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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}).

Uttar Pradesh UpmspUP Board (UPMSP) Intermediate 2023Subjective· 5mImportance★★★★★
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Apply the skew-lines formula; the distance is ∣(a⃗2−a⃗1)⋅(b⃗1×b⃗2)∣∣b⃗1×b⃗2∣=32\dfrac{|(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)|}{|\vec b_1\times\vec b_2|}=\dfrac{3}{\sqrt2}.

Concept. For 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{\left|(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)\right|}{|\vec b_1\times\vec b_2|}.

Data. a⃗1=(1,2,1), b⃗1=(1,−1,1)\vec a_1=(1,2,1),\ \vec b_1=(1,-1,1); a⃗2=(2,−1,−1), b⃗2=(2,1,2)\vec a_2=(2,-1,-1),\ \vec b_2=(2,1,2).

Cross product b⃗1×b⃗2\vec b_1\times\vec b_2:

∣i^j^k^1−11212∣=i^(−1⋅2−1⋅1)−j^(1⋅2−1⋅2)+k^(1⋅1−(−1)⋅2)=−3i^+0j^+3k^.\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)=-3\hat i+0\hat j+3\hat k.

∣b⃗1×b⃗2∣=9+0+9=18=32.|\vec b_1\times\vec b_2|=\sqrt{9+0+9}=\sqrt{18}=3\sqrt2.

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