May 27, 2015 · The distance between $ (1,2,-1)$ and $ (1,0,1)$ is way less than $57.6$ and it's very unlikely that $ (1,2,-1)$ is the closest point on the line. So you must have made a mistake.

Jun 6, 2016 · Compute the distance between the point R(1, 1, 1) and the line (0, -1, 1) + t(1, -2, 2) Can someone please show me how to answer this in full? It was in a test recently and is still …

Aug 28, 2016 · 2 Playing with the solutions, I built this simulation one of them: Using a vector equation of the line through $BC$. And Pythagoras's theorem to get the distance from $A$ to …

Oct 11, 2012 · The distance between two lines in $ \Bbb R^3 $ is equal to the distance between parallel planes that contain these lines. To find that distance first find the normal vector of …

May 5, 2018 · Yet another way takes advantage of the geometry of the situation to make a direct computation. If $\vec r_1$ and $\vec r_2$ are two points on the line, then the perpendicular …

Apr 21, 2017 · 7 Hint: The line and the plane (as you have noted) are parallel. The distance from the plane to the line is therefore the distance from the plane to any point on the line. So just …

Apr 28, 2018 · I have a point at (4,6) and a line defined by points (-7,9) and (10, 9). How would I find the shortest distance between the point and the line, without converting each into linear …

The distance from the point to the line is the height of this paralellogram when we consider $\vec {v}= (1,1,1)$ as basis. So the distance is the area divide by the basis.

Jan 30, 2022 · The correct distance can be found by setting the negative t to zero and putting it back in the half-line equation to get the first valid point to use for the distance

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