Let l_(1) be the line in xy-plane with x and y intercepts 1 / 8 and 1 / 4sqrt 2 respectively, and l_(2) be the line in zx-plane with x and z intercepts - 1 / 8 and - 1 / 6sqrt 3 respectively. If d is the shortest distance between the line l_(1) and l_(2), then d^(-2) is equal to _______________.

Equation of L1 fracx - (1 / 8)(1 / 8) = fracy-frac14sqrt(2) = (z / 0) ...... L_1 fracx - (1 / 8)1 = fracy-sqrt(2) = (z / 0) ...... (i) Equation of L2 fracx + (1 / 8)-6sqrt(3) = (y / 0) = (z / 8) ...... (ii) d = | frac(vec c - vec a) * (vec b x vec d)|vec b x vec d| | [12pt] = frac( (1 / 4)hat i ) * ( 4sqrt(2)hat i + 4hat j + 3sqrt(6)hat k )sqrt(4sqrt(2))^2 + 4^2 + (3sqrt(6))^2 [12pt] = fracsqrt(2)sqrt(32 + 16 + 54) [10pt] = frac1sqrt(51) d^-2 = 51

Let l1 be the line in xyplane with x and y intercepts 1... | null (Mathematics)

Equation of L1

\frac{x - \frac{1}{8}}{\frac{1}{8}} = \frac{y}{-\frac{1}{4\sqrt{2}}} = \frac{z}{0} \quad \text{...... } L_1 \\\\

\frac{x - \frac{1}{8}}{1} = \frac{y}{-\sqrt{2}} = \frac{z}{0} \quad \text{...... (i)} \\\\

Equation of L2

\frac{x + \frac{1}{8}}{-6\sqrt{3}} = \frac{y}{0} = \frac{z}{8} \quad \text{...... (ii)} \\\\

\begin{aligned} d &= \left| \frac{(\vec{c} - \vec{a}) \cdot (\vec{b} \times \vec{d})}{|\vec{b} \times \vec{d}|} \right| \\[12pt] &= \frac{\left( \frac{1}{4}\hat{i} \right) \cdot \left( 4\sqrt{2}\hat{i} + 4\hat{j} + 3\sqrt{6}\hat{k} \right)}{\sqrt{(4\sqrt{2})^2 + 4^2 + (3\sqrt{6})^2}} \\[12pt] &= \frac{\sqrt{2}}{\sqrt{32 + 16 + 54}} \\[10pt] &= \frac{1}{\sqrt{51}} \end{aligned} \\\\

d^{-2} = 51

Explanation

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