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UPSC 2015 Maths Optional Paper 1 Q3b — Step-by-Step Solution

13 marks · Section A

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

Question

Which point of the sphere $x^2+y^2+z^2=1$ is at the maximum distance from the point $(2,1,3)$?

Technique

Geometric: farthest point on a sphere from an external point lies on the line joining the external point to the centre, on the opposite side of the centre. Distance = $|PC|+r$.

Solution

Geometric fact. Let $S$ be a sphere with centre $C$ and radius $r$, and $P$ an external point. The line $PC$ meets the sphere in two antipodal points; the one on the far side from $P$ is the farthest point on $S$ from $P$, and the one on the near side is the nearest. So this is geometry, not Lagrange multipliers.

Step 1 — Identify centre and external point

Centre $C=(0,0,0)$, radius $r=1$. External point $P=(2,1,3)$.

$|PC|=\sqrt{4+1+9}=\sqrt{14}$. Since $\sqrt{14}>1$, $P$ is outside.

Step 2 — Unit vector from $P$ to $C$

$\vec{PC}=C-P=(-2,-1,-3)$. Unit: $\hat u=\dfrac{1}{\sqrt{14}}(-2,-1,-3)$.

Step 3 — Farthest point on sphere from $P$

Start at $C$ and step a distance $r=1$ in the direction away from $P$. Direction away from $P$ at the centre is $\hat v=-\hat u=\dfrac{1}{\sqrt{14}}(2,1,3)$.

Farthest point:

Equivalently, the line through $P$ and $C$ meets the sphere where $\hat v$ exits, i.e. at $C+r\hat v$.

Step 4 — Maximum distance (sanity)

$|PQ|=|PC|+r=\sqrt{14}+1$. (Farthest along the line through $P$ and $C$.)

Verify: $|Q-P|=\bigl|\tfrac{1}{\sqrt{14}}(2,1,3)-(2,1,3)\bigr|=\bigl|(2,1,3)\bigl(\tfrac{1}{\sqrt{14}}-1\bigr)\bigr|=\sqrt{14}\cdot|1-1/\sqrt{14}|\cdot$ — careful with sign. Since $\sqrt{14}>1$, $1-1/\sqrt{14}>0$, so $|PQ|=\sqrt{14}(1-1/\sqrt{14})=\sqrt{14}-1$.

Wait — that contradicts the geometric claim. Let me recheck Step 3.

Recheck. $P=(2,1,3)$, $C=(0,0,0)$. The line $PC$ parametrised as $\vec r(t)=P+t(C-P)=(2,1,3)+t(-2,-1,-3)$. At $t=0$: $P$. At $t=1$: $C$. The line continues for $t>1$ to points beyond $C$ from $P$. The sphere is $\{|\vec r|=1\}$. Intersection: $|(2-2t,1-t,3-3t)|^2=1$, i.e. $(1-t)^2(4+1+9)=1$, $(1-t)^2\cdot 14=1$, $1-t=\pm 1/\sqrt{14}$, $t=1\mp 1/\sqrt{14}$.

The two intersections:

• $t_1=1-1/\sqrt{14}$ (near side, between $P$ and $C$): $\vec r=(2-2t_1,1-t_1,3-3t_1)=(2(1-t_1),1-t_1,3(1-t_1))=(1-t_1)(2,1,3)=(1/\sqrt{14})(2,1,3)$.
• $t_2=1+1/\sqrt{14}$ (far side, beyond $C$): $\vec r=(1-t_2)(2,1,3)=(-1/\sqrt{14})(2,1,3)$.

Distance from $P$ to each:

• To $t_1$: $|PC|\cdot t_1=\sqrt{14}(1-1/\sqrt{14})=\sqrt{14}-1$.
• To $t_2$: $|PC|\cdot t_2=\sqrt{14}(1+1/\sqrt{14})=\sqrt{14}+1$.

So the farther point is at $t_2$, which is at $\dfrac{-1}{\sqrt{14}}(2,1,3)=\left(\dfrac{-2}{\sqrt{14}},\dfrac{-1}{\sqrt{14}},\dfrac{-3}{\sqrt{14}}\right)$.

Corrected Step 3. The farthest point from $P$ on the sphere is on the opposite side of the centre from $P$ — direction from $C$ pointing away from $P$ is $-\hat u_{P\to C}$, but $\hat u_{P\to C}=\hat{PC}/|PC|=(-2,-1,-3)/\sqrt{14}$ points from $P$ to $C$. The direction from $C$ away from $P$ is the same direction (continuing past $C$), i.e. $\hat{PC}/|PC|=(-2,-1,-3)/\sqrt{14}$.

Farthest point: $Q=C+r\cdot\hat{PC}/|PC|=(0,0,0)+1\cdot\dfrac{(-2,-1,-3)}{\sqrt{14}}=\dfrac{1}{\sqrt{14}}(-2,-1,-3)$.

This matches $t_2$ above.

Verification

$|Q-P|=\bigl|\dfrac{(-2,-1,-3)}{\sqrt{14}}-(2,1,3)\bigr|=\bigl|(2,1,3)\bigl(-\dfrac{1}{\sqrt{14}}-1\bigr)\bigr|=\sqrt{14}\cdot\bigl(1+\dfrac{1}{\sqrt{14}}\bigr)=\sqrt{14}+1$ ✓.

$|Q|^2=(4+1+9)/14=14/14=1$, so $Q$ is on the sphere ✓.

Conclusion

Answer