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Q.Find the 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^−5j^+2k^)\vec{r} = 2\hat{i} + \hat{j} - \hat{k} + \mu(3\hat{i} - 5\hat{j} + 2\hat{k}).

Uttar Pradesh UpmspUP Board (UPMSP) Intermediate 2024Subjective· 5mImportance★★★★★
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Use d=∣(a⃗2−a⃗1)⋅(b⃗1×b⃗2)∣∣b⃗1×b⃗2∣d=\dfrac{|(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)|}{|\vec b_1\times\vec b_2|}. Here b⃗1×b⃗2=3i^−j^−7k^\vec b_1\times\vec b_2=3\hat i-\hat j-7\hat k, numerator =10=10, so d=1059d=\dfrac{10}{\sqrt{59}}.

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

Identify. a⃗1=i^+j^, b⃗1=2i^−j^+k^;a⃗2=2i^+j^−k^, b⃗2=3i^−5j^+2k^.\vec a_1=\hat i+\hat j,\ \vec b_1=2\hat i-\hat j+\hat k;\quad \vec a_2=2\hat i+\hat j-\hat k,\ \vec b_2=3\hat i-5\hat j+2\hat k.

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

∣i^j^k^2−113−52∣=i^(−2+5)−j^(4−3)+k^(−10+3)=3i^−j^−7k^.\begin{vmatrix}\hat i&\hat j&\hat k\\ 2&-1&1\\ 3&-5&2\end{vmatrix}=\hat i(-2+5)-\hat j(4-3)+\hat k(-10+3)=3\hat i-\hat j-7\hat k.

∣b⃗1×b⃗2∣=9+1+49=59.|\vec b_1\times\vec b_2|=\sqrt{9+1+49}=\sqrt{59}.

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