Consider a square ABCD of diagonal length 4 -3- The square is folded along the diagonal AC so that the plane of - ABC is perpendicular to the plane of - ADC- The shortest distance between AB and CD is

Take B as origin , BC amp; BA as x amp; y axis. Plane of Δ ADC is perpendicular to plane of Δ ABC. ⇒B⃗A⃗=2√(6)ĵ C⃗D⃗= √(6)î+√(6)ĵ+2√(3)k̂ n⃗=12(√(2 ...

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Standard X

Mathematics

Introduction to LOCI

Consider a sq...

Question

Consider a square ABCD of diagonal length 4√3. The square is folded along the diagonal AC so that the plane of ΔABC is perpendicular to the plane of ΔADC. The shortest distance between AB and CD is

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Solution

Take B as origin , BC & BA as x & y axis. Plane of $Δ A D C$ is perpendicular to plane of $Δ A B C .$

$\Rightarrow \to B A = 2 \sqrt{6}^j$

$\to C D = - \sqrt{6}^i + \sqrt{6}^j + 2 \sqrt{3}^k$

$\to n = 12 (\sqrt{2}^i +^k)$

$\to B C = 2 \sqrt{6}^i$

Shortest distance = $∣ ∣ \to B C . \frac{\to n}{| \to n |} ∣ ∣ = 4$

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[Kurukshetra CEE 1998]

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Consider a square ABCD of diagonal length 4√3. The square is folded along the diagonal AC so that the plane of ΔABC is perpendicular to the plane of ΔADC. The shortest distance between AB and CD is

Take B as origin , BC & BA as x & y axis. Plane of ΔADC is perpendicular to plane of ΔABC. ⇒→BA=2√6^j →CD=−√6^i+√6^j+2√3^k →n=12(√2^i+^k) →BC=2√6^i Shortest distance =∣∣→BC.→n|→n|∣∣=4