JEE Mains · Maths · STD 12 - 11. three dimension geometry
If the angle between the lines, \(\frac{x}{2} = \frac{y}{2} = \frac{z}{1}\) and \(\frac{{5 - x}}{{ - 2}} = \frac{{7y - 14}}{p} = \frac{{z - 3}}{4}\) is \({\cos ^{ - 1}}\,\left( {\frac{2}{3}} \right),\) then \(p\) is equal to
- A \( - \frac{7}{4}\)
- B \(\frac{2}{7}\)
- C \( - \frac{4}{7}\)
- D \(\frac{7}{2}\)
Answer & Solution
Correct Answer
(D) \(\frac{7}{2}\)
Step-by-step Solution
Detailed explanation
Let \(\theta\) be the angle between the two lines Here direction cosines of \(\frac{x}{2}=\frac{y}{2}=\frac{z}{1}\) are \(2,2,1\) Also second line can be written as: \(\frac{{\frac{{x - 5}}{2} = \frac{{y - 2}}{P} = \frac{{ - 3}}{4}}}{7}\) \(\therefore \) its direction cosines…
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