Question

Find the length of perpendicular from point $\left( 1,0,2 \right)$ to the line $\dfrac{x+1}{3}=\dfrac{y-2}{-2}=\dfrac{z+1}{-1}$

Answer 1

We solve this problem by using some standard condition and formula of points and lines in the 3D plane. Let us draw a rough figure that represents the given information

First we find the co – ordinates of point Q by assuming that $\dfrac{x+1}{3}=\dfrac{y-2}{-2}=\dfrac{z+1}{-1}=k$ because the point Q lies on line equation. Now, we find the directional ratios of line PQ using the formula of directional rations of two points $A\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right),B\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$ is given as
$DR=\left( \left( {{x}_{2}}-{{x}_{1}} \right),\left( {{y}_{2}}-{{y}_{1}} \right),\left( {{z}_{2}}-{{z}_{1}} \right) \right)$
Then we find the directional ratios of a given line. The directional ratios of line of the form $\dfrac{x+{{x}_{1}}}{a}=\dfrac{y+{{y}_{1}}}{b}=\dfrac{z+{{z}_{1}}}{c}$ are given as $\left( a,b,c \right)$
Then we use the condition for the directional ratios of two perpendicular lines that is if the two lines having the directional ratios $\left( {{a}_{1}}.{{b}_{1}},{{c}_{1}} \right),\left( {{a}_{2}},{{b}_{2}},{{c}_{2}} \right)$ is given as
$\Rightarrow {{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}=0$
By using these conditions we find the point Q then we use the distance formula that is the distance between two points $A\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right),B\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$ is given as
$\Rightarrow AB=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}+{{\left( {{z}_{1}}-{{z}_{2}} \right)}^{2}}}$

Complete step by step answer:
We are given that the equation of line as
$\dfrac{x+1}{3}=\dfrac{y-2}{-2}=\dfrac{z+1}{-1}$
We know that the directional ratios of line of the form $\dfrac{x+{{x}_{1}}}{a}=\dfrac{y+{{y}_{1}}}{b}=\dfrac{z+{{z}_{1}}}{c}$ is given as $\left( a,b,c \right)$
By using the above formula we get the directional ratios of given line as
$\Rightarrow {{\left( DR \right)}_{line}}=\left( 3,-2,-1 \right)$
Now, let us find the co – ordinates of point Q
Let us assume that the given line equation as
$\dfrac{x+1}{3}=\dfrac{y-2}{-2}=\dfrac{z+1}{-1}=k$
Now, by the first term to constant we get

& \Rightarrow \dfrac{x+1}{3}=k \\\ & \Rightarrow x=3k-1 \\\ \end{aligned}$$ Now, by taking the second term to constant we get $$\begin{aligned} & \Rightarrow \dfrac{y-2}{-2}=k \\\ & \Rightarrow y=-2k+2 \\\ \end{aligned}$$ Similarly, by taking the third term with the constant we get $$\begin{aligned} & \Rightarrow \dfrac{z+1}{-1}=k \\\ & \Rightarrow z=-k-1 \\\ \end{aligned}$$ So, we can say that the co – ordinates of point Q are $$\left( 3k-1,-2k+2,-k-1 \right)$$ Now, let us find the directional ratios of line PQ. We know that the directional rations of two points $$A\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right),B\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$$ is given as $$DR=\left( \left( {{x}_{2}}-{{x}_{1}} \right),\left( {{y}_{2}}-{{y}_{1}} \right),\left( {{z}_{2}}-{{z}_{1}} \right) \right)$$ By using the above formula we get the directional ratios of line PQ as $$\begin{aligned} & \Rightarrow {{\left( DR \right)}_{PQ}}=\left( \left( 3k-1-1 \right),\left( -2k+2-0 \right),\left( -k-1-2 \right) \right) \\\ & \Rightarrow {{\left( DR \right)}_{PQ}}=\left( 3k-2,-2k+2,-k-3 \right) \\\ \end{aligned}$$ We now that the condition of perpendicular lines that is if the two lines having the directional ratios $$\left( {{a}_{1}}.{{b}_{1}},{{c}_{1}} \right),\left( {{a}_{2}},{{b}_{2}},{{c}_{2}} \right)$$ is given as $$\Rightarrow {{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}=0$$ By using the above formula to given line and the line PQ we get $$\begin{aligned} & \Rightarrow 3\left( 3k-2 \right)-2\left( -2k+2 \right)-1\left( -k-3 \right)=0 \\\ & \Rightarrow 14k=+7 \\\ & \Rightarrow k=\dfrac{1}{2} \\\ \end{aligned}$$ Now, by substituting the value of $$'k'$$ in the point $$Q\left( 3k-1,-2k+2,-k-1 \right)$$ we get $$\Rightarrow Q=\left( \dfrac{1}{2},1,\dfrac{-3}{2} \right)$$ We know that the distance between two points $$A\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right),B\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$$ is given as $$\Rightarrow AB=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}+{{\left( {{z}_{1}}-{{z}_{2}} \right)}^{2}}}$$ By using the above formula to the points P and Q we get $$\begin{aligned} & \Rightarrow PQ=\sqrt{{{\left( 1-\dfrac{1}{2} \right)}^{2}}+{{\left( 0-1 \right)}^{2}}+{{\left( 2-\left( \dfrac{-3}{2} \right) \right)}^{2}}} \\\ & \Rightarrow PQ=\sqrt{\dfrac{1}{4}+1+\dfrac{49}{4}} \\\ & \Rightarrow PQ=\sqrt{\dfrac{54}{4}} \\\ \end{aligned}$$ Now by replacing the value of $$\sqrt{54}$$ by $$3\sqrt{6}$$ we get the length of PQ as $$\Rightarrow PQ=\dfrac{3\sqrt{6}}{2}$$ **Therefore the length of perpendicular from point $$\left( 1,0,2 \right)$$ to the line $$\dfrac{x+1}{3}=\dfrac{y-2}{-2}=\dfrac{z+1}{-1}$$ is $$\dfrac{3\sqrt{6}}{2}$$** **Note:** Here, we used so many results and formulas to solve this problem. There may be a possibility of taking any one of the conditions and formulas as wrong. We have the condition for perpendicular lines in terms of directional ratios that is the condition of perpendicular lines that is if the two lines having the directional ratios $$\left( {{a}_{1}}.{{b}_{1}},{{c}_{1}} \right),\left( {{a}_{2}},{{b}_{2}},{{c}_{2}} \right)$$ is given as $$\Rightarrow {{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}=0$$ But, we know that for the two perpendicular lines having the slopes $${{m}_{1}},{{m}_{2}}$$ is $$\Rightarrow {{m}_{1}}\times {{m}_{2}}=-1$$ Students may take both the slopes and directional ratios condition and take the above formula as $$\Rightarrow {{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}=-1$$ This gives the wrong answer because the DR’s and slopes are different.