Question

Point P(0, 1,−2) to the line given by $r\left(t\right)=\left(1+5t,-6+t,-4t\right)$

Short answer

The answer is $\sqrt{\frac{1084}{21}}$units.

Step-by-step solution

Given information

A point $P(0,1,-2)$ to the line by $r\left(t\right)=(1+5t,-6+t,-4t)$

Calculation

The goal is to calculate the distance between the point and the line.

The formula for the distance is $\frac{||d\times {\overrightarrow{P}}_{0}P||}{\Vert d\Vert }$

Let the point $P$is $P(0,1,-2)$

The point $P_0$ on the line equation $r\left(t\right)=(1+5t,-6+t,-4t)$ is $(1,-6,0)$

The direction vector $\overrightarrow{P_{0}P}=(0-1,1-(-6),-2-0)$

Then $\overrightarrow{P_{0}P}=(-1,7,-2)$

Take the direction vector of the equation $r\left(t\right)=(1+5t,-6+t,-4t)$is $d=(5,1,-4)$

Substitute the values $d=(5,1,-4)$and $\overrightarrow{P_{0}P}=(-1,7,-2)$in the formula $\underset{¯}{\Vert d\times \overrightarrow{P_{0}P}}$

Then the distance $=\frac{\Vert (5,1,-4)\times (-1,7,-2)\Vert }{\Vert (5,1,-4)\Vert }$

The cross product of $(5,1,-4)\times (-1,7,-2)$ is calculated as follows,

$(5,1,-4)\times (-1,7,-2)=\left|\begin{array}{c}ijk\\ 51-4\\ -17-2\end{array}\right|$

$$\begin{gathered} =i(-2+28)-j(-10-4)+k(35+1) \\ =26i+14j+36k \end{gathered}$$

Calculation

Thus,

$$\begin{gathered} \text{Distance}=\frac{\Vert 26i+14j+36k\Vert }{\Vert (5,1,-4)\Vert } \\ \text{Distance}=\frac{\sqrt{{26}^2+{14}^2+{36}^2}}{\sqrt{5^2+1^2+(-4{)}^2}} \\ \text{Distance}=\frac{\sqrt{676+196+1296}}{\sqrt{25+1+16}} \\ \text{Distance}=\frac{\sqrt{2168}}{\sqrt{42}} \end{gathered}$$

On further simplification,

Distance $=\sqrt{\frac{1084}{21}}$ units

Thus, the distance between the point $P(0,1,-2)$ and the line $r\left(t\right)=(1+5t,-6+t,-4t)$ is $\sqrt{\frac{1084}{21}}$ units.

Therefore, the answer is $\sqrt{\frac{1084}{21}}$ units.