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UPSC 2015 Maths Optional Paper 1 Q2d — Step-by-Step Solution 13 marks · Section A
Cone · Analytic Geometry · asked 14× in 13 yrs · Read the full method →
Question
If 6 x = 3 y = 2 z 6x=3y=2z 6 x = 3 y = 2 z represents one of the three mutually perpendicular generators of the cone 5 y z − 8 z x − 3 x y = 0 5yz-8zx-3xy=0 5 y z − 8 z x − 3 x y = 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 / n m/n m / 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 6 x = 3 y = 2 z 6x=3y=2z 6 x = 3 y = 2 z , i.e. x 1 = y 2 = z 3 \dfrac{x}{1}=\dfrac{y}{2}=\dfrac{z}{3} 1 x = 2 y = 3 z . Direction ℓ ⃗ 1 = ( 1 , 2 , 3 ) \vec\ell_1=(1,2,3) ℓ 1 = ( 1 , 2 , 3 ) .
A line through the origin r ⃗ = t ( l , m , n ) \vec r=t(l,m,n) r = t ( l , m , n ) lies on the cone 5 y z − 8 z x − 3 x y = 0 5yz-8zx-3xy=0 5 y z − 8 z x − 3 x y = 0 iff its direction ( l , m , n ) (l,m,n) ( l , m , n ) satisfies
5 m n − 8 n l − 3 l m = 0. ( ⋆ ) 5mn-8nl-3lm=0.\qquad(\star) 5 mn − 8 n l − 3 l m = 0. ( ⋆ )
Check ( 1 , 2 , 3 ) (1,2,3) ( 1 , 2 , 3 ) : 5 ( 6 ) − 8 ( 3 ) − 3 ( 2 ) = 30 − 24 − 6 = 0 5(6)-8(3)-3(2)=30-24-6=0 5 ( 6 ) − 8 ( 3 ) − 3 ( 2 ) = 30 − 24 − 6 = 0 ✓.
Strategy. We seek ℓ ⃗ 2 = ( l 2 , m 2 , n 2 ) \vec\ell_2=(l_2,m_2,n_2) ℓ 2 = ( l 2 , m 2 , n 2 ) and ℓ ⃗ 3 = ( l 3 , m 3 , n 3 ) \vec\ell_3=(l_3,m_3,n_3) ℓ 3 = ( l 3 , m 3 , n 3 ) , both on the cone, both perpendicular to ℓ ⃗ 1 = ( 1 , 2 , 3 ) \vec\ell_1=(1,2,3) ℓ 1 = ( 1 , 2 , 3 ) , and perpendicular to each other.
Step 1 — Perpendicularity to ℓ ⃗ 1 \vec\ell_1 ℓ 1
l + 2 m + 3 n = 0 ⟹ l = − 2 m − 3 n . ( 1 ) l+2m+3n=0\;\Longrightarrow\;l=-2m-3n.\qquad(1) l + 2 m + 3 n = 0 ⟹ l = − 2 m − 3 n . ( 1 )
Step 2 — Substitute into the cone equation ( ⋆ ) (\star) ( ⋆ )
5 m n − 8 n l − 3 l m = 0 5mn-8nl-3lm=0 5 mn − 8 n l − 3 l m = 0 .
l = − 2 m − 3 n l=-2m-3n l = − 2 m − 3 n gives n l = − 2 m n − 3 n 2 nl=-2mn-3n^2 n l = − 2 mn − 3 n 2 and l m = − 2 m 2 − 3 m n lm=-2m^2-3mn l m = − 2 m 2 − 3 mn .
5 m n − 8 ( − 2 m n − 3 n 2 ) − 3 ( − 2 m 2 − 3 m n ) = 0 5mn-8(-2mn-3n^2)-3(-2m^2-3mn)=0 5 mn − 8 ( − 2 mn − 3 n 2 ) − 3 ( − 2 m 2 − 3 mn ) = 0
5 m n + 16 m n + 24 n 2 + 6 m 2 + 9 m n = 0 5mn+16mn+24n^2+6m^2+9mn=0 5 mn + 16 mn + 24 n 2 + 6 m 2 + 9 mn = 0
6 m 2 + 30 m n + 24 n 2 = 0 6m^2+30mn+24n^2=0 6 m 2 + 30 mn + 24 n 2 = 0
m 2 + 5 m n + 4 n 2 = 0 m^2+5mn+4n^2=0 m 2 + 5 mn + 4 n 2 = 0
( m + n ) ( m + 4 n ) = 0. (m+n)(m+4n)=0. ( m + n ) ( m + 4 n ) = 0.
So either m = − n m=-n m = − n or m = − 4 n m=-4n m = − 4 n .
Step 3 — Case m = − n m=-n m = − n
From (1): l = − 2 ( − n ) − 3 n = 2 n − 3 n = − n l=-2(-n)-3n=2n-3n=-n l = − 2 ( − n ) − 3 n = 2 n − 3 n = − n . Direction ( l , m , n ) = ( − n , − n , n ) = n ( − 1 , − 1 , 1 ) (l,m,n)=(-n,-n,n)=n(-1,-1,1) ( l , m , n ) = ( − n , − n , n ) = n ( − 1 , − 1 , 1 ) , i.e. ℓ ⃗ 2 = ( − 1 , − 1 , 1 ) \vec\ell_2=(-1,-1,1) ℓ 2 = ( − 1 , − 1 , 1 ) (or equivalently ( 1 , 1 , − 1 ) (1,1,-1) ( 1 , 1 , − 1 ) ).
Step 4 — Case m = − 4 n m=-4n m = − 4 n
From (1): l = − 2 ( − 4 n ) − 3 n = 8 n − 3 n = 5 n l=-2(-4n)-3n=8n-3n=5n l = − 2 ( − 4 n ) − 3 n = 8 n − 3 n = 5 n . Direction ( l , m , n ) = ( 5 n , − 4 n , n ) = n ( 5 , − 4 , 1 ) (l,m,n)=(5n,-4n,n)=n(5,-4,1) ( l , m , n ) = ( 5 n , − 4 n , n ) = n ( 5 , − 4 , 1 ) , i.e. ℓ ⃗ 3 = ( 5 , − 4 , 1 ) \vec\ell_3=(5,-4,1) ℓ 3 = ( 5 , − 4 , 1 ) .
Step 5 — Verify all three are mutually perpendicular
ℓ ⃗ 1 ⋅ ℓ ⃗ 2 = ( 1 , 2 , 3 ) ⋅ ( − 1 , − 1 , 1 ) = − 1 − 2 + 3 = 0 \vec\ell_1\cdot\vec\ell_2=(1,2,3)\cdot(-1,-1,1)=-1-2+3=0 ℓ 1 ⋅ ℓ 2 = ( 1 , 2 , 3 ) ⋅ ( − 1 , − 1 , 1 ) = − 1 − 2 + 3 = 0 ✓.
ℓ ⃗ 1 ⋅ ℓ ⃗ 3 = ( 1 , 2 , 3 ) ⋅ ( 5 , − 4 , 1 ) = 5 − 8 + 3 = 0 \vec\ell_1\cdot\vec\ell_3=(1,2,3)\cdot(5,-4,1)=5-8+3=0 ℓ 1 ⋅ ℓ 3 = ( 1 , 2 , 3 ) ⋅ ( 5 , − 4 , 1 ) = 5 − 8 + 3 = 0 ✓.
ℓ ⃗ 2 ⋅ ℓ ⃗ 3 = ( − 1 , − 1 , 1 ) ⋅ ( 5 , − 4 , 1 ) = − 5 + 4 + 1 = 0 \vec\ell_2\cdot\vec\ell_3=(-1,-1,1)\cdot(5,-4,1)=-5+4+1=0 ℓ 2 ⋅ ℓ 3 = ( − 1 , − 1 , 1 ) ⋅ ( 5 , − 4 , 1 ) = − 5 + 4 + 1 = 0 ✓.
Verify each lies on the cone ( ⋆ ) (\star) ( ⋆ ) :
ℓ ⃗ 2 = ( − 1 , − 1 , 1 ) \vec\ell_2=(-1,-1,1) ℓ 2 = ( − 1 , − 1 , 1 ) : 5 ( − 1 ) ( 1 ) − 8 ( 1 ) ( − 1 ) − 3 ( − 1 ) ( − 1 ) = − 5 + 8 − 3 = 0 5(-1)(1)-8(1)(-1)-3(-1)(-1)=-5+8-3=0 5 ( − 1 ) ( 1 ) − 8 ( 1 ) ( − 1 ) − 3 ( − 1 ) ( − 1 ) = − 5 + 8 − 3 = 0 ✓.
ℓ ⃗ 3 = ( 5 , − 4 , 1 ) \vec\ell_3=(5,-4,1) ℓ 3 = ( 5 , − 4 , 1 ) : 5 ( − 4 ) ( 1 ) − 8 ( 1 ) ( 5 ) − 3 ( 5 ) ( − 4 ) = − 20 − 40 + 60 = 0 5(-4)(1)-8(1)(5)-3(5)(-4)=-20-40+60=0 5 ( − 4 ) ( 1 ) − 8 ( 1 ) ( 5 ) − 3 ( 5 ) ( − 4 ) = − 20 − 40 + 60 = 0 ✓.
Conclusion
The other two generators are the lines through the origin with direction ratios ( − 1 , − 1 , 1 ) (-1,-1,1) ( − 1 , − 1 , 1 ) and ( 5 , − 4 , 1 ) (5,-4,1) ( 5 , − 4 , 1 ) :
Answer
x − 1 = y − 1 = z 1 and x 5 = y − 4 = z 1 . \boxed{\;\dfrac{x}{-1}=\dfrac{y}{-1}=\dfrac{z}{1}\quad\text{and}\quad\dfrac{x}{5}=\dfrac{y}{-4}=\dfrac{z}{1}.\;} − 1 x = − 1 y = 1 z and 5 x = − 4 y = 1 z .