Let P(x, y, z) be a point in the first octant | vedclass

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(A) Given $P(x, y, z)$ in the first octant,$OP = \gamma = \sqrt{x^2+y^2+z^2}$.The projection of $P$ on the $xy$-plane is $Q(x, y, 0)$.Let $OQ = r = \s

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$\gamma \sqrt{1-\sin^2 \phi \cos^2 	heta}$

Vedclass · Answer

(A) Given $P(x, y, z)$ in the first octant,$OP = \gamma = \sqrt{x^2+y^2+z^2}$.The projection of $P$ on the $xy$-plane is $Q(x, y, 0)$.Let $OQ = r = \sqrt{x^2+y^2}$.Given the angle between $OQ$ and the positive $x$-axis is $	heta$,we have $x = r \cos 	heta$ and $y = r \sin 	heta$.Given the angle between $OP$ and the positive $z$-axis is $\phi$,we have $z = OP \cos \phi = \gamma \cos \phi$.Also,$r = OP \sin \phi = \gamma \sin \phi$.Thus,$x = \gamma \sin \phi \cos 	heta$,$y = \gamma \sin \phi \sin 	heta$,and $z = \gamma \cos \phi$.The distance of $P(x, y, z)$ from the $x$-axis is $\sqrt{y^2+z^2}$.Substituting the values: $\sqrt{(\gamma \sin \phi \sin 	heta)^2 + (\gamma \cos \phi)^2} = \gamma \sqrt{\sin^2 \phi \sin^2 	heta + \cos^2 \phi}$.Using $\cos^2 \phi = 1 - \sin^2 \phi$,we get $\gamma \sqrt{\sin^2 \phi \sin^2 	heta + 1 - \sin^2 \phi} = \gamma \sqrt{1 - \sin^2 \phi (1 - \sin^2 	heta)} = \gamma \sqrt{1 - \sin^2 \phi \cos^2 	heta}$.

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