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UPSC 2021 Maths Optional Paper 1 Q3b — Step-by-Step Solution

15 marks · Section A

Sphere · Analytic Geometry · asked 17× in 13 yrs · Read the full method →

Question

A sphere of constant radius $r$ passes through $O$ and cuts axes at $A,B,C$. Find the locus of foot of perpendicular from $O$ to plane $ABC$.

Technique

Sphere through $O$ + axes intercepts $(\alpha,\beta,\gamma)$; centre at $(\alpha/2,\beta/2,\gamma/2)$ with radius $\sqrt{\alpha^2+\beta^2+\gamma^2}/2=r$; foot of perpendicular from $O$ to plane $x/\alpha+y/\beta+z/\gamma=1$; eliminate $\alpha,\beta,\gamma$.

Solution

Setup. Sphere passes through $O$ and cuts axes at $A=(\alpha,0,0)$, $B=(0,\beta,0)$, $C=(0,0,\gamma)$. The sphere through these four points: from Q1(e) of 2022 P1, the sphere is $x^2+y^2+z^2-\alpha x-\beta y-\gamma z=0$, centre $(\alpha/2,\beta/2,\gamma/2)$, radius $\sqrt{\alpha^2+\beta^2+\gamma^2}/2$.

Given radius $=r$: $\sqrt{\alpha^2+\beta^2+\gamma^2}/2=r$, i.e., $\alpha^2+\beta^2+\gamma^2=4r^2.\qquad(*)$

Step 1 — Equation of plane $ABC$

Intercept form: $\dfrac{x}{\alpha}+\dfrac{y}{\beta}+\dfrac{z}{\gamma}=1$.

Step 2 — Foot of perpendicular from $O$ to plane $ABC$

The foot is along the normal $\vec n=(1/\alpha,1/\beta,1/\gamma)$ (or scaled $(\beta\gamma,\alpha\gamma,\alpha\beta)$).

Foot $P=t\vec n$ for some $t$. Substituting into the plane equation:

$\dfrac{t/\alpha}{\alpha}+\dfrac{t/\beta}{\beta}+\dfrac{t/\gamma}{\gamma}=1$,

$t\!\left(\dfrac{1}{\alpha^2}+\dfrac{1}{\beta^2}+\dfrac{1}{\gamma^2}\right)=1$,

$t=\dfrac{1}{1/\alpha^2+1/\beta^2+1/\gamma^2}$.

So $P=\!\left(\dfrac{t}{\alpha},\dfrac{t}{\beta},\dfrac{t}{\gamma}\right)$.

Let $(X,Y,Z)$ be $P$. Then $X=t/\alpha$, $Y=t/\beta$, $Z=t/\gamma$, i.e., $\alpha=t/X$, $\beta=t/Y$, $\gamma=t/Z$.

Step 3 — Apply constraint $(*)$

$\alpha^2+\beta^2+\gamma^2=4r^2$: $t^2/X^2+t^2/Y^2+t^2/Z^2=4r^2$, $t^2\!\left(\dfrac{1}{X^2}+\dfrac{1}{Y^2}+\dfrac{1}{Z^2}\right)=4r^2.\qquad(1)$

Also from $t=1/(1/\alpha^2+\ldots)$: $1/t=1/\alpha^2+1/\beta^2+1/\gamma^2=X^2/t^2+Y^2/t^2+Z^2/t^2=(X^2+Y^2+Z^2)/t^2$.

So $t^2/t=X^2+Y^2+Z^2$, $t=X^2+Y^2+Z^2$.

Substitute into $(1)$: $(X^2+Y^2+Z^2)^2\!\left(\dfrac{1}{X^2}+\dfrac{1}{Y^2}+\dfrac{1}{Z^2}\right)=4r^2.$

Replace $(X,Y,Z)\to(x,y,z)$:

Answer