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UPSC 2015 Maths Optional Paper 1 Q2d — Step-by-Step Solution

13 marks · Section A

Cone · Analytic Geometry · asked 15× in 14 yrs · Read the full method →

Question

If 6x=3y=2z6x=3y=2z represents one of the three mutually perpendicular generators of the cone 5yz−8zx−3xy=05yz-8zx-3xy=0, then obtain the equations of the other two generators.

Technique

Use the perpendicularity-of-three-mutually-perpendicular-generators standard recipe: one constraint from cone equation + one from perpendicularity to the given generator gives a quadratic in m/nm/n; the two roots are the directions of the other two generators; their mutual perpendicularity is then automatic (standard result for the three principal generators).

Solution

Setup. The given generator has direction ratios satisfying 6x=3y=2z6x=3y=2z, i.e. x1=y2=z3\dfrac{x}{1}=\dfrac{y}{2}=\dfrac{z}{3}. Direction ℓ⃗1=(1,2,3)\vec\ell_1=(1,2,3).

A line through the origin r⃗=t(l,m,n)\vec r=t(l,m,n) lies on the cone 5yz−8zx−3xy=05yz-8zx-3xy=0 iff its direction (l,m,n)(l,m,n) satisfies

5mn−8nl−3lm=0.(⋆)5mn-8nl-3lm=0.\qquad(\star)

Check (1,2,3)(1,2,3): 5(6)−8(3)−3(2)=30−24−6=05(6)-8(3)-3(2)=30-24-6=0 ✓.

Strategy. We seek ℓ⃗2=(l2,m2,n2)\vec\ell_2=(l_2,m_2,n_2) and ℓ⃗3=(l3,m3,n3)\vec\ell_3=(l_3,m_3,n_3), both on the cone, both perpendicular to ℓ⃗1=(1,2,3)\vec\ell_1=(1,2,3), and perpendicular to each other.

Step 1 — Perpendicularity to ℓ⃗1\vec\ell_1

l+2m+3n=0  ⟹  l=−2m−3n.(1)l+2m+3n=0\;\Longrightarrow\;l=-2m-3n.\qquad(1)

Step 2 — Substitute into the cone equation (⋆)(\star)

5mn−8nl−3lm=05mn-8nl-3lm=0.

l=−2m−3nl=-2m-3n gives nl=−2mn−3n2nl=-2mn-3n^2 and lm=−2m2−3mnlm=-2m^2-3mn.

5mn−8(−2mn−3n2)−3(−2m2−3mn)=05mn-8(-2mn-3n^2)-3(-2m^2-3mn)=0 5mn+16mn+24n2+6m2+9mn=05mn+16mn+24n^2+6m^2+9mn=0 6m2+30mn+24n2=06m^2+30mn+24n^2=0 m2+5mn+4n2=0m^2+5mn+4n^2=0 (m+n)(m+4n)=0.(m+n)(m+4n)=0.

So either m=−nm=-n or m=−4nm=-4n.

Step 3 — Case m=−nm=-n

From (1): l=−2(−n)−3n=2n−3n=−nl=-2(-n)-3n=2n-3n=-n. Direction (l,m,n)=(−n,−n,n)=n(−1,−1,1)(l,m,n)=(-n,-n,n)=n(-1,-1,1), i.e. ℓ⃗2=(−1,−1,1)\vec\ell_2=(-1,-1,1) (or equivalently (1,1,−1)(1,1,-1)).

Step 4 — Case m=−4nm=-4n

From (1): l=−2(−4n)−3n=8n−3n=5nl=-2(-4n)-3n=8n-3n=5n. Direction (l,m,n)=(5n,−4n,n)=n(5,−4,1)(l,m,n)=(5n,-4n,n)=n(5,-4,1), i.e. ℓ⃗3=(5,−4,1)\vec\ell_3=(5,-4,1).

Step 5 — Verify all three are mutually perpendicular

Verify each lies on the cone (⋆)(\star):

Conclusion

The other two generators are the lines through the origin with direction ratios (−1,−1,1)(-1,-1,1) and (5,−4,1)(5,-4,1):

Answer

  x−1=y−1=z1andx5=y−4=z1.  \boxed{\;\dfrac{x}{-1}=\dfrac{y}{-1}=\dfrac{z}{1}\quad\text{and}\quad\dfrac{x}{5}=\dfrac{y}{-4}=\dfrac{z}{1}.\;}
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