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Q.Find the shortest distance between the lines : 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 r⃗=(s+1)i^+(2s−1)j^−(2s+1)k^\vec{r} = (s+1)\hat{i} + (2s-1)\hat{j} - (2s+1)\hat{k}.

Jammu Kashmir JkboseJKBOSE Class 12 Annual Regular Examination 2025Subjective· 4mImportance★★★★★
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Write both lines in point + direction-vector form, then apply the skew-lines shortest-distance formula; the distance is 829\dfrac{8}{\sqrt{29}}.

Rewrite each line in the form r⃗=a⃗+λb⃗\vec r = \vec a + \lambda\vec b.

Line 1: r⃗=(1−t)i^+(t−2)j^+(3−2t)k^=(i^−2j^+3k^)+t(−i^+j^−2k^)\vec r = (1-t)\hat i+(t-2)\hat j+(3-2t)\hat k = (\hat i - 2\hat j+3\hat k) + t(-\hat i+\hat j-2\hat k)

So a⃗1=i^−2j^+3k^\vec a_1 = \hat i-2\hat j+3\hat k, b⃗1=−i^+j^−2k^\vec b_1 = -\hat i+\hat j-2\hat k.

Line 2: r⃗=(s+1)i^+(2s−1)j^−(2s+1)k^=(i^−j^−k^)+s(i^+2j^−2k^)\vec r = (s+1)\hat i+(2s-1)\hat j-(2s+1)\hat k = (\hat i-\hat j-\hat k)+s(\hat i+2\hat j-2\hat k)

So a⃗2=i^−j^−k^\vec a_2 = \hat i-\hat j-\hat k, b⃗2=i^+2j^−2k^\vec b_2 = \hat i+2\hat j-2\hat k.

For skew lines, the shortest distance is:

d=∣(a⃗2−a⃗1)⋅(b⃗1×b⃗2)∣b⃗1×b⃗2∣∣d = \left|\frac{(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)}{|\vec b_1\times\vec b_2|}\right|

a⃗2−a⃗1=(1−1)i^+(−1−(−2))j^+(−1−3)k^=0i^+j^−4k^\vec a_2-\vec a_1 = (1-1)\hat i+(-1-(-2))\hat j+(-1-3)\hat k = 0\hat i+\hat j-4\hat k.

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