Question

: The distance between the points (4, 3, 7) and (1, -1, -5) is?
(a) 7
(b) 12
(c) 13
(d) 25

Answer 1

Hint: Assume (4, 3, 7) as $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and (1, -1, 5) as $\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$. Use distance formula in 3-dimensional geometry given by: $d=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}+{{\left( {{z}_{1}}-{{z}_{2}} \right)}^{2}}}$, where ‘d’ is the distance between the two points $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$.

Complete step-by-step answer:
We know that a point lying in any plane is represented by the coordinates $\left( x,y,z \right)$. Now, we have been provided with two points (4, 3, 7) and (1, -1, -5) and we have to find the distance between these two.
Let us assume these points as $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$ respectively. Therefore,
$\left( 4,3,7 \right)=\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and $\left( 1,-1,-5 \right)=\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$.
Let us assume that the distance between these two points is $d$.
By distance formula, we know that, distance between two points is, $d=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}+{{\left( {{z}_{1}}-{{z}_{2}} \right)}^{2}}}$. Therefore,
$\begin{align}
  & d=\sqrt{{{\left( 4-1 \right)}^{2}}+{{\left( 3-(-1) \right)}^{2}}+{{\left( 7-(-5) \right)}^{2}}} \\
 & =\sqrt{{{3}^{2}}+{{4}^{2}}+{{12}^{2}}} \\
 & =\sqrt{9+16+144} \\
 & =\sqrt{169} \\
 & =13 \\
\end{align}$
Therefore, the distance between these two points is 13 units.
Hence, option (c) is the correct answer.

Note: One may note that in 2-dimension geometry, we have only two coordinates of a particular point, that is, (x, y) which lies in the x-y plane. The distance between any two points in 2-D space is given by: $d=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$. Similarly, in 3-D space we have another coordinate in addition to x and y, that is z. This z-coordinate represents that the required point is above or below the x-y plane. So, we use the distance formula, $d=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}+{{\left( {{z}_{1}}-{{z}_{2}} \right)}^{2}}}$ for the calculation of distance between two points, like we did in the above question.