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Mathematics Question on distance between two points

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

The distance between points P(x1, y1, z1) and P(x2, y2, z2) is given by
PQ = (x2−x1)2+(y2−y1)2+(z2−z1)2\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)
=(4−2)2+(3−3)2+(1−5)2\sqrt{(4-2)^2+(3-3)^2+(1-5)^2}
= (2)2+(0)2+(−4)2\sqrt{(2)^2+(0)^2+(-4)^2}
= 4+16\sqrt{4+16}
= 20\sqrt{20}
=252\sqrt5

(ii) Distance between points (-3, 7, 2) and (2, 4, -1)
=(2+3)2+(4−7)2+(−1−2)2\sqrt{(2+3)^2+(4-7)^2+(-1-2)^2}
=(5)2+(−3)2+(−3)2\sqrt{(5)^2+(-3)^2+(-3)^2}
= 25+9+9\sqrt{25+9+9}
= 43\sqrt{43}

(iii) Distance between points (-1, 3, -4) and (1, -3, 4)
=(1+1)2+(−3−3)2+(4+4)2\sqrt{(1+1)^2+(-3-3)^2+(4+4)^2}
=(2)2+(−6)2+(8)2\sqrt{(2)^2+(-6)^2+(8)^2}
=4+36+64\sqrt{4+36+64}
= 104\sqrt{104}
=2262\sqrt26

(iv) Distance between points (2, -1, 3) and (-2, 1, 3)
= (−2−2)2+(1+1)2+(3−3)2\sqrt{(-2-2)^2+(1+1)^2+(3-3)^2}
= (−4)2+(2)2+(0)2\sqrt{(-4)^2+(2)^2+(0)^2}
= 16+4\sqrt{16+4}
= 20\sqrt{20} =25\sqrt5