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UPSC 2022 Maths Optional Paper 1 Q3c — Step-by-Step Solution

15 marks · Section A

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

Question

Find the equation of the sphere of smallest possible radius which touches the straight lines:

Technique

Find common perpendicular of two skew lines (parameterise both, require perpendicularity to both direction vectors); centre at midpoint; radius is half the shortest distance.

Solution

Setup. Two skew lines (we’ll verify skewness). The smallest sphere tangent to both lines has diameter equal to the shortest distance between the lines (the common perpendicular segment).

The centre lies on the common perpendicular of the two lines, at its midpoint.

Step 1 — Find direction vectors and a point on each line

$L_1$: point $P_1=(3,8,3)$, direction $\vec d_1=(3,-1,1)$.

$L_2$: point $P_2=(-3,-7,6)$, direction $\vec d_2=(-3,2,4)$.

Step 2 — Verify the lines are skew

Compute $\vec d_1\times\vec d_2$:

$\hat\imath:\;(-1)(4)-(1)(2)=-4-2=-6$. $\hat\jmath:\;-[(3)(4)-(1)(-3)]=-[12+3]=-15$. $\hat k:\;(3)(2)-(-1)(-3)=6-3=3$.

$\vec d_1\times\vec d_2=(-6,-15,3)$. $|\vec d_1\times\vec d_2|=\sqrt{36+225+9}=\sqrt{270}=3\sqrt{30}$.

Step 3 — Shortest distance between $L_1$ and $L_2$

$\vec{P_1 P_2}=P_2-P_1=(-3-3,\;-7-8,\;6-3)=(-6,-15,3)$.

Shortest distance $d=\dfrac{|\vec{P_1 P_2}\cdot(\vec d_1\times\vec d_2)|}{|\vec d_1\times\vec d_2|}$.

$\vec{P_1 P_2}\cdot(\vec d_1\times\vec d_2)=(-6)(-6)+(-15)(-15)+(3)(3)=36+225+9=270$.

$d=\dfrac{270}{3\sqrt{30}}=\dfrac{90}{\sqrt{30}}=\dfrac{90\sqrt{30}}{30}=3\sqrt{30}$.

Hmm — interesting, $\vec{P_1 P_2}$ is parallel to $\vec d_1\times\vec d_2$ (both equal $(-6,-15,3)$). So $\vec{P_1 P_2}$ is itself along the common perpendicular direction!

Step 4 — Common perpendicular foot points

The common perpendicular is the line segment connecting the closest points $A\in L_1$ and $B\in L_2$. We have $\vec{AB}=\lambda\,\vec d_1\times\vec d_2/|\vec d_1\times\vec d_2|$ for $\lambda=d$.

Parameterise: $A=P_1+s\vec d_1$, $B=P_2+t\vec d_2$. Then $\vec{AB}=B-A=(P_2-P_1)+t\vec d_2-s\vec d_1=\vec{P_1 P_2}+t\vec d_2-s\vec d_1$.

Require $\vec{AB}\cdot\vec d_1=0$ and $\vec{AB}\cdot\vec d_2=0$:

$\vec{P_1 P_2}\cdot\vec d_1+t(\vec d_2\cdot\vec d_1)-s(\vec d_1\cdot\vec d_1)=0$ $\vec{P_1 P_2}\cdot\vec d_2+t(\vec d_2\cdot\vec d_2)-s(\vec d_1\cdot\vec d_2)=0$

Compute:

• $\vec{P_1 P_2}\cdot\vec d_1=(-6)(3)+(-15)(-1)+(3)(1)=-18+15+3=0$.
• $\vec{P_1 P_2}\cdot\vec d_2=(-6)(-3)+(-15)(2)+(3)(4)=18-30+12=0$.
• $\vec d_1\cdot\vec d_1=9+1+1=11$.
• $\vec d_2\cdot\vec d_2=9+4+16=29$.
• $\vec d_1\cdot\vec d_2=(3)(-3)+(-1)(2)+(1)(4)=-9-2+4=-7$.

System:

From second: $7s=-29t\Rightarrow s=-29t/7$. Substitute into first: $7t+11(-29t/7)=7t-319t/7=(49t-319t)/7=-270t/7=0\Rightarrow t=0$.

Then $s=0$.

So $A=P_1=(3,8,3)$ and $B=P_2=(-3,-7,6)$.

Centre of sphere: midpoint of $AB$ = $(0,1/2,9/2)$.

Step 5 — Radius

Radius = half of $|\vec{AB}|=|\vec{P_1 P_2}|/2$.

$|\vec{P_1 P_2}|=\sqrt{36+225+9}=\sqrt{270}=3\sqrt{30}$.

Radius $r=\dfrac{3\sqrt{30}}{2}$.

Step 6 — Equation of sphere

Answer