Finding the distance from point to line by space vector Note: point to line, not point to face or line to line

Finding the distance from point to line by space vector Note: point to line, not point to face or line to line

Let the direction vector of the line be V, and find the distance from M0 to the line
Take any point M1 on the line L,
d=|v*M1M0|/|v|;
V * m1m0 is the inner product of vectors;
M1m0 is a vector

How to find the distance from point to line by vector method?

Let a known point be a (1, 10), the perpendicular of a straight line through a, and the perpendicular foot be B (x, y, z), then there is a vector AB = (x-1, Y-1, z), so 1-x + 1-y + Z = 0, that is, x + Y-Z = 2, and because B is on the line, so - x = - y = Z, find out x = 2 / 3, y = 2 / 3, z = - 2 / 3, so the vector AB = (- 1 / 3, - 1 / 3, - 2 / 3), and then the formula finds out AB = root 6 / 3. In fact, this is faster without vector method, you can draw a space rectangular coordinate system, Then draw the line and the point. It is clear that the distance from the point to the line is the distance from a vertex of a cube to the diagonal of the cube. It is very easy to do it with geometric relations

In the plane xoz of the space rectangular coordinate system, the vector B is perpendicular to a = (1, - 1,2) and the module is twice of | vector a |=
Give the answer
Write the final number, don't want to calculate,

Let B = (x, y, z) be perpendicular to a, then x + (- y) + 2Z = 0, that is, X-Y + 2Z = 0
The module length of a is the square of root 1 plus the square of - 1 plus the square of 2, which is equal to root 6. The module length of B is 2 root 6, then the root x square plus y square plus Z square is equal to 2 root 6