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Q.Find the shortest distance between the skew lines (x + 1)/3 = (y − 1)/2 = (z − 9)/1 and (x − 2)/2 = (y + 1)/1 = (z + 1)/1.

Kerala DhseKerala DHSE Plus Two Board 2026Subjective· 4mImportance★★★★★
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For skew lines through points a1→, a2→ with directions b1→, b2→, the shortest distance is |((a2→ − a1→)·(b1→ × b2→))| / |b1→ × b2→|.

Line 1: x+13=y−12=z−91\dfrac{x+1}{3}=\dfrac{y-1}{2}=\dfrac{z-9}{1} passes through a1⃗=(−1,1,9)\vec{a_1}=(-1,1,9) with direction b1⃗=(3,2,1)\vec{b_1}=(3,2,1).

Line 2: x−22=y+11=z+11\dfrac{x-2}{2}=\dfrac{y+1}{1}=\dfrac{z+1}{1} passes through a2⃗=(2,−1,−1)\vec{a_2}=(2,-1,-1) with direction b2⃗=(2,1,1)\vec{b_2}=(2,1,1).

Step 1: a2⃗−a1⃗\vec{a_2}-\vec{a_1}.

a2⃗−a1⃗=(2−(−1), −1−1, −1−9)=(3,−2,−10).\vec{a_2}-\vec{a_1} = (2-(-1),\,-1-1,\,-1-9) = (3,-2,-10).

Step 2: b1⃗×b2⃗\vec{b_1}\times\vec{b_2}.

b1⃗×b2⃗=∣i^j^k^321211∣=i^(2⋅1−1⋅1)−j^(3⋅1−1⋅2)+k^(3⋅1−2⋅2)\vec{b_1}\times\vec{b_2} = \begin{vmatrix}\hat i&\hat j&\hat k\\3&2&1\\2&1&1\end{vmatrix} = \hat i(2\cdot1-1\cdot1)-\hat j(3\cdot1-1\cdot2)+\hat k(3\cdot1-2\cdot2)

=i^(2−1)−j^(3−2)+k^(3−4)=(1,−1,−1).= \hat i(2-1)-\hat j(3-2)+\hat k(3-4) = (1,-1,-1).

Step 3: dot product. …

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