Find the distance between the following pairs of points:

Solution:

The distance between points $\mathrm{P}\left(x_{1}, y_{1}, z_{1}\right)$ and $\mathrm{P}\left(x_{2}, y_{2}, z_{2}\right)$ is given by $\mathrm{PQ}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}}$

(i) Distance between points (2, 3, 5) and (4, 3, 1)

$=\sqrt{(4-2)^{2}+(3-3)^{2}+(1-5)^{2}}$

$=\sqrt{(2)^{2}+(0)^{2}+(-4)^{2}}$

$=\sqrt{4+16}$

$=\sqrt{20}$

$=2 \sqrt{5}$

(ii) Distance between points (–3, 7, 2) and (2, 4, –1)

$=\sqrt{(2+3)^{2}+(4-7)^{2}+(-1-2)^{2}}$

$=\sqrt{(5)^{2}+(-3)^{2}+(-3)^{2}}$

$=\sqrt{25+9+9}$

$=\sqrt{43}$

(iii) Distance between points $(-1,3,-4)$ and $(1,-3,4)$

$=\sqrt{(1+1)^{2}+(-3-3)^{2}+(4+4)^{2}}$

$=\sqrt{(2)^{2}+(-6)^{3}+(8)^{2}}$

$=\sqrt{4+36+64}=\sqrt{104}=2 \sqrt{26}$

(iv) Distance between points $(2,-1,3)$ and $(-2,1,3)$

$=\sqrt{(-2-2)^{2}+(1+1)^{2}+(3-3)^{2}}$

$=\sqrt{(-4)^{2}+(2)^{2}+(0)^{2}}$

$=\sqrt{(-4)}+(2)$

$=\sqrt{16+4}$

$=\sqrt{20}$

$=2 \sqrt{5}$