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(D) The shortest distance between two lines $\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z

Vedclass · Accepted Answer

$48$

Vedclass · Answer

(D) The shortest distance between two lines $\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$ is given by $d = \frac{|(\vec{r_2}-\vec{r_1}) \cdot (\vec{b_1} 	imes \vec{b_2})|}{|\vec{b_1} 	imes \vec{b_2}|}$.Here,$\vec{r_1} = (-2, -3, 5)$,$\vec{r_2} = (3, 2, -4)$,$\vec{b_1} = (2, 3, 4)$,and $\vec{b_2} = (1, -3, 2)$.$\vec{r_2}-\vec{r_1} = (5, 5, -9)$.$\vec{b_1} 	imes \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & 4 \ 1 & -3 & 2 \end{vmatrix} = \hat{i}(6+12) - \hat{j}(4-4) + \hat{k}(-6-3) = 18\hat{i} - 9\hat{k}$.$|\vec{b_1} 	imes \vec{b_2}| = \sqrt{18^2 + (-9)^2} = \sqrt{324 + 81} = \sqrt{405} = 9\sqrt{5}$.Shortest distance $d = \frac{|(5, 5, -9) \cdot (18, 0, -9)|}{9\sqrt{5}} = \frac{|90 + 0 + 81|}{9\sqrt{5}} = \frac{171}{9\sqrt{5}} = \frac{19}{\sqrt{5}}$.Given $d = \frac{38}{3\sqrt{5}}k$,so $\frac{19}{\sqrt{5}} = \frac{38}{3\sqrt{5}}k \Rightarrow k = \frac{19 	imes 3}{38} = \frac{3}{2}$.Now,$\int_0^{3/2} [x^2] dx = \int_0^1 [x^2] dx + \int_1^{\sqrt{2}} [x^2] dx + \int_{\sqrt{2}}^{3/2} [x^2] dx = 0 + \int_1^{\sqrt{2}} 1 dx + \int_{\sqrt{2}}^{3/2} 2 dx = (\sqrt{2}-1) + 2(\frac{3}{2}-\sqrt{2}) = \sqrt{2}-1+3-2\sqrt{2} = 2-\sqrt{2}$.Comparing with $\alpha-\sqrt{\alpha}$,we get $\alpha=2$.Thus,$6\alpha^3 = 6(2^3) = 6 	imes 8 = 48$.

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