Find the nearest point on a 3D parametric line

Question

Question: A 3D line is given by

$$r(t) = (8, 8, 9) + t(9, 4, 5), t ∈ R $$

Let $P(x, y, z)$ be the point on the line that is nearest to the the origin. Find the point $(x, y, z)$.

I know I can find the normal, n to vector v(9,4,5) by doing cross product of u x v where u is just origin to point (8,8,9). By doing cross product of u x v, I got my n to be (-5,41,-40). From here, I compare the ||n|| which is the vertical distance by finding another normal to the line which is OP x v.

If I equate this ||n|| = ||OP x v||, I would get

$$||(9y-8z, 9x+8z ,8x-8y)|| = ||(-5,41,-40)||$$

Finally, by simultaneous equation where by

$$9y-8z=-5,\>\>\> 9x+8z=41,\>\>\>8x-8y=-40$$

and solve it to be $y=5,\>\>\> x=-1,\>\>\> z= 25/4$

which is the wrong answer. The answer to the question is

$$x=-365/122,\>\>\> y= 190/61, \>\>\>z = 353/122$$

But,I simply can't figure this out.

Attached: (A sketch on this question) which I'm not sure is accurate or not.

Answer 1

Let $\vec{a}$ and $\vec{b}$ represent two points on the lines. Use the area formula and the dot product below,

$$\frac 12 d |\vec{a}-\vec{b}|=\frac 12 ab \sin\theta, \>\>\>\vec{a}\cdot \vec{b}=ab \cos\theta$$

to get the distance to the nearest point,

$$d^2= \frac{ (ab)^2-(\vec{a}\cdot \vec{b})^2}{(\vec{a}-\vec{b})^2}$$

Then, just pick two points from the curve above, say, (8,8,9) and (-1,4,4), and plug them into the above expression,

$$d^2=\frac{3297}{122}$$

The nearest point can be solved by letting,

$$\frac{3297}{122}=(8+9t)^2+(8+4t)^2+(9+5t)^2$$

which can be simplified to

$$t^2+\frac{298}{122}t+\frac{22201}{122^2}= \left(t+\frac{149}{122}\right)^2=0$$

Therefore, $t=-\frac{149}{122}$ and the nearest point is

$$(8, 8, 9) -(9, 4, 5)\frac{149}{122}=\left(\frac{365}{122}, \frac{ 190}{61}, \frac{353}{122} \right)$$

Answer 2

So $$OP=\sqrt{x^2+y^2+z^2}=\sqrt{(8+9t)^2+(8+4t)^2+(9+5t)^2}=f(t)$$ Determine the minimum of $$f(t)$$

Answer 3

I don’t entirely follow the reasoning in your attempt, but there’s the germ of an idea in there of looking at vectors perpendicular to the line’s direction. The nearest point to the origin will lie on some line through the origin that’s perpendicular to the given line, i.e., that $r(t)\cdot(9,4,5)=0$. This is a simple linear equation in $t$ to be solved.

Equivalently, find the point at which the given line intersects its perpendicular plane through the origin. You should be able to write down an equation for the latter directly from the parametric equation of the line. Substituting the parameterization into this equation produces the same equation in $t$ as above.

Answer 4

You have line $\vec{r}(t)$, $$\vec{r}(t) = (8, 8, 9) + t (9, 4, 5), ~ t \in \mathbb{R}$$ The distance between points $\vec{r}(t)$ and $\vec{p}$ is $\lVert \vec{p} - \vec{r}(t)\rVert$. Because it is nonnegative everywhere, it reaches a minimum whenever its square reaches a minimum.

A continuous function reaches its minimum or maximum when its derivative is zero. In this case, the function is the squared distance between origin and the point $t$ on the line $\vec{r}$: $$f(t) = \left\lVert \vec{r}(t) \right\rVert^2 = \vec{r}(t) \cdot \vec{r}(t) = (8 + 9 t)^2 + (8 + 4 t)^2 + (9 + 5 t)^2 = 122 t^2 + 298 t + 209$$ Its derivative is $$\frac{d f(t)}{d t} = 244 t + 298$$ This reaches zero at $$\frac{d f(t)}{d t} = 244 t + 298 = 0 ~ ~ \iff ~ ~ t = -\frac{149}{122}$$ thus line $\vec{r}(t)$ is closest to origin at $t = -149/122$, i.e. $$\vec{r}\left(-\frac{149}{122}\right) = (8, 8, 9) -\frac{149}{122}(9, 4, 5)$$ i.e. at $$\left(-\frac{365}{122}, \frac{190}{61}, \frac{353}{122}\right) = \frac{1}{122}(-365, 380, 353)$$

This method works for all single-variate curves, not just lines. However, only lines and quadratic curves (having $t^2$ but no higher powers of $t$) always have an algebraic solution; many/most higher order curves are only solvable numerically. Note that for each minimum of the maximum, you must check the actual distance to determine if it is a minimum or maximum; for lines, this is not necessary, because they can only have a minimum (and always have exactly one minimum for points not at infinity).