Here you will learn distance formula in 3d to calculate distance between two points with example.<\/p>\n
Let’s begin –<\/p>\n
The distance between the points P\\((x_1, y_1, z_1)\\) and Q\\((x_2, y_2, z_2)\\) is given by<\/p>\n
\nPQ = \\(\\sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 + (z_2 – z_1)^2}\\)<\/p>\n<\/blockquote>\n
Proof :<\/strong> Let O be the origin and let P\\((x_1, y_1, z_1)\\) and Q\\((x_2, y_2, z_2)\\) be two given points. Then,<\/p>\n
\\(\\vec{OP}\\) = \\(x_1\\hat{i} + y_1\\hat{j} + z_1\\hat{k}\\), \\(\\vec{OQ}\\) = \\(x_2\\hat{i} + y_2\\hat{j} + z_2\\hat{k}\\)<\/p>\n
Now, <\/p>\n
\\(\\vec{PQ}\\) = Position vector Q – Position vector of P<\/p>\n
\\(\\implies\\) \\(\\vec{PQ}\\) = \\(x_2\\hat{i} + y_2\\hat{j} + z_2\\hat{k}\\) – \\(x_1\\hat{i} + y_1\\hat{j} + z_1\\hat{k}\\)<\/p>\n
\\(\\implies\\) \\(\\vec{PQ}\\) = \\((x_2 – x_1)\\hat{i} + (y_2 – y_1)\\hat{j} + (z_2 – z_1)\\hat{k}\\)<\/p>\n
\\(\\therefore\\) PQ = |\\(\\vec{PQ}\\)| = \\(\\sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 + (z_2 – z_1)^2}\\)<\/p>\n
Hence, PQ = \\(\\sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 + (z_2 – z_1)^2}\\).<\/p>\n
Example<\/span><\/strong> : Find the distance between the points P (-2, 4, 1) and Q (1, 2, -5).<\/p>\n
Solution<\/span><\/strong> : We have, P (-2, 4, 1) and Q (1, 2, -5).<\/p>\n
Distance PQ = \\(\\sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 + (z_2 – z_1)^2}\\)<\/p>\n
\\(\\implies\\) PQ = \\(\\sqrt{(1 – (-2))^2 + (2 – 4)^2 + (-5 – 1)^2}\\)<\/p>\n
\\(\\implies\\) PQ = \\(\\sqrt{9 + 4 + 36}\\) = 7 units<\/p>\n
Example<\/span><\/strong> : Prove by using the distance formula that the points P(1, 2, 3), Q(-1, -1, -1) and R(3, 5, 7) are collinear.<\/p>\n
Solution<\/span><\/strong> : We have, P(1, 2, 3), Q(-1, -1, -1) and R(3, 5, 7)<\/p>\n
Distance Formula = \\(\\sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 + (z_2 – z_1)^2}\\)<\/p>\n
PQ = \\(\\sqrt{(-1 – 1)^2 + (-2 – 2)^2 + (-1 – 3)^2}\\) = \\(\\sqrt{4 + 9 + 16}\\) = \\(\\sqrt{29}\\) units<\/p>\n
QR = \\(\\sqrt{(3 + 1)^2 + (5 + 1)^2 + (7 + 1)^2}\\) = \\(\\sqrt{16 + 36 + 64}\\) = \\(\\sqrt{116}\\) units<\/p>\n
and, PR = \\(\\sqrt{(3 – 1)^2 + (5 – 2)^2 + (7 – 3)^2}\\) = \\(\\sqrt{4 + 9 + 16}\\) = \\(\\sqrt{29}\\) units<\/p>\n
Clearly, QR = PQ + PR.<\/p>\n
Therefore, Points P, Q and R are collinear.<\/p>\n\n\n