Question

Find the distance of the point (-2, 3, -4) from the line $\dfrac{{x + 2}}{3} = \dfrac{{2y + 3}}{4} = \dfrac{{3z + 4}}{5}$ measured parallel to the plane $4x + 12y - 3z + 1 = 0$ .

Answer 1

In order to find the distance of the point from the line parallel to the plane, first we will find the point which is at the shortest distance from the given point parallel to the given plane and lies on the given line. Then easily with the help of the distance formula for the three-dimensional coordinate system we will find the distance.

Complete step-by-step answer :
Given point is (-2, 3, -4)
We need to find the distance from the line $\dfrac{{x + 2}}{3} = \dfrac{{2y + 3}}{4} = \dfrac{{3z + 4}}{5}$
Parallel to the plane $4x + 12y - 3z + 1 = 0$

First, we will find the point which is at the shortest distance from the given point parallel to the given plane and lies on the given line.
For finding the general point on the line.
Let $\dfrac{{x + 2}}{3} = \dfrac{{2y + 3}}{4} = \dfrac{{3z + 4}}{5} = \lambda$
So, we have coordinates of general point as
$\because \dfrac{{x + 2}}{3} = \lambda \\\ \Rightarrow x + 2 = 3\lambda \\\ \Rightarrow x = 3\lambda - 2 \\\ \because \dfrac{{2y + 3}}{4} = \lambda \\\ \Rightarrow 2y + 3 = 4\lambda \\\ \Rightarrow 2y = 4\lambda - 3 \\\ \Rightarrow y = \dfrac{{4\lambda - 3}}{2} \\\ \because \dfrac{{3z + 4}}{5} = \lambda \\\ \Rightarrow 3z + 4 = 5\lambda \\\ \Rightarrow 3z = 5\lambda - 4 \\\ \Rightarrow z = \dfrac{{5\lambda - 4}}{3} \\\$
So, the general point on the line is:
$\left( {3\lambda - 2,\dfrac{{4\lambda - 3}}{2},\dfrac{{5\lambda - 4}}{3}} \right)$
Now the two points are:
$\left( { - 2,3, - 4} \right)$and $\left( {3\lambda - 2,\dfrac{{4\lambda - 3}}{2},\dfrac{{5\lambda - 4}}{3}} \right)$
As we know that for general points $\left( {{x_1},{y_1},{z_1}} \right)\& \left( {{x_2},{y_2},{z_2}} \right)$
The direction ratio between the points is given by:
$\left( {{x_1} - {x_2},{y_1} - {y_2},{z_1} - {z_2}} \right)$
So, for the given two points the direction ratio is given by
$= \left( {3\lambda - 2 - \left( { - 2} \right),\dfrac{{4\lambda - 3}}{2} - 3,\dfrac{{5\lambda - 4}}{3} - \left( { - 4} \right)} \right) \\\ = \left( {3\lambda ,\dfrac{{4\lambda - 9}}{2},\dfrac{{5\lambda + 8}}{3}} \right) \\\$
As we know that the distance is measure parallel to the plane $4x + 12y - 3z + 1 = 0$
So, using the above direction ration in the plane without the constant part, we can find the value of $\lambda$
$4x + 12y - 3z + 1 = 0 \\\ \Rightarrow 4\left( {3\lambda } \right) + 12\left( {\dfrac{{4\lambda - 9}}{2}} \right) - 3\left( {\dfrac{{5\lambda + 8}}{3}} \right) = 0 \\\$
Let us solve the above equation to find the value of $\lambda$
$\Rightarrow 12\lambda + 24\lambda - 54 - 5\lambda - 8 = 0 \\\ \Rightarrow 31\lambda - 62 = 0 \\\ \Rightarrow 31\lambda = 62 \\\ \Rightarrow \lambda = \dfrac{{62}}{{31}} \\\ \Rightarrow \lambda = 2 \\\$
Let us put the value of $\lambda$ in the general coordinate of the line to find the nearest point.
Since second point is $\left( {3\lambda - 2,\dfrac{{4\lambda - 3}}{2},\dfrac{{5\lambda - 4}}{3}} \right)$ for $\lambda = 2$
So, the point is:

= \left( {3\left( 2 \right) - 2,\dfrac{{4\left( 2 \right) - 3}}{2},\dfrac{{5\left( 2 \right) - 4}}{3}} \right) \\\ = \left( {4,\dfrac{5}{2},2} \right) \\\ \end{gathered} $$ Now as we know the distance formula for two points is given by: $D = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} $ Let use the above formula to find the distance: $ \Rightarrow D = \sqrt {{{\left( {4 - \left( { - 2} \right)} \right)}^2} + {{\left( {\dfrac{5}{2} - 3} \right)}^2} + {{\left( {2 - \left( { - 4} \right)} \right)}^2}} \\\ \Rightarrow D = \sqrt {{{\left( {4 + 2} \right)}^2} + {{\left( {\dfrac{5}{2} - 3} \right)}^2} + {{\left( {2 + 4} \right)}^2}} \\\ \Rightarrow D = \sqrt {{{\left( 6 \right)}^2} + {{\left( { - \dfrac{1}{2}} \right)}^2} + {{\left( 6 \right)}^2}} \\\ \Rightarrow D = \sqrt {36 + \dfrac{1}{4} + 36} \\\ \Rightarrow D = \sqrt {\dfrac{{289}}{4}} \\\ \Rightarrow D = \sqrt {\dfrac{{17 \times 17}}{{2 \times 2}}} \\\ \Rightarrow D = \dfrac{{17}}{2}units \\\ $ Hence, the distance between the given point and the line parallel to the given plane is $\dfrac{{17}}{2}units$ **Note** : In order to solve such problems related to three-dimensional geometry students must remember the formulas in terms of three-dimensional coordinates. In mathematics, analytic geometry (also called Cartesian geometry) describes every point in three-dimensional space by means of three coordinates. Three coordinate axes are given, each perpendicular to the other two at the origin, the point at which they cross. They are usually labeled x, y, and z.