JEE Mains · Maths · STD 11 - 11. introduction to three dimensional geometry
Let \(\mathrm{P}(\mathrm{x}, \mathrm{y}, \mathrm{z})\) be a point in the first octant, whose projection in the xy-plane is the point \(\mathrm{Q}\). Let \(\mathrm{OP}=\gamma\); the angle between \(OQ\) and the positive \(\mathrm{x}\)-axis be \(\theta\); and the angle between \(\mathrm{OP}\) and the positive \(\mathrm{z}\)-axis be \(\phi\), where \(\mathrm{O}\) is the origin. Then the distance of \(\mathrm{P}\) from the \(\mathrm{x}\)-axis is :
- A \(\gamma \sqrt{1-\sin ^2 \phi \cos ^2 \theta}\)
- B \(\gamma \sqrt{1+\cos ^2 \theta \sin ^2 \phi}\)
- C \(\gamma \sqrt{1-\sin ^2 \theta \cos ^2 \phi}\)
- D \(\gamma \sqrt{1+\cos ^2 \phi \sin ^2 \theta}\)
Answer & Solution
Correct Answer
(A) \(\gamma \sqrt{1-\sin ^2 \phi \cos ^2 \theta}\)
Step-by-step Solution
Detailed explanation
\( \mathrm{P}(\mathrm{x}, \mathrm{y}, \mathrm{z}), \mathrm{Q}(\mathrm{x}, \mathrm{y}, \mathrm{O}) ; \mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2=\gamma^2 \) \( \overline{\mathrm{OQ}}=\mathrm{xi}+\mathrm{y} \) \( \cos \theta=\frac{\mathrm{x}}{\sqrt{\mathrm{x}^2+\mathrm{y}^2}} \)…
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