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Q.Find shortest distance between the lines r⃗=i^+j^+λ(2i^−j^+k^)\vec{r} = \hat{i} + \hat{j} + \lambda(2\hat{i} - \hat{j} + \hat{k}) and r⃗=(2i^+j^−k^)+μ(3i^+j^+2k^)\vec{r} = (2\hat{i} + \hat{j} - \hat{k}) + \mu(3\hat{i} + \hat{j} + 2\hat{k}).

Uttar Pradesh UpmspUP Board (UPMSP) Intermediate 2025Subjective· 5mImportance★★★★★
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Apply d=∣(a⃗2−a⃗1)⋅(b⃗1×b⃗2)∣∣b⃗1×b⃗2∣=835d=\dfrac{|(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)|}{|\vec b_1\times\vec b_2|}=\dfrac{8}{\sqrt{35}}.

Concept. 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

d=∣(a⃗2−a⃗1)⋅(b⃗1×b⃗2)∣∣b⃗1×b⃗2∣.d=\frac{\bigl|(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)\bigr|}{|\vec b_1\times\vec b_2|}.

Here a⃗1=i^+j^, b⃗1=2i^−j^+k^, a⃗2=2i^+j^−k^, b⃗2=3i^+j^+2k^\vec a_1=\hat i+\hat j,\ \vec b_1=2\hat i-\hat j+\hat k,\ \vec a_2=2\hat i+\hat j-\hat k,\ \vec b_2=3\hat i+\hat j+2\hat k.

b⃗1×b⃗2=∣i^j^k^2−11312∣=i^(−1⋅2−1⋅1)−j^(2⋅2−1⋅3)+k^(2⋅1−(−1)⋅3)\vec b_1\times\vec b_2=\begin{vmatrix}\hat i&\hat j&\hat k\\2&-1&1\\3&1&2\end{vmatrix}=\hat i(-1\cdot2-1\cdot1)-\hat j(2\cdot2-1\cdot3)+\hat k(2\cdot1-(-1)\cdot3)

=i^(−2−1)−j^(4−3)+k^(2+3)=−3i^−j^+5k^.=\hat i(-2-1)-\hat j(4-3)+\hat k(2+3)=-3\hat i-\hat j+5\hat k. …

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