Exams Joint Entrance Examination Mathematics 3D Geometry Question 60 Question 60 of 521 Medium MCQ
Let P ( x , y , z ) P(x, y, z) P ( x , y , z ) be a point in the first octant, whose projection in the x y x y x y -plane is the point Q Q Q . Let \[O P=\gamma\]; the angle between O Q O Q O Q and the positive x x x -axis be \[\theta\]; and the angle between O P O P O P and the posi...
Solutions · 1 Post your solution Official solution
O P → ← = x i ^ + y j ^ + z k ^ O Q → ← = x i ^ + y j ^ ∣ O P ∣ = γ = x 2 + y 2 + z 2 cos θ = x x 2 + y 2 ⇒ cos 2 θ = x 2 γ 2 − z 2 = x 2 γ 2 − γ 2 cos 2 ϕ cos ϕ = z x 2 + y 2 + z 2 = z γ P x = y 2 + z 2 d = γ 2 − x 2 ⇒ x 2 = γ 2 sin 2 ϕ cos 2 θ ⇒ d = γ 2 − γ 2 sin 2 ϕ cos 2 θ = γ 1 − sin 2 ϕ cos 2 θ \begin{gathered} \overset{\to \leftarrow }{OP}=x\hat{i}+y\hat{j}+z\hat{k} \\ \overset{\to \leftarrow }{OQ}=x\hat{i}+y\hat{j} \\ |OP|=\gamma =\sqrt{{x}^{2}+{y}^{2}+{z}^{2}} \\ \cos \theta =\frac{x}{\sqrt{{x}^{2}+{y}^{2}}}\Rightarrow {\cos }^{2}\theta =\frac{{x}^{2}}{{\gamma }^{2}-{z}^{2}}=\frac{{x}^{2}}{{\gamma }^{2}-{\gamma }^{2}{\cos }^{2}\phi } \\ \cos \phi =\frac{z}{\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}}=\frac{z}{\gamma } \\ Px=\sqrt{{y}^{2}+{z}^{2}} \\ d=\sqrt{{\gamma }^{2}-{x}^{2}} \\ \Rightarrow {x}^{2}={\gamma }^{2}{\sin }^{2}\phi {\cos }^{2}\theta \\ \Rightarrow d=\sqrt{{\gamma }^{2}-{\gamma }^{2}{\sin }^{2}\phi {\cos }^{2}\theta } \\ =\gamma \sqrt{1-{\sin }^{2}\phi {\cos }^{2}\theta } \end{gathered} O P →← = x i ^ + y j ^ + z k ^ O Q →← = x i ^ + y j ^ ∣ O P ∣ = γ = x 2 + y 2 + z 2 cos θ = x 2 + y 2 x ⇒ cos 2 θ = γ 2 − z 2 x 2 = γ 2 − γ 2 cos 2 ϕ x 2 cos ϕ = x 2 + y 2 + z 2 z = γ z P x = y 2 + z 2 d = γ 2 − x 2 ⇒ x 2 = γ 2 sin 2 ϕ cos 2 θ ⇒ d = γ 2 − γ 2 sin 2 ϕ cos 2 θ = γ 1 − sin 2 ϕ cos 2 θ Answer the question to unlock