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Q.Find the shortest distance between the lines: 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 plane passing through the points (1,1,−1)(1, 1, -1), (6,4,−5)(6, 4, -5) and (−4,−2,3)(-4, -2, 3).

Haryana BsehBSEH Intermediate Board 2019Subjective· 6mImportance★★★★★
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For two skew lines r⃗=a1⃗+λb1⃗\vec{r}=\vec{a_1}+\lambda\vec{b_1} and r⃗=a2⃗+μb2⃗\vec{r}=\vec{a_2}+\mu\vec{b_2}, shortest distance =∣(a2⃗−a1⃗)⋅(b1⃗×b2⃗)∣b1⃗×b2⃗∣∣= \left|\dfrac{(\vec{a_2}-\vec{a_1})\cdot(\vec{b_1}\times\vec{b_2})}{|\vec{b_1}\times\vec{b_2}|}\right|.

Line 1: a1⃗=i^+j^\vec{a_1}=\hat{i}+\hat{j}, b1⃗=2i^−j^+k^\vec{b_1}=2\hat{i}-\hat{j}+\hat{k}

Line 2: a2⃗=2i^+j^−k^\vec{a_2}=2\hat{i}+\hat{j}-\hat{k}, b2⃗=3i^−5j^+2k^\vec{b_2}=3\hat{i}-5\hat{j}+2\hat{k}

a2⃗−a1⃗=i^+0j^−k^\vec{a_2}-\vec{a_1} = \hat{i}+0\hat{j}-\hat{k}

b1⃗×b2⃗=∣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^\hat{i}: (−1)(2)−(1)(−5)=−2+5=3(-1)(2)-(1)(-5) = -2+5=3

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

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

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