Given two points in space M_1(x_1,y_1,z_1) and…

Given two points in space $M_1(x_1,y_1,z_1)$ and $M_2(x_2,y_2,z_2)$. Prove that $\overrightarrow{M_1M_2} = (x_2-x_1)\vec{i}+(y_2-y_1)\vec{j}+(z_2-z_1)\vec{k}$ and that $|\overrightarrow{M_1M_2}| = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$.

Step 1: Position vectors

By Problem 27: \[\overrightarrow{OM_1} = x_1\vec{i}+y_1\vec{j}+z_1\vec{k}, \qquad \overrightarrow{OM_2} = x_2\vec{i}+y_2\vec{j}+z_2\vec{k}.\]

Step 2: Component formula

The triangle rule in triangle $OM_1M_2$ gives $\overrightarrow{OM_1} + \overrightarrow{M_1M_2} = \overrightarrow{OM_2}$, hence

\[ \overrightarrow{M_1M_2} = \overrightarrow{OM_2} - \overrightarrow{OM_1} = (x_2-x_1)\vec{i} + (y_2-y_1)\vec{j} + (z_2-z_1)\vec{k}. \]

\[\overrightarrow{M_1M_2} = (x_2-x_1)\vec{i}+(y_2-y_1)\vec{j}+(z_2-z_1)\vec{k}\]

Step 3: Distance formula

Write $\Delta x = x_2-x_1$, $\Delta y = y_2-y_1$, $\Delta z = z_2-z_1$. The unit vectors are orthonormal: \[\vec{i}\cdot\vec{i}=\vec{j}\cdot\vec{j}=\vec{k}\cdot\vec{k}=1, \qquad \vec{i}\cdot\vec{j}=\vec{j}\cdot\vec{k}=\vec{i}\cdot\vec{k}=0.\]

Expanding the dot product, all mixed terms vanish: \[|\overrightarrow{M_1M_2}|^2 = (\Delta x\,\vec{i}+\Delta y\,\vec{j}+\Delta z\,\vec{k})\cdot(\Delta x\,\vec{i}+\Delta y\,\vec{j}+\Delta z\,\vec{k}) = (\Delta x)^2+(\Delta y)^2+(\Delta z)^2.\]

\[|\overrightarrow{M_1M_2}| = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}\]

Step 4: Geometric interpretation

Step 1. The projections $M_1'(x_1,y_1,0)$ and $M_2'(x_2,y_2,0)$ onto the $xy$-plane are at distance

\[ d_{xy} = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}. \]

Step 2. $M_1M_2$ is the hypotenuse of a right triangle with legs $d_{xy}$ (horizontal) and $|z_2-z_1|$ (vertical):

\[ |M_1M_2| = \sqrt{d_{xy}^2+(z_2-z_1)^2}. \]

Step 5: Special cases

Step 6: Numerical example

For $M_1(1,2,-1)$ and $M_2(4,-1,3)$: \[\overrightarrow{M_1M_2} = 3\vec{i}-3\vec{j}+4\vec{k}, \qquad |\overrightarrow{M_1M_2}| = \sqrt{9+9+16} = \sqrt{34}.\]