Let l_{1} be the line in xy-plane with x and | vedclass

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(B) The equation of line $l_{1}$ in the $xy$-plane $(z=0)$ with $x$-intercept $\frac{1}{8}$ and $y$-intercept $\frac{1}{4 \sqrt{2}}$ is given by $\fra

Vedclass · Accepted Answer

$51$

Vedclass · Answer

(B) The equation of line $l_{1}$ in the $xy$-plane $(z=0)$ with $x$-intercept $\frac{1}{8}$ and $y$-intercept $\frac{1}{4 \sqrt{2}}$ is given by $\frac{x}{1/8} + \frac{y}{1/(4 \sqrt{2})} = 1$,which simplifies to $8x + 4 \sqrt{2}y = 1$.The direction vector of $l_{1}$ is $\vec{v}_{1} = (4 \sqrt{2}, -8, 0)$,which can be simplified to $(1, -\sqrt{2}, 0)$. The line passes through $A = (\frac{1}{8}, 0, 0)$.The equation of line $l_{2}$ in the $zx$-plane $(y=0)$ with $x$-intercept $-\frac{1}{8}$ and $z$-intercept $-\frac{1}{6 \sqrt{3}}$ is given by $\frac{x}{-1/8} + \frac{z}{-1/(6 \sqrt{3})} = 1$,which simplifies to $-8x - 6 \sqrt{3}z = 1$.The direction vector of $l_{2}$ is $\vec{v}_{2} = (-6 \sqrt{3}, 0, 8)$,which can be simplified to $(3 \sqrt{3}, 0, -4)$. The line passes through $B = (-\frac{1}{8}, 0, 0)$.The shortest distance $d$ between two skew lines is given by $d = \frac{|(\vec{B}-\vec{A}) \cdot (\vec{v}_{1} 	imes \vec{v}_{2})|}{|\vec{v}_{1} 	imes \vec{v}_{2}|}$.First,calculate $\vec{v}_{1} 	imes \vec{v}_{2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -\sqrt{2} & 0 \ 3 \sqrt{3} & 0 & -4 \end{vmatrix} = 4 \sqrt{2} \hat{i} + 4 \hat{j} + 3 \sqrt{6} \hat{k}$.The magnitude $|\vec{v}_{1} 	imes \vec{v}_{2}| = \sqrt{(4 \sqrt{2})^2 + 4^2 + (3 \sqrt{6})^2} = \sqrt{32 + 16 + 54} = \sqrt{102}$.$\vec{B}-\vec{A} = (-\frac{1}{8} - \frac{1}{8}, 0, 0) = (-\frac{1}{4}, 0, 0)$.$|(\vec{B}-\vec{A}) \cdot (\vec{v}_{1} 	imes \vec{v}_{2})| = |(-\frac{1}{4}, 0, 0) \cdot (4 \sqrt{2}, 4, 3 \sqrt{6})| = |-\sqrt{2}| = \sqrt{2}$.$d = \frac{\sqrt{2}}{\sqrt{102}} = \frac{1}{\sqrt{51}}$.Therefore,$d^{-2} = 51$.

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