The Perpendicular Distance between a given Point and a given Line |
Problem: |
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A diagram of this is shown on the right |
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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 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] ![]() = 1×1 + -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 |
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