UPSC 2016 Maths Optional Paper 1 Q4b — Step-by-Step Solution

10 marks · Section A

Cone · Analytic Geometry · asked 14× in 13 yrs · Read the full method →

Question

Show that the cone $3yz-2zx-2xy=0$ has an infinite set of three mutually perpendicular generators. If $\dfrac{x}{1}=\dfrac{y}{1}=\dfrac{z}{2}$ is a generator belonging to one such set, find the other two.

Technique

Existence via the trace-zero (sum of square-coefficients $=0$ ) condition; the other two generators are found by intersecting the cone equation with the plane through the origin perpendicular to the given generator.

Solution

Step 1 — The “sum of coefficients of squares” criterion

Write the cone as $F(x,y,z)=0$ . A second-degree cone $\sum a_{ij}x_ix_j=0$ admits a triad of mutually perpendicular generators iff the sum of the coefficients of $x^2,y^2,z^2$ is zero (equivalently $\operatorname{tr}$ of the associated symmetric matrix is $0$ ). Here

3yz-2zx-2xy=0

has no $x^2,y^2,z^2$ terms, so that sum is $0+0+0=0$ . Hence infinitely many mutually perpendicular triads of generators exist. $\;\blacksquare$

Step 2 — Set up equations for the other two generators

A generator direction $(\ell,m,n)$ lies on the cone if

3mn-2n\ell-2\ell m=0.\tag{cone}

The given generator is $\mathbf u=(1,1,2)$ . The other two directions must each be perpendicular to $\mathbf u$ :

\ell+m+2n=0.\tag{perp}

Solve (cone) and (perp) simultaneously for the directions of the remaining two generators.

Step 3 — Solve the two equations

From (perp), $\ell=-m-2n$ . Substitute into (cone):

3mn-2n(-m-2n)-2(-m-2n)m=0\Rightarrow 3mn+2mn+4n^2+2m^2+4mn=0\Rightarrow 2m^2+9mn+4n^2=0.

Factor: $2m^2+9mn+4n^2=(2m+n)(m+4n)=0$ , giving $n=-2m$ or $m=-4n$ .

Case $n=-2m$ (take $m=1,n=-2$ ): $\ell=-1-2(-2)=3$ . Direction $(3,1,-2)$ .
Case $m=-4n$ (take $n=1,m=-4$ ): $\ell=4-2=2$ . Direction $(2,-4,1)$ .

Answer

\boxed{\;\text{Other two generators: }\ \frac{x}{3}=\frac{y}{1}=\frac{z}{-2}\quad\text{and}\quad\frac{x}{2}=\frac{y}{-4}=\frac{z}{1}.\;}