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Question

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

CBSECBSE Class XII Board 2020Subjective· 4mImportance★★★★★
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Using d=∣(a⃗2−a⃗1)⋅(b⃗1×b⃗2)∣∣b⃗1×b⃗2∣d = \dfrac{\left|\left(\vec{a}_2 - \vec{a}_1\right)\cdot\left(\vec{b}_1 \times \vec{b}_2\right)\right|}{\left|\vec{b}_1 \times \vec{b}_2\right|}, the shortest distance is 13=33\dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}}{3} units.

Solution

Here a⃗1=2i^−j^+k^\vec{a}_1 = 2\hat{i} - \hat{j} + \hat{k}, b⃗1=3i^−2j^+5k^\vec{b}_1 = 3\hat{i} - 2\hat{j} + 5\hat{k} and a⃗2=3i^+2j^−4k^\vec{a}_2 = 3\hat{i} + 2\hat{j} - 4\hat{k}, b⃗2=4i^−j^+3k^\vec{b}_2 = 4\hat{i} - \hat{j} + 3\hat{k}.

Cross product of the directions:

b⃗1×b⃗2=∣i^j^k^3−254−13∣=(−6+5)i^−(9−20)j^+(−3+8)k^=−i^+11j^+5k^.\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\3 & -2 & 5 \\4 & -1 & 3 \end{vmatrix} = (-6+5)\hat{i} - (9-20)\hat{j} + (-3+8)\hat{k} = -\hat{i} + 11\hat{j} + 5\hat{k}.

∣b⃗1×b⃗2∣=(−1)2+112+52=147=73.\left|\vec{b}_1 \times \vec{b}_2\right| = \sqrt{(-1)^2 + 11^2 + 5^2} = \sqrt{147} = 7\sqrt{3}.

Vector joining the points: …

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