To find the length of the projection of the line segment joining the points $(5, -1, 4)$ and $(4, -1, 3)$ on the plane $x + y + z = 7$, we will follow the steps outlined below.
First, calculate the vector of the line segment from point $(5, -1, 4)$ to point $(4, -1, 3)$. The vector is found by subtracting the corresponding coordinates:
$$ \mathbf{a} = (4 - 5, -1 + 1, 3 - 4) = (-1, 0, -1) $$.
Next, find the normal vector of the plane $x + y + z = 7$. The coefficients of $x, y, z$ in the plane equation form the normal vector $\mathbf{n} = (1, 1, 1)$.
The projection of vector $\mathbf{a}$ on the plane can be calculated using the formula for the projection of a vector $\mathbf{a}$ onto another vector $\mathbf{n}$ given by:
$$ \text{proj}_{\mathbf{n}} \, \mathbf{a} = \frac{\mathbf{a} \cdot \mathbf{n}}{\mathbf{n} \cdot \mathbf{n}} \, \mathbf{n} $$.
Substitute these into the projection formula to find:
$$ \text{proj}_{\mathbf{n}} \, \mathbf{a} = \frac{-2}{3} \, (1, 1, 1) = \left(\frac{-2}{3}, \frac{-2}{3}, \frac{-2}{3}\right) $$.
The magnitude (length) of this projection vector is calculated as:
$$ \left\| \left( \frac{-2}{3}, \frac{-2}{3}, \frac{-2}{3} \right) \right\| = \sqrt{\left(\frac{-2}{3}\right)^2 + \left(\frac{-2}{3}\right)^2 + \left(\frac{-2}{3}\right)^2} $$
$$ = \sqrt{3 \times \left(\frac{4}{9}\right)} = \sqrt{\frac{12}{9}} = \sqrt{\frac{4}{3}} = \sqrt{\frac{4}{3}} = \sqrt{\frac{2}{3}} $$.
Thus, the length of the projection of the line segment on the plane $x + y + z = 7$ is $\sqrt{\frac{2}{3}}$.
