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Q.Find the shortest distance between the lines r⃗=i^+2j^+3k^+λ(2i^+3j^+4k^)\vec{r} = \hat{i} + 2\hat{j} + 3\hat{k} + \lambda(2\hat{i} + 3\hat{j} + 4\hat{k}) and r⃗=2i^+4j^+5k^+t(3i^+4j^+5k^)\vec{r} = 2\hat{i} + 4\hat{j} + 5\hat{k} + t(3\hat{i} + 4\hat{j} + 5\hat{k}).

Jharkhand JacJAC Intermediate Board 2019Subjective· 6mImportance★★★★★
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For two skew lines r⃗=a⃗1+λb⃗1\vec r=\vec a_1+\lambda\vec b_1 and r⃗=a⃗2+tb⃗2\vec r=\vec a_2+t\vec b_2, the shortest distance is ∣(a⃗2−a⃗1)⋅(b⃗1×b⃗2)∣b⃗1×b⃗2∣∣\left|\dfrac{(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)}{|\vec b_1\times\vec b_2|}\right|.

Given lines:

r⃗=(i^+2j^+3k^)+λ(2i^+3j^+4k^),r⃗=(2i^+4j^+5k^)+t(3i^+4j^+5k^)\vec r = (\hat i+2\hat j+3\hat k) + \lambda(2\hat i+3\hat j+4\hat k), \qquad \vec r = (2\hat i+4\hat j+5\hat k) + t(3\hat i+4\hat j+5\hat k)

So a⃗1=(1,2,3)\vec a_1=(1,2,3), b⃗1=(2,3,4)\vec b_1=(2,3,4), a⃗2=(2,4,5)\vec a_2=(2,4,5), b⃗2=(3,4,5)\vec b_2=(3,4,5).

a⃗2−a⃗1=(1,2,2)\vec a_2-\vec a_1 = (1,2,2)

b⃗1×b⃗2=∣i^j^k^234345∣=i^(3⋅5−4⋅4)−j^(2⋅5−4⋅3)+k^(2⋅4−3⋅3)\vec b_1\times\vec b_2 = \begin{vmatrix}\hat i&\hat j&\hat k\\2&3&4\\3&4&5\end{vmatrix} = \hat i(3\cdot5-4\cdot4)-\hat j(2\cdot5-4\cdot3)+\hat k(2\cdot4-3\cdot3)

=i^(15−16)−j^(10−12)+k^(8−9)=−i^+2j^−k^= \hat i(15-16)-\hat j(10-12)+\hat k(8-9) = -\hat i+2\hat j-\hat k

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