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CBSE Class 12 Math 2018 Solved Paper

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Question : 21 of 29
Marks: +1, -0
Find the shortest distance between the lines
r⃗\vec{r} = (4i^−j^)(\hat{4i} - \hat{j}) + λ (i^+2i^−3k^)(\hat{i} + \hat{2i} - \hat{3k}) and r⃗\vec{r} = (i^−j^+2k^)(\hat{i} - \hat{j} + \hat{2k}) + µ (2i^+4j^−5k^)(\hat{2i} + \hat{4j} - \hat{5k})
Solution:  
The shortest distance is given by,
= ∣(A2−A1)⋅(B1×B2)∣B1×B2∣∣\left| \frac{(A_2 - A_1) \cdot (B_1 \times B_2)}{|B_1 \times B_2|} \right|
A2−A1A_2 - A_1 = (i^−j^−2k^)(\hat{i} - \hat{j} - \hat{2k}) - (4i^−j^)(\hat{4i} - \hat{j}) = - 3i^−2k^\hat{3i} - \hat{2k}
B1×B2B_1 \times B_2 = ∣i^j^k^12−324−5∣\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 2 & 4 & -5 \end{vmatrix} = 2i^−j^\hat{2i} - \hat{j}
(A2−A1)(A_2 - A_1) . (B1×B2)(B_1 \times B_2) = (−3i^−2k^)(-\hat{3i} - \hat{2k}) . (2i^−j^)(\hat{2i} - \hat{j}) = - 6
∣B1×B2∣|B_1 \times B_2| = 22+12\sqrt{2^2 + 1^2} = 5\sqrt{5}
so shortest distance between two lines = ∣−65∣\left| \frac{-6}{\sqrt{5}} \right| = 65\frac{6}{\sqrt{5}} units
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