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Question of 68

Q.Find the shortest distance between the lines whose vector equations are:
[!FORMULA] r⃗=i^+2j^+3k^+λ(i^−3j^+2k^)\vec{r} = \hat{i} + 2\hat{j} + 3\hat{k} + \lambda(\hat{i} - 3\hat{j} + 2\hat{k})
and
[!FORMULA] r⃗=4i^+5j^+6k^+μ(2i^+3j^+k^)\vec{r} = 4\hat{i} + 5\hat{j} + 6\hat{k} + \mu(2\hat{i} + 3\hat{j} + \hat{k})
OR Find the shortest distance between the lines whose vector equations are:
[!FORMULA] r⃗=(1−t)i^+(t−2)j^+(3−2t)k^\vec{r} = (1-t)\hat{i} + (t-2)\hat{j} + (3-2t)\hat{k}
and
[!FORMULA] r⃗=(s+1)i^+(2s−1)j^−(2s+1)k^\vec{r} = (s+1)\hat{i} + (2s-1)\hat{j} - (2s+1)\hat{k}

Haryana BsehBSEH Intermediate Board 2024Subjective· 5mImportance★★★★★
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Shortest distance =319=\dfrac{3}{\sqrt{19}}.

r⃗1=i^+2j^+3k^+λ(i^−3j^+2k^)\vec r_1=\hat i+2\hat j+3\hat k+\lambda(\hat i-3\hat j+2\hat k), r⃗2=4i^+5j^+6k^+μ(2i^+3j^+k^)\vec r_2=4\hat i+5\hat j+6\hat k+\mu(2\hat i+3\hat j+\hat k).

So b⃗1−a⃗1=(4−1)i^+(5−2)j^+(6−3)k^=3i^+3j^+3k^\vec b_1-\vec a_1 = (4-1)\hat i+(5-2)\hat j+(6-3)\hat k = 3\hat i+3\hat j+3\hat k, and direction vectors d⃗1=(1,−3,2)\vec d_1=(1,-3,2), d⃗2=(2,3,1)\vec d_2=(2,3,1).

d⃗1×d⃗2=∣i^j^k^1−32231∣=i^[(−3)(1)−(2)(3)]−j^[(1)(1)−(2)(2)]+k^[(1)(3)−(−3)(2)]\vec d_1\times\vec d_2 = \begin{vmatrix}\hat i&\hat j&\hat k\\1&-3&2\\2&3&1\end{vmatrix} = \hat i[(-3)(1)-(2)(3)] - \hat j[(1)(1)-(2)(2)] + \hat k[(1)(3)-(-3)(2)]

=i^(−3−6)−j^(1−4)+k^(3+6)=−9i^+3j^+9k^= \hat i(-3-6) - \hat j(1-4) + \hat k(3+6) = -9\hat i+3\hat j+9\hat k

∣d⃗1×d⃗2∣=81+9+81=171=319|\vec d_1\times\vec d_2| = \sqrt{81+9+81} = \sqrt{171} = 3\sqrt{19}

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