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Q.Find the shortest distance between the 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 2024Subjective· 5mImportance★★★★★
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Compute b1×b2 for the two direction vectors, then apply the vector formula for shortest distance between skew lines.

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=(2−1)i^+(−1−2)j^+(−1−1)k^=i^−3j^−2k^\vec a_2-\vec a_1=(2-1)\hat i+(-1-2)\hat j+(-1-1)\hat k=\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^(−3)−j^(0)+k^(3)=−3i^+3k^=\hat i(-3)-\hat j(0)+\hat k(3)=-3\hat i+3\hat k

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