Let P(x, y, z) be a point in the first octant, whose projection in the x y -plane is the point Q . Let O P=\gamma ; the angle between O Q and the positive x -axis be \theta ; and the angle between O P and the positive z -axis be \phi , where O is the origin. Then the distance of P from the x -axis is

\gamma \sqrt1+\cos ^2 \theta \sin ^2 \phi

\gamma \sqrt1+\cos ^2 \phi \sin ^2 \theta

\gamma \sqrt1-\sin ^2 \theta \cos ^2 \phi

3D Geometry PYQ — JEE Main 2024 · 8 Apr · Shift 1 · Q42 with Solution

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Correct answer: A

Interpret the given angles and length

Let $P(x,y,z)$ be in the first octant.

$OP=\gamma$
$\angle( OP, \text{positive } z\text{-axis})=\phi$
The projection of $P$ on the $xy$ -plane is $Q$ , and the angle between $OQ$ and the positive $x$ -axis is $\theta$ .

So in spherical-coordinate style:

The component along the $z$ -axis is $z=\gamma \cos\phi$
The length of the projection on the $xy$ -plane is $OQ=\gamma \sin\phi$

Since $OQ$ makes angle $\theta$ with the positive $x$ -axis, $x=(OQ)\cos\theta=\gamma \sin\phi\cos\theta$ $y=(OQ)\sin\theta=\gamma \sin\phi\sin\theta$

Hence, $P\big(\gamma\sin\phi\cos\theta,\ \gamma\sin\phi\sin\theta,\ \gamma\cos\phi\big).$

Distance of $P$ from the $x$ -axis

The distance of a point $(x,y,z)$ from the $x$ -axis is $\sqrt{y^2+z^2}.$

Therefore, $d=\sqrt{y^2+z^2}$ $=\sqrt{\gamma^2\sin^2\phi\sin^2\theta+\gamma^2\cos^2\phi}$ $=\gamma\sqrt{\sin^2\phi\sin^2\theta+\cos^2\phi}.$

Now simplify:

=1-\sin^2\phi+\sin^2\phi\sin^2\theta$$ $$=1-\sin^2\phi(1-\sin^2\theta)$$ $$=1-\sin^2\phi\cos^2\theta.$$ So, $$d=\gamma\sqrt{1-\sin^2\phi\cos^2\theta}.$$ 3. **Match with the options** This is exactly **Option A**: $$\boxed{\gamma \sqrt{1-\sin ^2 \phi \cos ^2 \theta}}$$ 4. **Comparison with stored correct answer** Stored correct answer: **A** Our derived answer: **A** So they agree.

More from 3D Geometry

View written solutionFree

Correct answer: A

Interpret the given angles and length

Let $P(x,y,z)$ be in the first octant.

$OP=\gamma$
$\angle( OP, \text{positive } z\text{-axis})=\phi$
The projection of $P$ on the $xy$ -plane is $Q$ , and the angle between $OQ$ and the positive $x$ -axis is $\theta$ .

So in spherical-coordinate style:

The component along the $z$ -axis is $z=\gamma \cos\phi$
The length of the projection on the $xy$ -plane is $OQ=\gamma \sin\phi$

Since $OQ$ makes angle $\theta$ with the positive $x$ -axis, $x=(OQ)\cos\theta=\gamma \sin\phi\cos\theta$ $y=(OQ)\sin\theta=\gamma \sin\phi\sin\theta$

Hence, $P\big(\gamma\sin\phi\cos\theta,\ \gamma\sin\phi\sin\theta,\ \gamma\cos\phi\big).$

Distance of $P$ from the $x$ -axis

The distance of a point $(x,y,z)$ from the $x$ -axis is $\sqrt{y^2+z^2}.$

Therefore, $d=\sqrt{y^2+z^2}$ $=\sqrt{\gamma^2\sin^2\phi\sin^2\theta+\gamma^2\cos^2\phi}$ $=\gamma\sqrt{\sin^2\phi\sin^2\theta+\cos^2\phi}.$

Now simplify:

=1-\sin^2\phi+\sin^2\phi\sin^2\theta$$ $$=1-\sin^2\phi(1-\sin^2\theta)$$ $$=1-\sin^2\phi\cos^2\theta.$$ So, $$d=\gamma\sqrt{1-\sin^2\phi\cos^2\theta}.$$ 3. **Match with the options** This is exactly **Option A**: $$\boxed{\gamma \sqrt{1-\sin ^2 \phi \cos ^2 \theta}}$$ 4. **Comparison with stored correct answer** Stored correct answer: **A** Our derived answer: **A** So they agree.

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