Let Q be the cube with the set of vertices x 1 , x 2 , x 3 ∈ ℝ 3 : x 1 , x 2 , x 3 ∈ 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

Question

Let Q be the cube with the set of vertices x 1 , x 2 , x 3 ∈ ℝ 3 : x 1 , x 2 , x 3 ∈ 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

TestHub · Accepted Answer

Plotting the diagram of cube we get,
Now equation of $O D$ line will be,
$\vec{r} = \vec{0} + λ (\hat{i} + \hat{j})$
And equation of diagonal $B E$ will be,
${\vec{r}}_{1} = \hat{j} + μ (\hat{i} - \hat{j} + \hat{k})$
Now finding the shortest distance between line $O D & B E$ we get,
$S . D = |\frac{(\hat{j} - 0) \cdot [(\hat{i} + \hat{j}) \times (\hat{i} - \hat{j} + \hat{k})]}{\sqrt{1^{2} + 1^{2} + 0^{2}} \cdot \sqrt{1^{2} + 1^{2} + 1^{2}}}|$
$\Rightarrow S . D = |\frac{\hat{j} \cdot (\hat{i} - \hat{j} - 2 \hat{k})}{\sqrt{6}}| = \frac{1}{\sqrt{6}}$
Now in other case shortest distance will be zero.

Mathematics - Vector Question with Solution | TestHub

Options:

Doubts & Discussion