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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}).

Jammu Kashmir JkboseJKBOSE Class 12 Annual Regular Examination 2024Subjective· 4mImportance★★★★★
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For skew lines r⃗=a⃗1+λb⃗1\vec r=\vec a_1+\lambda\vec b_1 and r⃗=a⃗2+μb⃗2\vec r=\vec a_2+\mu\vec b_2, the shortest distance is ∣(a⃗2−a⃗1)⋅(b⃗1×b⃗2)∣b⃗1×b⃗2∣∣\left|\dfrac{(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)}{|\vec b_1\times\vec b_2|}\right|.

Here 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; 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\cdot2-1\cdot1)-\hat j(1\cdot2-1\cdot2)+\hat k(1\cdot1-(-1)\cdot2)

=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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