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Q.Find the shortest distance between the lines l1l_1 and l2l_2: l1:r⃗=i^+j^+λ(2i^−j^+k^)l_1 : \vec{r} = \hat{i}+\hat{j}+\lambda(2\hat{i}-\hat{j}+\hat{k}) and l2:r⃗=2i^+j^−k^+μ(3i^−5j^+2k^)l_2 : \vec{r} = 2\hat{i}+\hat{j}-\hat{k}+\mu(3\hat{i}-5\hat{j}+2\hat{k}) OR Find the image of the point (1,6,3)(1, 6, 3) in the line x1=y−12=z−23\frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3}.

Haryana BsehBSEH Intermediate Board 2025Subjective· 5mImportance★★★★★
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For skew lines, 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|.

Here a⃗1=i^+j^\vec a_1=\hat i+\hat j, b⃗1=2i^−j^+k^\vec b_1=2\hat i-\hat j+\hat k (line l1l_1), and 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 (line l2l_2).

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