which of the following pairs of points have a distance  $\sqrt{43}$ ?

Solution:

(a) Let the given points are A(2,3,5) and B(4,3,1).

      $\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{x}}_1=2,{\mathrm{y}}_1=3,{\mathrm{z}}_1=5\\ \text{ and }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{x}}_2=4,{\mathrm{y}}_2=3,{\mathrm{z}}_2=1\end{array}$

      Required distance

      $\begin{array}{c}\mathrm{AB}=\sqrt{{\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)}^2+{\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)}^2+{\left({\mathrm{z}}_2-{\mathrm{z}}_1\right)}^2}\\ \Rightarrow \mathrm{AB}=\sqrt{(4-2{)}^2+(3-3{)}^2+(1-5{)}^2}\\ =\sqrt{4+0+16}=\sqrt{20}=2\sqrt{5}\end{array}$

(b) Let the given points are A(- 3,7,2) and B(2,4, -1).

        Here,

      $\begin{array}{r}{\mathrm{x}}_1=-3,{\mathrm{y}}_1=7,{\mathrm{z}}_1=2\\ {\mathrm{x}}_2=2,{\mathrm{y}}_2=4,{\mathrm{z}}_2=-1\end{array}$

         Required distance

         $\begin{array}{c}\mathrm{AB}=\sqrt{{\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)}^2+{\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)}^2+{\left({\mathrm{z}}_2-{\mathrm{z}}_1\right)}^2}\\ =\sqrt{[2-(-3){]}^2+(4-7{)}^2+(-1-2{)}^2}\\ =\sqrt{(2+3{)}^2+(4-7{)}^2+(-1-2{)}^2}\\ =\sqrt{25+9+9}=\sqrt{43}\end{array}$

(c)  Let the given ,points  (-1, 3, -4) and (1, -3, 4)

       $\begin{array}{l}{\mathrm{x}}_1=-1,{\mathrm{y}}_1=3,{\mathrm{z}}_1=-4\\ {\mathrm{x}}_2=1,{\mathrm{y}}_2=-3,{\mathrm{z}}_2=4\end{array}$

       Required distance,

       $\begin{array}{r}\mathrm{AB}=\sqrt{(1+1{)}^2+(-3-3{)}^2+(4+4{)}^2}\\ =\sqrt{4+36+64}=\sqrt{104}=2\sqrt{26}\end{array}$

(d) Let the given points are A(2, -1,3) and B(-2,1,3).

       $\begin{array}{r}{\mathrm{x}}_1=2,{\mathrm{y}}_1=-1,{\mathrm{z}}_1=3\\ {\mathrm{x}}_2=-2,{\mathrm{y}}_2=1,{\mathrm{z}}_2=3\end{array}$

         Required distance,

        $\begin{array}{r}\mathrm{AB}=\sqrt{(-2-2{)}^2+(1+1{)}^2+(3-3{)}^2}=\sqrt{4^2+2^2+0}\\ =\sqrt{16+4}=\sqrt{20}=2\sqrt{5}\end{array}$