The Perpendicular Distance between a given Point and a given Line

Problem: Find the perpendicular distance between the point (6, 7, 10) and a line which is parallel to the vector [2, 1, 1] and passing through the point (5, 9, 4)
A diagram of this is shown on the right		P is the given point. A is the given point through which the line passes. F is the foot of the perpendicular from P to the line. v is the vector to which the line is parallel.
Joining the points A and P together forms a rightangle triangle AFP.		The length of the side FP is the required length.
By the Pythagorean Theorem; = +	Now, the scalar (dot) product of a vector with itself is the square of its length; so, = AP . AP AP is the vector from point A (5, 9, 4) to P (6, 7, 10) AP = [6, 7, 10] - [5, 9, 4] = [6-5, 7-9, 10-4] = [1, -2, 6] = [1, -2, 6]. [1, -2, 6] = 1×1 + -2×-2 + 6×6 = 41 AF is the projection of the vector AP onto the line parallel to v = [2, 1, 1] AF is then the scalar (dot) product of AP and a unit vector, = ; AF = [1, -2, 6] . = = 2 We can now, finally, calculate FP using = + FP = = 3

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