AP EAMCET · Maths · Three Dimensional Geometry
The direction cosines of the line which is perpendicular to the lines with direction cosines proportional to \(\langle 1,-2,-2\rangle\) and \(\langle 0,2,1\rangle\) is given by
- A \(\left\langle\frac{2}{3}, \frac{1}{3}, \frac{2}{3}\right\rangle\)
- B \(\left\langle\frac{-2}{3}, \frac{-1}{3}, \frac{-2}{3}\right\rangle\)
- C \(\left\langle\frac{2}{3}, \frac{1}{3}, \frac{-2}{3}\right\rangle\)
- D \(\left\langle\frac{2}{3}, \frac{-1}{3}, \frac{2}{3}\right\rangle\)
Answer & Solution
Correct Answer
(D) \(\left\langle\frac{2}{3}, \frac{-1}{3}, \frac{2}{3}\right\rangle\)
Step-by-step Solution
Detailed explanation
Let \(l, m, n\) be the direction cosines of the required line. Since, it is perpendicular to the lines whose direction cosines are proportional to \(1,-2,-2\) and \(0,2,1\) respectively Thus, \(l-2 m-2 n=0\) and \(2 m+n=0\) On solving \(\frac{1}{2}=\frac{m}{-1}=\frac{n}{2}\)…
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