First slide

Introduction to 3-D Geometry

Question

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

Easy

A

(2,3, 5) and (4,3,1)
B

(-3,7,2) and (2, 4, -1)
C

(-1, 3, -4) and (1, -3, 4)
D

(2, -1, 3) and (-2,1, 3)

Solution

(a) Let the given points are A(2,3,5) and B(4,3,1).
   $\begin{array}{ll} ∴ x_{1} = 2, y_{1} = 3, z_{1} = 5 \\ and x_{2} = 4, y_{2} = 3, z_{2} = 1 \end{array}$
Required distance
   $\begin{matrix} AB = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2} + {(z_{2} - z_{1})}^{2}} \\ \Rightarrow AB = \sqrt{(4 - 2)^{2} + (3 - 3)^{2} + (1 - 5)^{2}} \\ = \sqrt{4 + 0 + 16} = \sqrt{20} = 2 \sqrt{5} \end{matrix}$
(b) Let the given points are A(- 3,7,2) and B(2,4, -1).
Here,
   $\begin{aligned} x_{1} = - 3, y_{1} = 7, z_{1} = 2 \\ x_{2} = 2, y_{2} = 4, z_{2} = - 1 \end{aligned}$
Required distance
$\begin{matrix} AB = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2} + {(z_{2} - z_{1})}^{2}} \\ = \sqrt{[2 - (- 3)]^{2} + (4 - 7)^{2} + (- 1 - 2)^{2}} \\ = \sqrt{(2 + 3)^{2} + (4 - 7)^{2} + (- 1 - 2)^{2}} \\ = \sqrt{25 + 9 + 9} = \sqrt{43} \end{matrix}$
(c) Let the given ,points (-1, 3, -4) and (1, -3, 4)
$\begin{array}{l} x_{1} = - 1, y_{1} = 3, z_{1} = - 4 \\ x_{2} = 1, y_{2} = - 3, z_{2} = 4 \end{array}$
Required distance,
$\begin{aligned} AB = \sqrt{(1 + 1)^{2} + (- 3 - 3)^{2} + (4 + 4)^{2}} \\ = \sqrt{4 + 36 + 64} = \sqrt{104} = 2 \sqrt{26} \end{aligned}$
(d) Let the given points are A(2, -1,3) and B(-2,1,3).
$\begin{aligned} x_{1} = 2, y_{1} = - 1, z_{1} = 3 \\ x_{2} = - 2, y_{2} = 1, z_{2} = 3 \end{aligned}$
Required distance,
   $\begin{aligned} AB = \sqrt{(- 2 - 2)^{2} + (1 + 1)^{2} + (3 - 3)^{2}} = \sqrt{4^{2} + 2^{2} + 0} \\ = \sqrt{16 + 4} = \sqrt{20} = 2 \sqrt{5} \end{aligned}$

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