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Q.Find the shortest distance between the lines →r = î + ĵ + λ(2î − ĵ + k̂) and →r = 2î + ĵ − k̂ + μ(3î − 5ĵ + 2k̂).

Kerala DhseKerala DHSE Plus Two Board 2024Subjective· 4mImportance★★★★★
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For two skew lines r⃗=a⃗1+λb⃗1\vec r=\vec a_1+\lambda\vec b_1 and r⃗=a⃗2+μb⃗2\vec r=\vec a_2+\mu\vec b_2, the shortest distance is ∣(a⃗2−a⃗1)⋅(b⃗1×b⃗2)∣b⃗1×b⃗2∣∣\left|\dfrac{(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)}{|\vec b_1\times\vec b_2|}\right|.

Given lines:

r⃗=i^+j^+λ(2i^−j^+k^),r⃗=2i^+j^−k^+μ(3i^−5j^+2k^)\vec r=\hat i+\hat j+\lambda(2\hat i-\hat j+\hat k), \qquad \vec r=2\hat i+\hat j-\hat k+\mu(3\hat i-5\hat j+2\hat k)

So a⃗1=(1,1,0)\vec a_1=(1,1,0), b⃗1=(2,−1,1)\vec b_1=(2,-1,1); a⃗2=(2,1,−1)\vec a_2=(2,1,-1), b⃗2=(3,−5,2)\vec b_2=(3,-5,2).

Step 1: a⃗2−a⃗1\vec a_2-\vec a_1.

a⃗2−a⃗1=(2−1, 1−1, −1−0)=(1,0,−1)\vec a_2-\vec a_1=(2-1,\,1-1,\,-1-0)=(1,0,-1)

Step 2: b⃗1×b⃗2\vec b_1\times\vec b_2.

b⃗1×b⃗2=∣i^j^k^2−113−52∣\vec b_1\times\vec b_2=\begin{vmatrix}\hat i&\hat j&\hat k\\2&-1&1\\3&-5&2\end{vmatrix}

i^:(−1)(2)−(1)(−5)=−2+5=3\hat i: (-1)(2)-(1)(-5)=-2+5=3

j^:−[(2)(2)−(1)(3)]=−(4−3)=−1\hat j: -[(2)(2)-(1)(3)]=-(4-3)=-1

k^:(2)(−5)−(−1)(3)=−10+3=−7\hat k: (2)(-5)-(-1)(3)=-10+3=-7

b⃗1×b⃗2=3i^−j^−7k^\vec b_1\times\vec b_2=3\hat i-\hat j-7\hat k

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