Q.Find the distance between the following pairs of points:

Concept: Distance between two points in 3D space — the formula is a direct extension of the 2D distance formula using the three coordinate differences.

For points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$:

(i) $(2, 3, 5)$ and $(4, 3, 1)$

Differences: $(2, 0, -4)$

$d = \sqrt{4 + 0 + 16} = \sqrt{20} = 2\sqrt{5}$

(ii) $(-3, 7, 2)$ and $(2, 4, -1)$

Differences: $(5, -3, -3)$

$d = \sqrt{25 + 9 + 9} = \sqrt{43}$

(iii) $(-1, 3, -4)$ and $(1, -3, 4)$

Differences: $(2, -6, 8)$

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

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

Differences: $(-4, 2, 0)$

$d = \sqrt{16 + 4 + 0} = \sqrt{20} = 2\sqrt{5}$

The distance formula in three dimensions is a direct extension of the distance formula in the plane. If you know how to find the distance between two points on a flat sheet of paper, you already know the core idea — just add one more dimension.

Think of it this way: to go from point $A$ to point $B$ in space, you move along three independent directions (x, y, z). The straight-line distance is the length of the diagonal of a rectangular box whose sides are the differences in each coordinate. That diagonal length comes straight from the Pythagorean theorem applied twice — once to get the diagonal of the base, then again to include the height.

Let’s work through each pair step by step.

(i) $(2, 3, 5)$ and $(4, 3, 1)$

1. Find the differences in each coordinate:

   • $x$: $4 - 2 = 2$
   • $y$: $3 - 3 = 0$
   • $z$: $1 - 5 = -4$

2. Square each difference:

   • $2^2 = 4$
   • $0^2 = 0$
   • $(-4)^2 = 16$

3. Sum the squares: $4 + 0 + 16 = 20$

4. Take the square root: $\sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5}$

(ii) $(-3, 7, 2)$ and $(2, 4, -1)$

1. Differences:

   • $x$: $2 - (-3) = 5$
   • $y$: $4 - 7 = -3$
   • $z$: $-1 - 2 = -3$

2. Squares:

   • $5^2 = 25$
   • $(-3)^2 = 9$
   • $(-3)^2 = 9$

3. Sum: $25 + 9 + 9 = 43$

4. Square root: $\sqrt{43}$

Since 43 is prime, this cannot be simplified further.

(iii) $(-1, 3, -4)$ and $(1, -3, 4)$

1. Differences:

   • $x$: $1 - (-1) = 2$
   • $y$: $-3 - 3 = -6$
   • $z$: $4 - (-4) = 8$

2. Squares:

   • $2^2 = 4$
   • $(-6)^2 = 36$
   • $8^2 = 64$

3. Sum: $4 + 36 + 64 = 104$

4. Square root: $\sqrt{104} = \sqrt{4 \times 26} = 2\sqrt{26}$

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

1. Differences:

   • $x$: $-2 - 2 = -4$
   • $y$: $1 - (-1) = 2$
   • $z$: $3 - 3 = 0$

2. Squares:

   • $(-4)^2 = 16$
   • $2^2 = 4$
   • $0^2 = 0$

3. Sum: $16 + 4 + 0 = 20$

4. Square root: $\sqrt{20} = 2\sqrt{5}$

Notice that the $z$-coordinates are the same (both 3), so the points lie in a horizontal plane. The distance is purely in the $xy$-plane.