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UPSC Maths 2013 Paper 1 Q4a — Solution

15 marks · Section A

Question

Show that three mutually perpendicular tangent lines can be drawn to the sphere $x^{2}+y^{2}+z^{2}=r^{2}$ from any point on the sphere $2(x^{2}+y^{2}+z^{2})=3r^{2}$.

Technique

Tangent-cone semi-angle $\sin\alpha=r/|OP|$; orthonormality $\Rightarrow$ squared-dot-products sum to 1; the constraint $3\cos^{2}\alpha=1$ pins down the locus.

Solution

Strategy. A line through external point $P$ is tangent to a sphere centred at $O$ with radius $r$ iff it makes angle $\alpha$ with $\overrightarrow{OP}$ where $\sin\alpha=r/|\overrightarrow{OP}|$. Three mutually perpendicular tangent lines all on this tangent cone (semi-angle $\alpha$, axis $\overrightarrow{OP}$) impose a numerical constraint on $\alpha$ — yielding the director-sphere locus.

Step 1 — Geometric setup

Let $P=(x_0,y_0,z_0)$ be on the sphere $2(x^{2}+y^{2}+z^{2})=3r^{2}$, i.e., $|\overrightarrow{OP}|^{2}=\tfrac{3r^{2}}{2}$.

Since $\tfrac{3r^{2}}{2}>r^{2}$, $P$ lies outside the unit-radius sphere $S:\,x^{2}+y^{2}+z^{2}=r^{2}$, so a tangent cone from $P$ exists. Its semi-vertex angle $\alpha$ satisfies

Hence $\sin^{2}\alpha=\tfrac{2}{3}$ and $\cos^{2}\alpha=\tfrac{1}{3}$.

Step 2 — Three mutually perpendicular lines through $P$ making a fixed angle with $\overrightarrow{OP}$

Suppose $\vec u,\vec v,\vec w$ are unit vectors forming an orthonormal frame (three mutually perpendicular unit vectors). Let $\hat n=\overrightarrow{OP}/|\overrightarrow{OP}|$ be the unit vector along $OP$.

The squared cosines of angles between $\hat n$ and each frame axis sum to $|\hat n|^{2}=1$:

(This is the standard fact that any unit vector’s components in an orthonormal basis have squared sum 1.)

For each line through $P$ in direction $\vec u$ to be tangent to $S$, the line must make angle $\alpha$ with $\hat n$ — equivalently $|\hat n\cdot\vec u|=\cos\alpha$ (i.e. $(\hat n\cdot\vec u)^{2}=\cos^{2}\alpha$). Same for $\vec v,\vec w$.

Imposing this on all three:

Substituting into $(\ast)$: $3\cos^{2}\alpha=1$, i.e. $\cos^{2}\alpha=\tfrac{1}{3}$.

Step 3 — Match the constraint

From Step 1, $\cos^{2}\alpha=\tfrac{1}{3}$ for any $P$ on $2(x^{2}+y^{2}+z^{2})=3r^{2}$. The geometric requirement from Step 2 is exactly this — so such an orthonormal frame of tangent directions exists.

Step 4 — Construct the frame explicitly

Given $\cos^{2}\alpha=\tfrac{1}{3}$, an orthonormal frame $\{\vec u,\vec v,\vec w\}$ with $(\hat n\cdot\vec u)^{2}=(\hat n\cdot\vec v)^{2}=(\hat n\cdot\vec w)^{2}=\tfrac{1}{3}$ can be built as follows.

Rotate coordinates so $\hat n=\bigl(\tfrac{1}{\sqrt 3},\tfrac{1}{\sqrt 3},\tfrac{1}{\sqrt 3}\bigr)^{T}$ (in a rotated frame where $\hat n$ has equal components). Then take $\vec u=\hat e_1,\;\vec v=\hat e_2,\;\vec w=\hat e_3$ — the standard basis in this rotated frame. Each satisfies $\hat n\cdot\vec u=\tfrac{1}{\sqrt 3}$, so $(\hat n\cdot\vec u)^{2}=\tfrac{1}{3}$ ✓.

The lines through $P$ in these three directions are mutually perpendicular (by orthonormality) and each tangent to $S$ (by the angle calculation in Step 2).

Step 5 — Conclude

Answer