Question

intercepts made by the line of shortest distance.

In the following problem,
The line of shortest distance of the lines (x/2)=(-y/3)=(z) and (x-2)/3=(y-1)/-5=(z+2)/2
intersects these two lines in P andQ respectively.
1.find the coordinates of P
2.find the coordinates of Q.
i can find the shortest distance between these two lines.but how to find the points of intersection with the line of shortest distance?

Answer 1

Question#6411. The line of shortest distance of the lines $(x/2) = (-y/3) = (z)$ and $(x-2)/3 = (y-1)/5 = (z+2)/2$ intersects these two lines in $P$ and $Q$ respectively.

Find the shortest distance between these two lines, find the coordinates of $P$ and of $Q$.

Solution

$$l_1: \frac{x}{2} = \frac{y}{-3} = \frac{z}{1}$$ $$l_2: \frac{x - 2}{3} = \frac{y - 1}{-5} = \frac{z + 2}{2}$$

The shortest distance between these two lines we will find by the formula:

$$d = \frac{|M_1 M_2 * \vec{a}|}{|\vec{a}|}$$ $$\vec{a} = \vec{s}_1 \times \vec{s}_2 = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 2 & -3 & 1 \\ 3 & -5 & 2 \end{vmatrix} = -\vec{i} - \vec{j} - \vec{k};$$ $$M_1 M_2 = (-2 - 1; -1 - 0; 2 - 0) = (-3; -1; 2)$$ $$|\vec{a}| = \sqrt{3}$$ $$|M_1 M_2 * \vec{a}| = (-3) * (-1) + (-1) * (-1) + (2) * (-1) = 2$$ $$d = \frac{|M_1 M_2 * \vec{a}|}{|\vec{a}|} = \frac{2}{\sqrt{3}} \quad \text{— the shortest distance between the lines } l_1 \text{ and } l_2$$

The canonical equation of the line, which is the shortest distance between the lines $l_1$ and $l_2$:

$$\frac{x - a}{m} = \frac{y - b}{n} = \frac{z - c}{k}, \text{ where } \vec{a} = (m; n; k) = (-1; -1; -1)$$ $$\frac{x - a}{-1} = \frac{y - b}{-1} = \frac{z - c}{-1}$$