A particle is moving on a plane x + 2y + 3z = 13. Find the point on this plane which is closest to the point (1, 1, 1). (Hint: use Lagrange's multiplier method).

Steps to Solve

1. Formulate the distance function To find the point on the plane that is closest to (1, 1, 1), we start by defining the distance squared (to avoid dealing with square roots) as: $$ D = (x - 1)^2 + (y - 1)^2 + (z - 1)^2 $$

2. Define the constraint The constraint for the plane is given by the equation: $$ g(x, y, z) = x + 2y + 3z - 13 = 0 $$

3. Set up the Lagrange function We use Lagrange multipliers to incorporate the constraint into our distance function: $$ \mathcal{L}(x, y, z, \lambda) = D + \lambda g(x, y, z) $$ Substituting in the expressions, we have: $$ \mathcal{L}(x, y, z, \lambda) = (x - 1)^2 + (y - 1)^2 + (z - 1)^2 + \lambda (x + 2y + 3z - 13) $$

4. Find the gradients We now find the partial derivatives of $\mathcal{L}$ and set them to zero: [ \frac{\partial \mathcal{L}}{\partial x} = 2(x - 1) + \lambda = 0 ] [ \frac{\partial \mathcal{L}}{\partial y} = 2(y - 1) + 2\lambda = 0 ] [ \frac{\partial \mathcal{L}}{\partial z} = 2(z - 1) + 3\lambda = 0 ] [ \frac{\partial \mathcal{L}}{\partial \lambda} = x + 2y + 3z - 13 = 0 ]

5. Solve the system of equations From the equations, we have:

• $x - 1 = -\frac{\lambda}{2}$
• $y - 1 = -\lambda$
• $z - 1 = -\frac{3\lambda}{2}$

Substituting these into the constraint $x + 2y + 3z = 13$: [ \left(-\frac{\lambda}{2} + 1\right) + 2\left(-\lambda + 1\right) + 3\left(-\frac{3\lambda}{2} + 1\right) = 13 ]

Expanding this gives: [ -\frac{\lambda}{2} + 1 - 2\lambda + 2 - \frac{9\lambda}{2} + 3 = 13 ] Combining like terms yields: [ -\frac{12\lambda}{2} + 6 = 13 ] Simplifying further: [ -6\lambda + 6 = 13 \implies -6\lambda = 7 \implies \lambda = -\frac{7}{6} ]

6. Back substitute to find coordinates Now substituting $\lambda$ back into the equations: [ x = 1 - \frac{\lambda}{2} = 1 + \frac{7}{12} ] [ y = 1 - \lambda = 1 + \frac{7}{6} ] [ z = 1 - \frac{3\lambda}{2} = 1 + \frac{21}{12} ]

Calculating these values: [ x = \frac{19}{12}, \quad y = \frac{13}{6}, \quad z = \frac{33}{12} ]