Find the distance between the origin and the line given by $\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z}{1}$

Question

Find the distance between the origin and the line given by $$\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z}{1}$$

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The parametric equation of the line is given by $$l=(1,-1,0)+t(2,-3,1)$$

So every point on the line is in the for $$(1+2t,-1-3t,t)$$

The distance between a point on the line to the origin is :$$\sqrt{\left(1+2t\right)^{2}+\left(-1-3t\right)^{2}+\left(t\right)^{2}}=\sqrt{14t^{2}+10t+2}$$

So the distance (the miniumum distance) happens when $14t^{2}+10t+2$ is minimum which happens at $t=-5/14$ so the desired answer is:$$\sqrt{14\left(-\frac{5}{14}\right)^{2}+10\left(-\frac{5}{14}\right)+2}\approx 0.462910049886$$

That was what I thought, I want to know how much of this work is correct.

Answer 1

Alternatively, you can intersect the line with the orthogonal plane by the origin,

$$2x-3y+z=0.$$

Plugging $x=2z+1$ and $y=-3z-1$, we obtain the point

$$\left(\frac4{14}, \frac1{14}, -\frac5{14}\right)$$ and the distance is

$$\frac{\sqrt{42}}{14}.$$

Answer 2

Your working is correct. But i would suggest you an alternative way. The minimum distance is always the perpendicular distance, which occurs when the dot product of direction ratios of the line and the vector joining origin to a point on the line is 0. From that you can directly find $t$ in your equation without optimising the function using derivatives.

Answer 3

By eliminating $x,y$, we minimize

$$x^2+y^2+z^2=(2z+1)^2+(-3z-1)+z^2=14\left(z+\frac5{14}\right)^2+2-\frac{5^2}{14}$$

and the constant term gives you the squared shortest distance.

Answer 4

Even more alternatively, you may compute the length of the cross product of $(1,-1,0)$ and the unit vector in direction $(2,-3,1)$.