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

Jharkhand JacJAC Intermediate Board 2023Subjective· 5mImportance★★★★★
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For two skew lines r⃗=a⃗1+tb⃗1\vec r=\vec a_1+t\vec b_1 and r⃗=a⃗2+λb⃗2\vec r=\vec a_2+\lambda\vec b_2, the shortest distance is ∣(a⃗2−a⃗1)⋅(b⃗1×b⃗2)∣∣b⃗1×b⃗2∣\dfrac{|(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)|}{|\vec b_1\times\vec b_2|}.

Here 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)=i^(15−16)−j^(10−12)+k^(8−9)=(−1,2,−1)\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) = \hat i(15-16)-\hat j(10-12)+\hat k(8-9) = (-1,2,-1).

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