Vector distances
Objective. Find geometric distances in three dimensions.
- 1. Learn
- 2. Worked example
- 3. Practice
- 4. Feedback
- 5. Continue
Learn: the key idea
Vector distances measure the shortest separation between geometric objects in three dimensions, such as the distance from a point to a line, found by minimising |AP| over all points P on the line, or equivalently using the perpendicular from the point to the line. For a point P and a line r = a + tb, the foot of the perpendicular occurs when (P−a−tb)·b = 0, which is solved for t before computing the resulting distance. For a point (0,0,0) and line r = (1,0,0)+t(0,1,0), the closest point is (1,0,0), giving distance 1. Always solve for the parameter at the perpendicular foot before calculating the final distance.
Worked example
Find the shortest distance from the point P(0,0,0) to the line r = (1,1,0) + t(1,−1,0).
- 1Write a general point on the line as (1+t, 1−t, 0)
- 2Form the vector from P: (1+t, 1−t, 0) and set its dot product with direction (1,−1,0) to zero
- 3Solve (1+t)(1) + (1−t)(−1) = 0, giving 1+t−1+t = 0, so t = 0
- 4Substitute t = 0 to get closest point (1,1,0), with distance √(1²+1²) = √2
Answer: √2
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Practice
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Start practisingCommon mistake. A common mistake is calculating the distance from the point to the line's given position vector a, rather than to the true closest point on the line found via the perpendicular condition. Always find the foot of the perpendicular first.
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