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Question 67 of 68

Q.Find the shortest distance between the lines r⃗=(4i^−j^)+λ(i^+2j^−3k^)\vec{r} = (4\hat{i} - \hat{j}) + \lambda(\hat{i} + 2\hat{j} - 3\hat{k}) and r⃗=(i^−j^+2k^)+μ(2i^+4j^−5k^)\vec{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \mu(2\hat{i} + 4\hat{j} - 5\hat{k}).

Yanam CbseCBSE Class XII Board 2018Subjective· 4mImportance★★★★★
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The shortest distance is 65=655\dfrac{6}{\sqrt5}=\dfrac{6\sqrt5}{5}.

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

Why this method. The common perpendicular direction is b⃗1×b⃗2\vec b_1\times\vec b_2.

Working. a⃗1=(4,−1,0), b⃗1=(1,2,−3), a⃗2=(1,−1,2), b⃗2=(2,4,−5).\vec a_1=(4,-1,0),\ \vec b_1=(1,2,-3),\ \vec a_2=(1,-1,2),\ \vec b_2=(2,4,-5).

b⃗1×b⃗2=∣i^j^k^12−324−5∣=(2)i^−(1)j^+(0)k^=2i^−j^.\vec b_1\times\vec b_2=\begin{vmatrix}\hat i&\hat j&\hat k\\1&2&-3\\2&4&-5\end{vmatrix}=(2)\hat i-(1)\hat j+(0)\hat k=2\hat i-\hat j. …

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