Faster ways to calculate distance between a point and a line in 3D

[Krushnaraj Pandya]

Homework Statement

Suppose a line is given, say (x-3)=(y-2)=z in Cartesian form and we are supposed to find a point on it which is at a distance say, 6 units from a given arbitrary point (1,2,3).

2. Relevant tedious methods
1) write line in vector form then equate modulus of vector between point on line and given point to 6
2)use given equation and distance formula to get the point.

The Attempt at a Solution

Using vector form of line as well using distance formula is very tedious and has high chances of calculation mistakes and its a really off-putting thing especially in exam pressure when you put a lot of time into it and still get a wrong answer, I would be glad if the people on PF could brainstorm a shorter, faster way to arrive at this-perhaps calculus could help in some way. Thank you for your help.

 

[Mark44]

Using vector form of line as well using distance formula is very tedious and has high chances of calculation mistakes and its a really off-putting thing especially in exam pressure when you put a lot of time into it and still get a wrong answer

Any technique you choose will entail a number of calculations, so if you make a mistake on one of them, it will affect your answer.

There is another method that you didn't mention. Let's say that you are given a line L and a point P not on the line, and that $\vec v$ is a vector in the same direction as the line.
Pick any arbitrary point Q on the line.
Find the vector projection of $\overrightarrow{QP}$ in the direction of $\vec v$. Some textbooks denote this as $\overrightarrow{Proj_v} \vec{QP}$. This is equal to $\frac{\vec v \cdot \overrightarrow{QP}}{|\vec v|} \frac{\vec v}{|\vec v|} = \left( \frac {\vec v \cdot \overrightarrow{QP}} {|\vec v|^2} \right)\vec v$.
The vector projection of $\overrightarrow{QP}$ in the direction of $\vec v$ can be thought of as the leg of a right triangle that lies on the line, that extends from the point Q to the point R that is closest to the given point P. The hypotentuse of the right triangle is of length $|\overrightarrow{QP}|$. The other leg of the right triangle is the vector from R to the given point P.
If you know one leg of a right triangle and the hypotenuse, you can get the other leg (the distance you need to find) in a straightforward fashion.

 

[Krushnaraj Pandya]

Any technique you choose will entail a number of calculations, so if you make a mistake on one of them, it will affect your answer.

There is another method that you didn't mention. Let's say that you are given a line L and a point P not on the line, and that $\vec v$ is a vector in the same direction as the line.
Pick any arbitrary point Q on the line.
Find the vector projection of $\overrightarrow{QP}$ in the direction of $\vec v$. Some textbooks denote this as $\overrightarrow{Proj_v} \vec{QP}$. This is equal to $\frac{\vec v \cdot \overrightarrow{QP}}{|\vec v|} \frac{\vec v}{|\vec v|} = \left( \frac {\vec v \cdot \overrightarrow{QP}} {|\vec v|^2} \right)\vec v$.
The vector projection of $\overrightarrow{QP}$ in the direction of $\vec v$ can be thought of as the leg of a right triangle that lies on the line, that extends from the point Q to the point R that is closest to the given point P. The hypotentuse of the right triangle is of length $|\overrightarrow{QP}|$. The other leg of the right triangle is the vector from R to the given point P.
If you know one leg of a right triangle and the hypotenuse, you can get the other leg (the distance you need to find) in a straightforward fashion.

That seems like a much better alternative to the methods I mentioned. Thank you very much.