# 12.3.63 Determine the smallest distance between point and a line

• MHB
• karush
In summary, the distance between the point $(1,1,1)$ and the line $L$ with direction $\langle -4,-5,8 \rangle$ can be found using the formula $D_{\min}=\frac{\left|\left(\vec{P_2}-\vec{P_1}\right)\times\left(\vec{P_1}-\vec{P_0}\right)\right|}{\left|\vec{P_2}-\vec{P_1}\right|}$, where $\vec{P_0}=\langle 1,1,1 \rangle$, $\vec{P_1}=\langle 0,0,0 \rangle$, and $\vec karush Gold Member MHB$\text{Determine the smallest distance between point}$$$P(1,1,1)$$$\textsf{ and the line $L$ through the origin $L$ has the direction}$$$\langle -4,-5,8 \rangle$$ ok just barely had time to post this but was ?? about direction presume going off of this$\textit{Distance between point and line }\textit{ Where A,B, and C are coeficients of line equation And M, N are coordinates of a point}\begin{align*}\displaystyle d&=\frac{|Am+Bn+C|}{\sqrt{A^2+B^2}} \end{align*} Last edited: A vector along the line is given by: $$\displaystyle \vec{v}=\left[\begin{array}{c}-4t \\ -5t \\ 8t \end{array}\right]$$ So, what is the square of the distance between a point on the line with parametert$and the point$(1,1,1)$? MarkFL said: A vector along the line is given by: $$\displaystyle \vec{v}=\left[\begin{array}{c}-4t \\ -5t \\ 8t \end{array}\right]$$ So, what is the square of the distance between a point on the line with parameter$t$and the point$(1,1,1)$? ok not real sure what you mean by this$\sqrt{(-4)^2+(-5)^2+(8)^2}$? karush said: ok not real sure what you mean by this$\sqrt{(-4)^2+(-5)^2+(8)^2}$? Let's back up a bit...a point on the line is: $$\displaystyle (-4t,-5t,8t)$$ And the given point is: $$\displaystyle (1,1,1)$$ What is the square of the distance between these two points? sorry this looks like duplicate post done much earlier somehow it never got recorded on the HW mod can delete :( karush said: sorry this looks like duplicate post done much earlier somehow it never got recorded on the HW mod can delete :( Yes, this is a duplicate of the thread posted here: http://mathhelpboards.com/calculus-10/231-12-3-65-determine-smallest-distance-between-point-line-22060.html The method used there was what I had in mind for finding the required distance. However, let's take this opportunity to develop a general formula. Suppose we have a line described by the two points: $$\displaystyle P_1\left(x_1,y_1,z_1\right),\,P_2\left(x_2,y_2,z_2\right)$$ And the point: $$\displaystyle P_0\left(x_0,y_0,z_0\right)$$ And so a vector along the line can be given as: $$\displaystyle \left[\begin{array}{c}x_1+\left(x_2-x_1\right)t \\ y_1+\left(y_2-y_1\right)t \\ z_1+\left(z_2-z_1\right)t \end{array}\right]$$ And thus, the square of the distance$D$between a point on the line with parameter$t$and the given point is: $$\displaystyle f(t)=D^2(t)=\left(\left(x_1-x_0\right)+\left(x_2-x_1\right)t\right)^2+\left(\left(y_1-y_0\right)+\left(y_2-y_1\right)t\right)^2+\left(\left(z_1-z_0\right)+\left(z_2-z_1\right)t\right)^2$$ Expanding, we get: $$\displaystyle f(t)=\left(\left(x_1-x_0\right)^2+\left(y_1-y_0\right)^2+\left(z_1-z_0\right)^2\right)+2\left(\left(x_1-x_0\right)\left(x_2-x_1\right)+\left(y_1-y_0\right)\left(y_2-y_1\right)+\left(z_1-z_0\right)\left(z_2-z_1\right)\right)t+\left(\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2\right)t^2$$ Using vector notation, this becomes: $$\displaystyle f(t)=\left|\vec{P_1}-\vec{P_0}\right|^2+2\left(\vec{P_1}-\vec{P_0}\right)\cdot\left(\vec{P_2}-\vec{P_1}\right)t+\left|\vec{P_2}-\vec{P_1}\right|^2t^2$$ Differentiating w.r.t$t$and equating the result to 0 and solving for$t$, we find: $$\displaystyle t=-\frac{\left(\vec{P_1}-\vec{P_0}\right)\cdot\left(\vec{P_2}-\vec{P_1}\right)}{\left|\vec{P_2}-\vec{P_1}\right|^2}$$ Observing that$f\$ is a parabolic function opening upwards, we can then state:

$$\displaystyle f_{\min}=f\left(-\frac{\left(\vec{P_1}-\vec{P_0}\right)\cdot\left(\vec{P_2}-\vec{P_1}\right)}{\left|\vec{P_2}-\vec{P_1}\right|^2}\right)=\left|\vec{P_1}-\vec{P_0}\right|^2-\frac{\left(\left(\vec{P_1}-\vec{P_0}\right)\cdot\left(\vec{P_2}-\vec{P_1}\right)\right)^2}{\left|\vec{P_2}-\vec{P_1}\right|^2}$$

$$\displaystyle f_{\min}=\frac{\left|\vec{P_1}-\vec{P_0}\right|^2\left|\vec{P_2}-\vec{P_1}\right|^2-\left(\left(\vec{P_1}-\vec{P_0}\right)\cdot\left(\vec{P_2}-\vec{P_1}\right)\right)^2}{\left|\vec{P_2}-\vec{P_1}\right|^2}$$

Using the vector quadruple product, we may write:

$$\displaystyle f_{\min}=\left(\frac{\left|\left(\vec{P_2}-\vec{P_1}\right)\times\left(\vec{P_1}-\vec{P_0}\right)\right|}{\left|\vec{P_2}-\vec{P_1}\right|}\right)^2$$

Hence:

$$\displaystyle D_{\min}=\frac{\left|\left(\vec{P_2}-\vec{P_1}\right)\times\left(\vec{P_1}-\vec{P_0}\right)\right|}{\left|\vec{P_2}-\vec{P_1}\right|}$$

Now, in the given problem, we may take:

$$\displaystyle \vec{P_0}=\langle 1,1,1 \rangle$$

$$\displaystyle \vec{P_1}=\langle 0,0,0 \rangle$$

$$\displaystyle \vec{P_2}=\langle -4,-5,8 \rangle$$

And so we have:

$$\displaystyle \vec{P_2}-\vec{P_1}=\langle -4,-5,8 \rangle$$

$$\displaystyle \vec{P_1}-\vec{P_0}=\langle -1,-1,-1 \rangle$$

And we then find:

$$\displaystyle \left(\vec{P_2}-\vec{P_1}\right)\times\left(\vec{P_1}-\vec{P_0}\right)=\langle 13,-12,-1 \rangle$$

So, we have:

$$\displaystyle D_{\min}=\sqrt{\frac{13^2+12^2+1^2}{4^2+5^2+8^2}}=\sqrt{\frac{314}{105}}\quad\checkmark$$

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## 1. What does the number 12.3.63 represent in "12.3.63 Determine the smallest distance between point and a line"?

The number 12.3.63 represents a specific problem or question that needs to be solved in regards to determining the smallest distance between a point and a line.

## 2. What is the purpose of determining the smallest distance between a point and a line?

The purpose of determining the smallest distance between a point and a line is to find the shortest distance between a given point and a given line. This can be useful in various mathematical and scientific applications, such as calculating the shortest route between two points or finding the closest approximation to a line.

## 3. How is the smallest distance between a point and a line calculated?

The smallest distance between a point and a line is calculated by using the formula d = |ax + by + c| / √(a² + b²), where a, b, and c represent the coefficients of the line's equation and x and y represent the coordinates of the given point. This formula is derived from the distance formula in coordinate geometry.

## 4. Can the smallest distance between a point and a line ever be negative?

No, the smallest distance between a point and a line cannot be negative. The distance between two objects is always a positive value, representing the length of the shortest path between the two points or objects. Therefore, the smallest distance between a point and a line will always be a positive number or zero if the point lies on the line.

## 5. What are some real-world applications of determining the smallest distance between a point and a line?

Determining the smallest distance between a point and a line has many real-world applications, such as finding the shortest distance between two cities on a map, calculating the closest approach of a comet to a planet, or determining the shortest path for a robot to navigate between two points. It can also be used in various fields of engineering, such as designing efficient road networks or optimizing flight paths for aircraft.

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