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If shortest distance between the lines r⃗=i^−j^+λ(2i^+k^)\vec{r} = \hat{i} - \hat{j} + \lambda(2\hat{i}+\hat{k}) and

r⃗=2i^−j^+μ(i^−j^−k^)\vec{r} = 2\hat{i} - \hat{j} + \mu(\hat{i}-\hat{j}-\hat{k}) is 1414m\dfrac{\sqrt{14}}{14}m units, the value of mm is

Solution

✅ Correct Option: 1

Here a2⃗−a1⃗=i^\vec{a_2}-\vec{a_1} = \hat{i} and b1⃗×b2⃗=i^+3j^−2k^\vec{b_1}\times\vec{b_2} = \hat{i}+3\hat{j}-2\hat{k}, whose magnitude is 14\sqrt{14}.

Shortest distance =∣(a2⃗−a1⃗)⋅(b1⃗×b2⃗)∣∣b1⃗×b2⃗∣=114=1414= \dfrac{|(\vec{a_2}-\vec{a_1})\cdot(\vec{b_1}\times\vec{b_2})|}{|\vec{b_1}\times\vec{b_2}|} = \dfrac{1}{\sqrt{14}} = \dfrac{\sqrt{14}}{14}.

Comparing with 1414m\dfrac{\sqrt{14}}{14}m gives m=1m = 1.

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