We can find distance between any two given points in space in terms of the coordinates of the points. The formula used to find distance between the two points is known as **distance formula**.

Let $P(x_1,y_1,z_1)$ and $Q(x_2,y_2,z_2)$ be the two given points with reference to the rectangular axes $OX$, $OY$ and $OZ$. Draw $PL$ and $QM$ perpendiculars to the xy-plane. Then, the coordinates of the points $L$ and $M$ in the xy-plane ($z=0$ plane) are $(x_1,y_1)$ and $(x_2,y_2)$ respectively.

The distance between the points $L$ and $M$ is given by \[LM^2=(x_2-x_1)^2+(y_2-y_1)^2\] Now, from $P$ draw $PR\perp MQ$. Then, $PR$ is parallel and equal to $LM$ and the coordinates of $R$ are $(x_2,y_2,z_1)$. Hence, from right angled $\Delta PQR$, \[PQ^2=PR^2+RQ^2\] \[=LM^2+(MQ-MR)^2\] \[=(x_2-x_1)^2+(y_2-y_1)^2+(z_1-z_1)^2\] \[\therefore PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}\]

**Cor.** if one point say $Q(x_2,y_2,z_2)$ be at the origin, then \[OP=\sqrt{x_1^2+y_1^2+z_1^2}\]

### Find the distance between the points $(-1,4,3)$ and $(2,2,-3)$ using distance formula.

Given points are $A(-1,4,3)$ and $B(2,2,-3)$. The distance between $A$ and $B$ is given by \[AB=\sqrt{(2+1)^2+(2-4)^2+(-3-3)^2}\] \[=\sqrt{3^2+(-2)^2+(-6)^2}\] \[=\sqrt{49}=7\text{ units}\]

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