Question

Find the distance between the following pairs of points:
(i) (2, 3, 5) and (4, 3, 1) (ii) (–3, 7, 2) and (2, 4, –1)
(iii) (–1, 3, – 4) and (1, –3, 4) (iv) (2, –1, 3) and (–2, 1, 3)

Answer 1

The distance between points P(x₁, y₁, z₁) and P(x₂, y₂, z₂) is given by
PQ = \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}\)

(i) Distance between points (2, 3, 5) and (4, 3, 1)
=\(\sqrt{(4-2)^2+(3-3)^2+(1-5)^2}\)
= \(\sqrt{(2)^2+(0)^2+(-4)^2}\)
= \(\sqrt{4+16}\)
= \(\sqrt{20}\)
=\(2\sqrt5\)

(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)
=\(\sqrt{(1+1)^2+(-3-3)^2+(4+4)^2}\)
=\(\sqrt{(2)^2+(-6)^2+(8)^2}\)
=\(\sqrt{4+36+64}\)
= \(\sqrt{104}\)
=\(2\sqrt26\)

(iv) Distance between points (2, -1, 3) and (-2, 1, 3)
= \(\sqrt{(-2-2)^2+(1+1)^2+(3-3)^2}\)
= \(\sqrt{(-4)^2+(2)^2+(0)^2}\)
= \(\sqrt{16+4}\)
= \(\sqrt{20}\) =2\(\sqrt5\)