MHT CET · Maths · Three Dimensional Geometry
The direction cosines of the line, which is perpendicular to the lines with direction rations \(-1,2,2\) and \(0,2,1\), are respectively
- A \(\frac{1}{3}, \frac{-2}{3}, \frac{-2}{3}\)
- B \(\frac{2}{3}, \frac{-1}{3}, \frac{2}{3}\)
- C \(\frac{-1}{3}, \frac{2}{3}, \frac{2}{3}\)
- D \(\frac{1}{3}, \frac{2}{3}, \frac{2}{3}\)
Answer & Solution
Correct Answer
(B) \(\frac{2}{3}, \frac{-1}{3}, \frac{2}{3}\)
Step-by-step Solution
Detailed explanation
The d.r.s can be obtained by
\(\frac{a}{2 \times 1-2 \times 2}=\frac{b}{0 \times 2-(-1) \times 1}=\) \(\frac{c}{(-1) \times 2-0 \times 2}\)
\(\Rightarrow\) \(< 2,-1,2>\)
\(\Rightarrow \text { d.c's are } < \frac{2}{3}, \frac{-1}{3}, \frac{2}{3}>\)
\(\frac{a}{2 \times 1-2 \times 2}=\frac{b}{0 \times 2-(-1) \times 1}=\) \(\frac{c}{(-1) \times 2-0 \times 2}\)
\(\Rightarrow\) \(< 2,-1,2>\)
\(\Rightarrow \text { d.c's are } < \frac{2}{3}, \frac{-1}{3}, \frac{2}{3}>\)
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