You mean sin θ, right?uart said:Use u x v = |u| |v| cos(theta)
vela said:You mean sin θ, right?
No and no. The norm of a vector v has a perfectly good geometric interpretation - it's the length of the vector. A line has infinite length - maybe you meant line segment?aero_zeppelin said:Ok, so we know that the norm of uxv is the area of the parallelogram formed by u and v...
The norm of v has no geometric interpretation (its just the length of a line)
aero_zeppelin said:How can we relate this so it gives the distance between P and the line...
any more ideas??
Mark44 said:No and no. The norm of a vector v has a perfectly good geometric interpretation - it's the length of the vector. A line has infinite length - maybe you meant line segment?
The distance between a point and a line is the shortest distance between the point and any point on the line. It is the length of the perpendicular line segment drawn from the point to the line.
The formula for calculating the distance between a point and a line is:
d = |Ax + By + C| / √(A² + B²)
where A, B, and C are the coefficients of the line's equation, and (x, y) are the coordinates of the point.
The coefficients of a line's equation can be determined using the slope-intercept form:
y = mx + b
where m is the slope of the line and b is the y-intercept. The coefficients can also be determined using the point-slope form:
y - y1 = m(x - x1)
where (x1, y1) is a point on the line and m is the slope of the line.
No, the distance between a point and a line cannot be negative. It is always a positive value, as it represents the length of a line segment.
Yes, the distance between a point and a line can be visualized as a right triangle, with the point being one of the vertices and the line being the hypotenuse. The distance is equal to the length of the altitude drawn from the point to the line.