Let Q be the cube with the set of vertices . Let F be the set of all twelve lines containing the diagonals of the six faces of the cube Q.

Question

Let Q be the cube with the set of vertices ${(x_{1}, x_{2}, x_{3}) \in R^{3} : x_{1}, x_{2}, x_{3} \in {0, 1}}$ . Let F be the set of all twelve lines containing the diagonals of the six faces of the cube Q. Let S be the set of all four lines containing the main diagonals of the cube Q; for instance, the line passing through the vertices (0,0,0) and (1,1,1) is in S. For lines $l_{1}$ and $l_{2}$ , let $d (l_{1}, l_{2})$ denote the shortest distance between them. Then the maximum value of $d (l_{1}, l_{2})$ , as $l_{1}$ varies over F and $l_{2}$ varies over S, is $λ$ . Find the value of $λ^{- 2}$ .

Smart Achievers · Accepted Answer

(6)

$Equation of face diagonal OD line is$ $l_{1} : \vec{r} = λ (\hat{i} + \hat{j})$

$Equation of main diagonal BE is$ $l_{2} : \vec{r} = \hat{j} + μ (\hat{i} - \hat{j} + \hat{k})$

$Shortest distance =$ $| \frac{\hat{j} \cdot (\hat{i} + \hat{j}) \times (\hat{i} - \hat{j} + \hat{k})}{(\hat{i} + \hat{j}) \times (\hat{i} - \hat{j} + \hat{k})} |$

$=$ $| \frac{\hat{j} \cdot (\hat{i} - \hat{j} - 2 \hat{k})}{\hat{i} - \hat{j} - 2 \hat{k}} |$ $= \frac{1}{\sqrt{6}}$

$In other case S.D is zero.$

Solution

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Solution

Ans. (6) Equation of face diagonal OD line is l1:r→=λ(i^+j^) Equation of main diagonal BE is l2:r→=j^+μ(i^-j^+k^) Shortest distance=|j^·(i^+j^)×(i^-j^+k^)(i^+j^)×(i^-j^+k^)| =|j^·(i^-j^-2k^)i^-j^-2k^|=16 In other case S.D is zero.

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