Distance from a point to a line - 3-Dimensional.
Distance from a point to a line in space formula
If M_{0}(x_{0}, y_{0}, z_{0}) point coordinates, s = {m; n; p} directing vector of line l, M_{1}(x_{1}, y_{1}, z_{1}) - coordinates of point on line l, then distance between point M_{0}(x_{0}, y_{0}, z_{0}) and line l can be found using the following formula:
d = | |M_{0}M_{1}×s| |
|s| |
Proof of the formula of distance from a point to a line for the space problem
If l is line equation then s = {m; n; p} is directing vector of line, M_{1}(x_{1}, y_{1}, z_{1})is coordinates of point on line. From properties of cross product it is known that the module of cross product of vectors is equal to the area of a parallelogramme constructed on these vectors
A = |M_{0}M_{1}×s|.
On the other hand parallelogramme area is equal to product of its side on height spent to this side
A = |s|d.
Having equated the areas it is simple to receive the formula of distance from a point to a line.
Examples of tasks with from a point to a line in space
x - 3 | = | y - 1 | = | z + 1 |
2 | 1 | 2 |
Solution.
From line equation find:
s = {2; 1; 2} - directing vector of line;
M_{1}(3; 1; -1) - coordinates of point on line.
Then
M_{0}M_{1} = {3 - 0; 1 - 2; -1 - 3} = {3; -1; -4}
M_{0}M_{1}×s = | i | j | k | = |
3 | -1 | -4 | ||
2 | 1 | 2 |
= i ((-1)·2 - (-4)·1) - j (3·2 - (-4)·2) + k (3·1 -(-1)·2) = {2; -14; 5}
d = | |M_{0}M_{1}×s| | = | √2^{2} + (-14)^{2} + 5^{2} | = | √225 | = | 15 | = 5 |
|s| | √2^{2} + 1^{2} + 2^{2} | √9 | 3 |
Answer: distance from point to line is equal to 5.
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