← 2017 Paper 1

UPSC 2017 Maths Optional Paper 1 Q2c — Step-by-Step Solution

10 marks · Section A

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

Question

Show that the plane 2x−2y+z+12=02x-2y+z+12=0 touches the sphere x2+y2+z2−2x−4y+2z−3=0x^2+y^2+z^2-2x-4y+2z-3=0. Find the point of contact.

Technique

Tangency   ⟺  \iff distance(centre, plane) == radius; contact point == centre −s∣n∣2n-\dfrac{s}{|\mathbf n|^2}\mathbf n where ss is the plane expression evaluated at the centre.

Solution

Step 1 — Centre and radius of the sphere

Complete squares in x2+y2+z2−2x−4y+2z−3=0x^2+y^2+z^2-2x-4y+2z-3=0:

(x−1)2+(y−2)2+(z+1)2=3+1+4+1=9.(x-1)^2+(y-2)^2+(z+1)^2=3+1+4+1=9.

Centre C=(1,2,−1)C=(1,2,-1), radius R=3R=3.

Step 2 — Distance from the centre to the plane

For the plane 2x−2y+z+12=02x-2y+z+12=0,  n=(2,−2,1)\ \mathbf n=(2,-2,1), ∣n∣=4+4+1=3|\mathbf n|=\sqrt{4+4+1}=3:

d=∣2(1)−2(2)+(−1)+12∣3=∣2−4−1+12∣3=93=3.d=\frac{|2(1)-2(2)+(-1)+12|}{3}=\frac{|2-4-1+12|}{3}=\frac{9}{3}=3.

Since d=3=Rd=3=R, the plane is tangent to the sphere.

Step 3 — Point of contact (foot of the perpendicular from CC)

The contact point is CC moved along n\mathbf n by the signed distance. With s=2(1)−2(2)+(−1)+12=9s=2(1)-2(2)+(-1)+12=9 and ∣n∣2=9|\mathbf n|^2=9:

contact=C−s∣n∣2 n=(1,2,−1)−99(2,−2,1)=(1−2, 2+2, −1−1).\text{contact}=C-\frac{s}{|\mathbf n|^2}\,\mathbf n=(1,2,-1)-\frac{9}{9}(2,-2,1)=(1-2,\ 2+2,\ -1-1).

Answer

  The plane touches the sphere; point of contact =(−1, 4, −2).  \boxed{\;\text{The plane touches the sphere; point of contact }=(-1,\,4,\,-2).\;}
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