Q. 56 - Calculus

Question

Let ${\alpha }_i, {\beta }_i, and {\gamma }_i$ be constants for i = 1, 2, and 3. A transformation $T: R^3\to R^3$  defined by,

$$\begin{gathered} x={\alpha }_{1}u+{\beta }_{1}v+{\gamma }_{1}w \\ y={\alpha }_{2}u+{\beta }_{2}v+{\gamma }_{2}w \\ z={\alpha }_{3}u+{\beta }_{3}v+{\gamma }_{3}w \end{gathered}$$ 

is called a linear transformation of $R^3$. Prove that this transformation takes a plane ax + by + cz = d in the xyz coordinate system to a plane in the uvw-coordinate system if the Jacobian of the transformation is nonzero.

Step-by-Step Solution

Answer

It is proven that the coefficients of the plane in uvw- system are non zero, and hence represent a plane, if Jacobian is non-zero.

Step 1: Given information

The equations of transformations are,

$$\begin{gathered} x={\alpha }_{1}u+{\beta }_{1}v+{\gamma }_{1}w \\ y={\alpha }_{2}u+{\beta }_{2}v+{\gamma }_{2}w \\ z={\alpha }_{3}u+{\beta }_{3}v+{\gamma }_{3}w \end{gathered}$$

The objective is to determine the transformation of a line $ax+by+cz=d$ in xyz- coordinate system to uvw- system.

Step 2: Find the Jacobian

The Jacobian of a transformation using partial derivatives is computed as,

$$\begin{gathered} \frac{\partial (x, y, z)}{\partial (u, v, w)}=\left|\begin{array}{ccc}\frac{\partial x}{\partial u}& \frac{\partial y}{\partial u}& \frac{\partial z}{\partial u}\\ \frac{\partial x}{\partial v}& \frac{\partial y}{\partial v}& \frac{\partial z}{\partial v}\\ \frac{\partial x}{\partial w}& \frac{\partial y}{\partial w}& \frac{\partial z}{\partial w}\end{array}\right| \\ \frac{\partial (x, y, z)}{\partial (u, v, w)}=\left|\begin{array}{ccc}{\alpha }_1& {\alpha }_2& {\alpha }_3\\ {\beta }_1& {\beta }_2& {\beta }_3\\ {\gamma }_1& {\gamma }_2& {\gamma }_3\end{array}\right| \end{gathered}$$

To transform the equation of plane in uvw- system , substitute the equations of transformations in the equation of line.

$$\begin{gathered} ax+by+cz=0 \\ (a{\alpha }_1+b{\alpha }_2+c{\alpha }_3)u+(a{\beta }_1+b{\beta }_2+c{\beta }_3)v+(a{\gamma }_1+b{\gamma }_2+c{\gamma }_3)w=d \end{gathered}$$

Step 3: Proof

For the purpose of proving the statement, let us assume the converse.

Let us consider that both coefficients are zero.

Write the system of equations in matrix form.

$\left|\begin{array}{ccc}{\alpha }_1& {\alpha }_2& {\alpha }_3\\ {\beta }_1& {\beta }_2& {\beta }_3\\ {\gamma }_1& {\gamma }_2& {\gamma }_3\end{array}\right|\left[\begin{array}{c}a\\ b\\ c\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

The determinant of the matrix is the Jacobian of the transformation.

Hence, the system is consistent if Jacobian is non-zero.

Since the system has only trivial solution. This is not an accepted solution for a given plane.

Hence, the system has not any real solution, if the Jacobian is non-zero.

This means the system of equations does not exist.

Thus, our assumption is proved to be incorrect.

Therefore, the coefficients of the plane in uvw- system are non zero, and hence represent a plane, if Jacobian is non-zero.