Question

What would be the length of the perpendicular from the point (2,-1, 4) on the straight line $\dfrac{x+3}{10}=\dfrac{y-2}{-7}=\dfrac{z}{1}$?
A. less than 2
B. greater than 3 but less than 4
C. greater than 4
D. greater than 2 but less than 3

Answer 1

Here, we can start by equating a given equation of line with constant r. Then, we will get coordinates of a point, A, in the terms of r. We can also get the direction ratios of the given line as (10, -7, 1). Then, we will also get the direction ratios of the perpendicular drawn to the given line by subtracting P(2, -1, 4) and A. We know the condition for perpendicular lines is given by ${{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}=0$, where $\left( {{a}_{1}},{{b}_{1}},{{c}_{1}} \right)\text{ and }\left( {{a}_{2}},{{b}_{2}},{{c}_{2}} \right)$ are direction ratios. Applying this, we will get the value of r and hence the coordinates of point A. Then, we can find the length of perpendicular PA using distance formula $PA=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{z}_{2}}-{{z}_{1}} \right)}^{2}}}$.

Complete step by step answer:
Equation of given straight line is
$\dfrac{x+3}{10}=\dfrac{y-2}{7}=\dfrac{z}{1}=r$...(i)
Now, we can solve this equation to get values of x, y and z. Here, we get equation for x as follows,

& \Rightarrow \dfrac{x+3}{10}=r \\\ & \Rightarrow x=10r-3 \\\ \end{aligned}$$ Now, we get the equation for y as $$\begin{aligned} & \Rightarrow \dfrac{y-2}{7}=r \\\ & \Rightarrow y=7r+2 \\\ \end{aligned}$$ Now, we will get the equation for z as $$\begin{aligned} & \Rightarrow \dfrac{z}{1}=r \\\ & \Rightarrow z=r \\\ \end{aligned}$$ Thus, we know the direction ratios of the line (i) are $$\left( 10,-7,1 \right)$$ and any point lying on this line will have coordinates as $$A\left( 10r-3,-7r+2,r \right)$$ .  Now, we know that the line joining the points $$P\left( 2,-1,4 \right)$$ and $$A\left( 10r-3,-7r+2,r \right)$$ is perpendicular to line (i). We can get the direction ratios of this line by subtracting the direction ratios of line containing P and A. Direction ratio of perpendicular line will be $$\begin{aligned} & \left( 10r-3-2,-7r+2+1,r-4 \right) \\\ & \Rightarrow \left( 10r-5,-7r+3,r-4 \right) \\\ \end{aligned}$$ Now, we know that the direction ratio of the given line (i) is (10, -7, 1). We also know the condition for perpendicular lines when direction ratios $\left( {{a}_{1}},{{b}_{1}},{{c}_{1}} \right)\text{ and }\left( {{a}_{2}},{{b}_{2}},{{c}_{2}} \right)$ are known is ${{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}=0$ . Now, applying this, we get $\begin{aligned} & \left[ \left( 10r-5 \right)\times 10 \right]+\left[ \left( -7r+3 \right)\times -7 \right]+\left[ \left( r-4 \right)\times 1 \right]=0 \\\ & \Rightarrow 100r-50+49r-21+r-4=0 \\\ & \Rightarrow 150r-75=0 \\\ & \Rightarrow 150r=75 \\\ & \Rightarrow r=\dfrac{1}{2} \\\ \end{aligned}$ Now, substituting r in A, we get $$A\left( 10\times \dfrac{1}{2}-3,-7\times \dfrac{1}{2}+2,\dfrac{1}{2} \right)\Rightarrow A\left( 2,-\dfrac{3}{2},\dfrac{1}{2} \right)$$ . So, the foot of perpendicular is $$A\left( 2,-\dfrac{3}{2},\dfrac{1}{2} \right)$$. Now, we can find the perpendicular distance of point P(2,−1,4) from the line (i), i.e $$A\left( 2,-\dfrac{3}{2},\dfrac{1}{2} \right)$$ as $$\begin{aligned} & PA=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{z}_{2}}-{{z}_{1}} \right)}^{2}}} \\\ & \Rightarrow \sqrt{{{\left( 2-2 \right)}^{2}}+{{\left( -\dfrac{3}{2}+1 \right)}^{2}}+{{\left( \dfrac{1}{2}-4 \right)}^{2}}} \\\ & \Rightarrow \sqrt{0+{{\left( \dfrac{-1}{2} \right)}^{2}}+{{\left( \dfrac{-7}{2} \right)}^{2}}} \\\ \end{aligned}$$ On simplification, we get $$\begin{aligned} & PA=\sqrt{\dfrac{1}{4}+\dfrac{49}{4}}=\sqrt{\dfrac{50}{4}} \\\ & =\dfrac{\sqrt{50}}{2} \\\ & =\dfrac{5\sqrt{2}}{2} \\\ & =\dfrac{5}{2}\times 1.414 \\\ & =3.535 \\\ \end{aligned}$$ We can see that it lies between 3 and 4. **So, the correct answer is “Option B”.** **Note:** Students should know how to calculate direction ratios of a given line and a perpendicular. We can also use the fact that the dot product of perpendicular lines will be 0. So, we could have obtained the vector of given line as $10\hat{i}-7\hat{j}+\hat{k}$ and for the perpendicular line as $$\left( 10r-5 \right)\hat{i}+\left( -7r+3 \right)\hat{j}+\left( r-4 \right)\hat{k}$$. Applying the dot product and simplifying, we get $\begin{aligned} & \left[ \left( 10r-5 \right)\times 10 \right]+\left[ \left( -7r+3 \right)\times -7 \right]+\left[ \left( r-4 \right)\times 1 \right]=0 \\\ & \Rightarrow 100r-50+49r-21+r-4=0 \\\ & \Rightarrow 150r-75=0 \\\ & \Rightarrow 150r=75 \\\ & \Rightarrow r=\dfrac{1}{2} \\\ \end{aligned}$ Thus, we have got the value of r and then we can proceed as usual.