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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 2018Subjective· 6mImportance★★★★★
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For two skew lines given in vector form, the shortest distance is ∣(a2⃗−a1⃗)⋅(b1⃗×b2⃗)∣b1⃗×b2⃗∣∣\left|\dfrac{(\vec{a_2}-\vec{a_1})\cdot(\vec{b_1}\times\vec{b_2})}{|\vec{b_1}\times\vec{b_2}|}\right|.

Line 1: r⃗=i^+2j^+k^+λ(i^−j^+k^)\vec{r}=\hat{i}+2\hat{j}+\hat{k}+\lambda(\hat{i}-\hat{j}+\hat{k}), so a1⃗=(1,2,1)\vec{a_1}=(1,2,1), b1⃗=(1,−1,1)\vec{b_1}=(1,-1,1).

Line 2: 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}), so a2⃗=(2,−1,−1)\vec{a_2}=(2,-1,-1), b2⃗=(2,1,2)\vec{b_2}=(2,1,2).

a2⃗−a1⃗=(1,−3,−2)\vec{a_2}-\vec{a_1} = (1,-3,-2)

b1⃗×b2⃗=∣i^j^k^1−11212∣=i^(−2−1)−j^(2−2)+k^(1+2)=−3i^+0j^+3k^\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}(-2-1)-\hat{j}(2-2)+\hat{k}(1+2) = -3\hat{i}+0\hat{j}+3\hat{k}

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