Question

The perpendicular distance of point (2, -1,4) from the line $\dfrac{x+3}{10}=\dfrac{y-2}{-7}=\dfrac{z}{1}$
lies between, choose the correct option.
A . (2. 3)
B . (3, 4)
C . (4, 5)
D . (1, 2)

Answer 1

Hint : First find the coordinates of the point A that lies of the line in terms of the constant k, where we have $\dfrac{x+3}{10}=\dfrac{y-2}{-7}=\dfrac{z}{1}=k$. Next, we will find the direction ratio of the line joining the point P (2, -1,4) and A. Also, this direction ratio is 10, -7 and 1. Now, the dot product of the two direction ratios is zero, as they are perpendicular lines. So, this gives the ration and can be used to find the value of k. Next, obtain the coordinates of the point A. Next use the distance formula $d=\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}+{{({{z}_{2}}-{{z}_{1}})}^{2}}}$between the points $P\left( 2,\text{ }-1,4 \right)$and point A, to find the distance and get the result.

Complete step-by-step answer :
In the question, we have to find the perpendicular distance of point $P\left( 2,\text{ }-1,4 \right)$from the line $\dfrac{x+3}{10}=\dfrac{y-2}{-7}=\dfrac{z}{1}$ . So, this line has the direction ratios as 10, -7 and 1. Also, we have the line
$\dfrac{x+3}{10}=\dfrac{y-2}{-7}=\dfrac{z}{1}=k$, which is then used to find any point let it be A, lying on the line. So the coordinates of the point A will be $A\left( 10k-3,\text{ }-7k+2,\text{ }k \right)$. Now, let the point $P\left( 2,\text{ }-1,4 \right)$and the point $A\left( 10k-3,\text{ }-7k+2,\text{ }k \right)$, are such that the line joining PA is perpendicular to the line$\dfrac{x+3}{10}=\dfrac{y-2}{-7}=\dfrac{z}{1}$ . So, the direction ratio of the line PA is $\left( 10k-3-2,\text{ }-7k+2+1,\text{ }k-4 \right)$
Next, the direction ratio of the line PA and the given line $\dfrac{x+3}{10}=\dfrac{y-2}{-7}=\dfrac{z}{1}$will have the dot product as zero, so we have:

& \Rightarrow \left\langle 10k-3-2,\text{ }-7k+2+1,\text{ }k-4 \right\rangle \bullet \left\langle 10,\text{ }-7\text{ , }1 \right\rangle =0 \\\ & \Rightarrow \left\langle 10k-5,\text{ }-7k+3,\text{ }k-4 \right\rangle \bullet \left\langle 10,\text{ }-7\text{ , }1 \right\rangle =0 \\\ & \Rightarrow \left( 10k-5 \right)10+\text{ (-7)}\left( -7k+3 \right)\text{+ }\left( k-4 \right)=0 \\\ & \Rightarrow 150k-75=0 \\\ & \Rightarrow k=0.5 \\\ \end{aligned}$$ So the coordinates of the point A will be: $$\begin{aligned} & \Rightarrow A\left( 10k-3,\text{ }-7k+2,\text{ }k \right) \\\ & \Rightarrow A\left( 10(0.5)-3,\text{ }-7(0.5)+2,\text{ }(0.5) \right) \\\ & \Rightarrow A\left( 2,-1.5,\text{ 0}\text{.5} \right) \\\ \end{aligned}$$ Next, we have to find the distance between the two points A and P using the distance formula. S we know that the distance between the point $$X({{x}_{1}},{{y}_{1}},{{z}_{1}})$$ and the point $$Y({{x}_{2}},{{y}_{2}},{{z}_{2}})$$ is given by $$d=\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}+{{({{z}_{2}}-{{z}_{1}})}^{2}}}$$. So here the distance between the points $$A\left( 2,-1.5,\text{ 0}\text{.5} \right)$$ and $$P\left( 2,\text{ }-1,4 \right)$$is given as: $$\begin{aligned} & \Rightarrow d=\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}+{{({{z}_{2}}-{{z}_{1}})}^{2}}} \\\ & \Rightarrow d=\sqrt{{{(2-2)}^{2}}+{{(-1+1.5)}^{2}}+{{(4-0.5)}^{2}}} \\\ & \Rightarrow d=\sqrt{(0.25)+12.25} \\\ & \Rightarrow d\approx 3.53 \\\ \end{aligned}$$ So, here we have the perpendicular distance as $$d\approx 3.53$$which lies between (3, 4). Hence, the correct answer is option B. **Note** : Here, we have to use the distance formula in three dimensions and not the two dimensions, so we have to be careful in that. Also, in the distance formula we can interchange $${{x}_{2}}$$and $${{x}_{1}}$$, where the overall result will not change.