda}{1} = \frac{z-\lambda}{1}$.The points on the lines are $a_1 = (\lambda, 1/2, 0)$ and $a_2 = (0, -2\lambda, \lambda)$.The direction vectors are $b_1 = (1, 1/2, -1/2)$ and $b_2 = (1, 1, 1)$.The cross product $b_1 	imes b_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1/2 & -1/2 \ 1 & 1 & 1 \end{vmatrix} = \hat{i}(1/2 + 1/2) - \hat{j}(1 + 1/2) + \hat{k}(1 - 1/2) = \hat{i} - \frac{3}{2}\hat{j} + \frac{1}{2}\hat{k} = \frac{1}{2}(2\hat{i} - 3\hat{j} + \hat{k})$.The magnitude $|b_1 	imes b_2| = \frac{1}{2}\sqrt{2^2 + (-3)^2 + 1^2} = \frac{\sqrt{14}}{2}$.The vector $a_2 - a_1 = (0-\lambda, -2\lambda-1/2, \lambda-0) = (-\lambda, -2\lambda-1/2, \lambda)$.The shortest distance $d = \frac{|(a_2 - a_1) \cdot (b_1 	imes b_2)|}{|b_1 	imes b_2|} = \frac{|-\lambda(1) - (2\lambda+1/2)(-3/2) + \lambda(1/2)|}{\sqrt{14}/2} = \frac{|-\lambda + 3\lambda + 3/4 + \lambda/2|}{\sqrt{14}/2} = \frac{|5\lambda/2 + 3/4|}{\sqrt{14}/2} = \frac{|10\lambda + 3|}{2\sqrt{14}}$.Given $d = \frac{\sqrt{7}}{2\sqrt{2}} = \frac{\sqrt{14}}{4}$.So,$\frac{|10\lambda + 3|}{2\sqrt{14}} = \frac{\sqrt{14}}{4} \implies |10\lambda + 3| = \frac{14}{2} = 7$.Case $1$: $10\lambda + 3 = 7 \implies 10\lambda = 4 \implies \lambda = 0.4$ (Not an integer).Case $2$: $10\lambda + 3 = -7 \implies 10\lambda = -10 \implies \lambda = -1$.Thus,$|\lambda| = |-1| = 1$.

Let $\lambda$ be an integer. If the shortest distance between the lines $x - \lambda = 2y - 1 = -2z$ and $x = y + 2\lambda = z - \lambda$ is $\frac{\sqrt{7}}{2\sqrt{2}}$,then the value of $|\lambda|$ is ...... .

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