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Q.Find the shortest distance between the skew lines r⃗ = î + 2ĵ + k̂ + λ(î − ĵ + k̂) and r⃗ = 2î − ĵ − k̂ + μ(2î + ĵ + 2k̂)

Kerala DhseKerala DHSE Plus Two Board 2021Subjective· 4mImportance★★★★★
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For two skew lines, the shortest distance is |(a₂-a₁)·(b₁×b₂)| / |b₁×b₂|.

Line 1: a⃗1=(1,2,1)\vec a_1=(1,2,1), b⃗1=(1,−1,1)\vec b_1=(1,-1,1). Line 2: a⃗2=(2,−1,−1)\vec a_2=(2,-1,-1), b⃗2=(2,1,2)\vec b_2=(2,1,2).

a⃗2−a⃗1=(1,−3,−2)\vec a_2-\vec a_1 = (1,-3,-2).

b⃗1×b⃗2=∣i^j^k^1−11212∣=i^(−1(2)−1(1))−j^(1(2)−1(2))+k^(1(1)−(−1)(2))\vec b_1\times\vec b_2 = \begin{vmatrix}\hat i&\hat j&\hat k\\1&-1&1\\2&1&2\end{vmatrix} = \hat i(-1(2)-1(1)) - \hat j(1(2)-1(2)) + \hat k(1(1)-(-1)(2))

=i^(−2−1)−j^(2−2)+k^(1+2)=−3i^+0j^+3k^=(−3,0,3)= \hat i(-2-1) - \hat j(2-2) + \hat k(1+2) = -3\hat i+0\hat j+3\hat k = (-3,0,3).

∣b⃗1×b⃗2∣=9+0+9=18=32|\vec b_1\times\vec b_2| = \sqrt{9+0+9}=\sqrt{18}=3\sqrt2.

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