AP EAMCET · Maths · Three Dimensional Geometry
If \(\bar{i}+\bar{j}-\bar{k},-\bar{i}+2 \bar{j}+\bar{k}, \bar{j}+2 \bar{k}, 2 \bar{i}-\bar{j}+2 \bar{k}\) are the position vectors of four points \(A, B, C, D\) respectively, then the shortest distance between the lines \(A B\) and \(C D\) is
- A \(\frac{1}{6}\)
- B \(\frac{7}{3}\)
- C \(\frac{1}{3}\)
- D \(\frac{7}{6}\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{3}\)
Step-by-step Solution
Detailed explanation
\(\vec{b_1} = (-\bar{i}+2\bar{j}+\bar{k}) - (\bar{i}+\bar{j}-\bar{k}) = -2\bar{i}+\bar{j}+2\bar{k}\) \(\vec{b_2} = (2\bar{i}-\bar{j}+2\bar{k}) - (\bar{j}+2\bar{k}) = 2\bar{i}-2\bar{j}\) \(\vec{a_2}-\vec{a_1} = (\bar{j}+2\bar{k}) - (\bar{i}+\bar{j}-\bar{k}) = -\bar{i}+3\bar{k}\)…
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