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Q.Find the shortest distance between the lines l1l_1 and l2l_2 whose vector equations are r⃗=i^+j^+λ(2i^−j^+k^)\vec{r} = \hat{i} + \hat{j} + \lambda(2\hat{i} - \hat{j} + \hat{k}) and r⃗=2i^+j^−k^+μ(3i^−5j^+2k^)\vec{r} = 2\hat{i} + \hat{j} - \hat{k} + \mu(3\hat{i} - 5\hat{j} + 2\hat{k}). OR Find the equation of the line in vector and in Cartesian form that passes through the point with position vector 2i^−j^+4k^2\hat{i} - \hat{j} + 4\hat{k} and is in the direction i^+2j^−k^\hat{i} + 2\hat{j} - \hat{k}.

Rajasthan RbseRajasthan Board Senior Secondary Examination 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∣\dfrac{|(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)|}{|\vec b_1\times\vec b_2|}.

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

a⃗2=2i^+j^−k^\vec a_2 = 2\hat i+\hat j-\hat k, b⃗2=3i^−5j^+2k^\vec b_2 = 3\hat i-5\hat j+2\hat k

a⃗2−a⃗1=i^+0j^−k^\vec a_2 - \vec a_1 = \hat i + 0\hat j - \hat k

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

=i^(−2+5)−j^(4−3)+k^(−10+3)=3i^−j^−7k^= \hat i(-2+5) - \hat j(4-3) + \hat k(-10+3) = 3\hat i - \hat j - 7\hat k

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