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Q.Find the shortest distance between the pair of the following lines: r⃗=(i^+2j^+k^)+λ(i^−j^+k^)\vec{r} = (\hat{i}+2\hat{j}+\hat{k}) + \lambda(\hat{i}-\hat{j}+\hat{k}) and r⃗=(2i^−j^−k^)+μ(2i^+j^+2k^)\vec{r} = (2\hat{i}-\hat{j}-\hat{k}) + \mu(2\hat{i}+\hat{j}+2\hat{k}).

Jharkhand JacJAC Intermediate Board 2026Subjective· 5mImportance★★★★★
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Use the skew-lines shortest-distance formula d=∣(a⃗2−a⃗1)⋅(b⃗1×b⃗2)∣∣b⃗1×b⃗2∣d = \dfrac{|(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)|}{|\vec b_1\times\vec b_2|}.

Line 1: a⃗1=i^+2j^+k^\vec a_1=\hat i+2\hat j+\hat k, b⃗1=i^−j^+k^\vec b_1=\hat i-\hat j+\hat k

Line 2: a⃗2=2i^−j^−k^\vec a_2=2\hat i-\hat j-\hat k, b⃗2=2i^+j^+2k^\vec b_2=2\hat i+\hat j+2\hat k

a⃗2−a⃗1=i^−3j^−2k^\vec a_2-\vec a_1 = \hat i-3\hat j-2\hat k

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

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