UPSC 2025 Maths Optional Paper 1 Q5e — Step-by-Step Solution

10 marks · Section B

Gradient: definition, geometric meaning, computation · Vector Analysis · asked 6× in 14 yrs · Read the full method →

Question

If $u = x + y + z$ , $v = x^2 + y^2 + z^2$ and $w = xy + yz + zx$ , then show that $\operatorname{grad} u$ , $\operatorname{grad} v$ and $\operatorname{grad} w$ are coplanar.

Technique

Three vectors are coplanar iff their scalar triple product is zero, i.e. iff the determinant whose rows are the three vectors (here the Jacobian $\dfrac{\partial(u,v,w)}{\partial(x,y,z)}$ ) vanishes.

Solution

Compute the gradients:

\nabla u = (1,\,1,\,1),

\nabla v = (2x,\,2y,\,2z),

\nabla w = (y+z,\,z+x,\,x+y).

The three vectors are coplanar iff the scalar triple product $\nabla u\cdot(\nabla v\times\nabla w)=0$ , i.e.

\begin{vmatrix} 1 & 1 & 1 \\ 2x & 2y & 2z \\ y+z & z+x & x+y \end{vmatrix} = 0.

Take out the factor $2$ from the middle row:

2\begin{vmatrix} 1 & 1 & 1 \\ x & y & z \\ y+z & z+x & x+y \end{vmatrix}.

Add the second row to the third ( $R_3 \to R_2 + R_3$ ):

R_2 + R_3 = \bigl(x+y+z,\; x+y+z,\; x+y+z\bigr) = (x+y+z)(1,1,1).

So the determinant becomes

2\begin{vmatrix} 1 & 1 & 1 \\ x & y & z \\ x+y+z & x+y+z & x+y+z \end{vmatrix} = 2(x+y+z)\begin{vmatrix} 1 & 1 & 1 \\ x & y & z \\ 1 & 1 & 1 \end{vmatrix} = 0,

since the first and third rows are identical.

Therefore the scalar triple product is zero, so $\nabla u,\ \nabla v,\ \nabla w$ are coplanar.

(Geometric reason: $v = u^2 - 2w$ , so $\nabla v = 2u\,\nabla u - 2\,\nabla w$ — the three gradients satisfy a linear relation, hence are linearly dependent, i.e. coplanar.)

Answer

\boxed{\;\nabla u\cdot(\nabla v\times\nabla w) = 0\ \Rightarrow\ \nabla u,\ \nabla v,\ \nabla w\ \text{are coplanar},\quad\text{indeed } \nabla v = 2u\,\nabla u - 2\,\nabla w.\;}