Question

How can I find the shortest distance between the point $(0,1, - 1)$ and the line $(x,y,z) = (2,1,3) + t(3, - 1, - 2)$ ?

Answer 1

Hint : We are given a point and a line in the form of a direction vector equation. First we will use the direction of the line to find the general equation of the plane. Then with the help of the point we will find the specific equation. Then we will be having the point and the value of the parametric equation. Then using the distance formula we will find the distance between the line and point.

Complete step by step solution:
Given that, point $(0,1, - 1)$ and the line $(x,y,z) = (2,1,3) + t(3, - 1, - 2)$
We are given two points that are already on the line.
The direction of the given line vector is $(3, - 1, - 2)$.
We know that general form of plane is $3x - y - 2z = 0$
Now substituting the values of the coordinates of the points $(0,1, - 1)$ so given we will find the exact or specific equation of the plane.
$3\left( 0 \right) - 1 - 2\left( { - 1} \right) = c$
On solving we get the value of c as,

\- 1 + 2 = c \\\ c = 1 \\\ \end{gathered} $$ Now the plane that contains the point very near to the point so given is, $$3x - y - 2z = 1$$ Now we need to find the parametric equation for the line so given, $$(x,y,z) = (2,1,3) + t(3, - 1, - 2)$$ So , $$\begin{gathered} x = 3t + 2 \\\ y = 1 - t \\\ z = 3 - 2t \\\ \end{gathered} $$ These are the three parametric equations. Now we need to find the value of t. Substituting the values in the equation of the plane above. $$3\left( {3t + 2} \right) - \left( {1 - t} \right) - 2\left( {3 - 2t} \right) = 1$$ On solving the brackets, $$9t + 6 - 1 + t - 6 + 4t = 1$$ Taking t terms on one side, $$9t + t + 4t + 6 - 1 - 6 - 1 = 0$$ On adding we get, $$14t - 2 = 0$$ To find the value of t transpose 2 on other side, $$14t = 2$$ $$t = \dfrac{2}{{14}}$$ On dividing we get. $$t = \dfrac{1}{7}$$ Now using the value of t we can find the exact value of the parametric equation, $$x = \dfrac{3}{7} + 2$$ Taking LCM we get, $$\begin{gathered} x = \dfrac{{3 + 14}}{7} \\\ x = \dfrac{{17}}{7} \\\ \end{gathered} $$ For y , $$\begin{gathered} y = 1 - t \\\ y = 1 - \dfrac{1}{7} \\\ \end{gathered} $$ Taking LCM, $$\begin{gathered} y = \dfrac{{7 - 1}}{7} \\\ y = \dfrac{6}{7} \\\ \end{gathered} $$ Lats for the value of z, $$\begin{gathered} z = 3 - 2t \\\ z = 3 - \dfrac{2}{7} \\\ \end{gathered} $$ Taking LCM, $$\begin{gathered} z = \dfrac{{21 - 2}}{7} \\\ z = \dfrac{{19}}{7} \\\ \end{gathered} $$ Now we have the point and the value of the parametric equation $$(0,1, - 1)\& \left( {\dfrac{{17}}{7},\dfrac{6}{7},\dfrac{{19}}{7}} \right)$$ also. Now we will use distance formula to find the shortest distance, $$d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} $$ Putting the values, $$d = \sqrt {{{\left( {\dfrac{{17}}{7} - 0} \right)}^2} + {{\left( {\dfrac{6}{7} - 1} \right)}^2} + {{\left( {\dfrac{{19}}{7} - \left( { - 1} \right)} \right)}^2}} $$ On solving with LCM, $$d = \sqrt {{{\left( {\dfrac{{17}}{7}} \right)}^2} + {{\left( {\dfrac{{ - 1}}{7}} \right)}^2} + {{\left( {\dfrac{{19}}{7} + 1} \right)}^2}} $$ $$d = \sqrt {{{\left( {\dfrac{{17}}{7}} \right)}^2} + {{\left( {\dfrac{{ - 1}}{7}} \right)}^2} + {{\left( {\dfrac{{26}}{7}} \right)}^2}} $$ Taking the square terms, $$d = \sqrt {\dfrac{{{{17}^2} + {{\left( { - 1} \right)}^2} + {{26}^2}}}{{{7^2}}}} $$ Now denominator is under root but is squared, $$d = \dfrac{{\sqrt {289 + 1 + 676} }}{7}$$ On adding, $$d = \dfrac{{\sqrt {966} }}{7}$$ This value is almost equals to, $$d \approx 4.44$$ This is the required answer. **So, the correct answer is “$$d \approx 4.44$$”.** **Note** : Note that though we have to find the distance between the line and point we have given the formula. We will proceed with the method mentioned above. We also can take help of vector methods. But that will be lengthy and complicated to understand. Only don’t forget to find the value of the parametric equation that is having the initial and end points.