Let the shortest distance between the lines L | vedclass

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(A) The line $L$ passes through $(5, \lambda, -\lambda)$ with direction vector $\vec{b_1} = (-2, 0, 1)$.The line $L_1$ can be written as $\frac{x+1}{1

Vedclass · Accepted Answer

$\alpha+2\gamma=24$

Vedclass · Answer

(A) The line $L$ passes through $(5, \lambda, -\lambda)$ with direction vector $\vec{b_1} = (-2, 0, 1)$.The line $L_1$ can be written as $\frac{x+1}{1} = \frac{y-1}{1} = \frac{z-4}{-1}$,passing through $(-1, 1, 4)$ with direction vector $\vec{b_2} = (1, 1, -1)$.The cross product $\vec{b_1} 	imes \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -2 & 0 & 1 \ 1 & 1 & -1 \end{vmatrix} = -\hat{i} - \hat{j} - 2\hat{k}$.The shortest distance is given by $d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} 	imes \vec{b_2})|}{ |\vec{b_1} 	imes \vec{b_2}| }$.Here $\vec{a_2} - \vec{a_1} = (-1-5, 1-\lambda, 4-(-\lambda)) = (-6, 1-\lambda, 4+\lambda)$.$d = \frac{|(-6)(-1) + (1-\lambda)(-1) + (4+\lambda)(-2)|}{\sqrt{(-1)^2 + (-1)^2 + (-2)^2}} = \frac{|6 - 1 + \lambda - 8 - 2\lambda|}{\sqrt{6}} = \frac{|-\lambda - 3|}{\sqrt{6}}$.Given $d = 2\sqrt{6}$,so $|\lambda+3| = 12$. Since $\lambda \geq 0$,$\lambda = 9$.For $L$,$(\alpha, \beta, \gamma) = (5-2k, \lambda, k-\lambda) = (5-2k, 9, k-9)$.Then $\alpha = 5-2k$ and $\gamma = k-9$. Thus $k = \gamma+9$.Substituting $k$: $\alpha = 5-2(\gamma+9) = 5-2\gamma-18 = -2\gamma-13$,so $\alpha+2\gamma = -13$.Checking options: $\alpha+2\gamma=24$ is not possible.

Let the shortest distance between the lines $L : \frac{x-5}{-2} = \frac{y-\lambda}{0} = \frac{z+\lambda}{1}, \lambda \geq 0$ and $L_1 : x+1 = y-1 = 4-z$ be $2\sqrt{6}$. If $(\alpha, \beta, \gamma)$ lies on $L$,then which of the following is $NOT$ possible?

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