Question

If the line $\dfrac{x-2}{3}=\dfrac{y+1}{2}=\dfrac{z-1}{-1}$ intersects the plane 2x+3y-z-13=0 at a point P and the plane 3x+y+4z=16 at the point Q, then PQ is equal to
(a) $2\sqrt{14}$
(b) $\sqrt{14}$
(c) $2\sqrt{7}$
(d) 14

Answer 1

Hint: First, before proceeding for this, we must take the proportionality constant for the given line as r and then we get $\dfrac{x-2}{3}=\dfrac{y+1}{2}=\dfrac{z-1}{-1}=r$. Then, we are given in the question with the condition that P lies on the plane with equation 2x+3y-z-13=0 and also given in the question with the condition that Q lies on the plane with equation 3x+y+4z=16 which gives the coordinates of P and Q. Then, by using the distance formula for the points P(-1, -3, 2) and Q(5, 1, 0), we get the distance between PQ.

Complete step-by-step answer:
In this question, we are supposed to find the length of PQ when the line $\dfrac{x-2}{3}=\dfrac{y+1}{2}=\dfrac{z-1}{-1}$ intersects the plane 2x+3y-z-13=0 at a point P and the plane 3x+y+4z=16 at the point Q.
So, we must draw a diagram of the planes with the equations 2x+3y-z-13=0 with point P and the plane 3x+y+4z=16 with point Q intersected by line as:

So, before proceeding for this, we must take the proportionality constant for the given line as r and then we get:
$\dfrac{x-2}{3}=\dfrac{y+1}{2}=\dfrac{z-1}{-1}=r$
Then, we get the coordinates of the above line as:
x=3r+2, y=2r-1 and z=-r+1
Now, let us assume for the points P and Q for the line be ${{r}_{1}}$and ${{r}_{2}}$respectively.
Then, we get the coordinates of the point P and Q as:
For P, points are $\left( 3{{r}_{1}}+2,2{{r}_{1}}-1,-{{r}_{1}}+1 \right)$
For Q, points are $\left( 3{{r}_{2}}+2,2{{r}_{2}}-1,-{{r}_{2}}+1 \right)$
Now, we are given the question with the condition that P lies on the plane with equation 2x+3y-z-13=0.
So, by substituting the coordinates of P found above in the equation of plane, we get:
$\begin{align}
  & 2\left( 3{{r}_{1}}+2 \right)+3\left( 2{{r}_{1}}-1 \right)-\left( -{{r}_{1}}+1 \right)+13=0 \\
 & \Rightarrow 6{{r}_{1}}+4+6{{r}_{1}}-3+{{r}_{1}}-1+13=0 \\
 & \Rightarrow 13{{r}_{1}}=-13 \\
 & \Rightarrow {{r}_{1}}=\dfrac{-13}{13} \\
 & \Rightarrow {{r}_{1}}=-1 \\
\end{align}$
So, we get the coordinates of point P by substituting the value of ${{r}_{1}}$as -1 in the $\left( 3{{r}_{1}}+2,2{{r}_{1}}-1,-{{r}_{1}}+1 \right)$, we get:
$\begin{align}
  & \left( 3\left( -1 \right)+2,2\left( -1 \right)-1,-\left( -1 \right)+1 \right) \\
 & \Rightarrow \left( -3+2,-2-1,1+1 \right) \\
 & \Rightarrow \left( -1,-3,2 \right) \\
\end{align}$
So, we get the coordinates of point P as (-1, -3, 2).
Similarly, we are given the question with the condition that Q lies on the plane with equation 3x+y+4z=16.
So, by substituting the coordinates of Q found above in the equation of plane, we get:
$\begin{align}
  & 3\left( 3{{r}_{2}}+2 \right)+\left( 2{{r}_{2}}-1 \right)+4\left( -{{r}_{2}}+1 \right)=16 \\
 & \Rightarrow 9{{r}_{2}}+6+2{{r}_{2}}-1-4{{r}_{2}}+4=16 \\
 & \Rightarrow 7{{r}_{2}}=7 \\
 & \Rightarrow {{r}_{2}}=\dfrac{7}{7} \\
 & \Rightarrow {{r}_{2}}=1 \\
\end{align}$
So, we get the coordinates of point P by substituting the value of ${{r}_{2}}$as 1 in the $\left( 3{{r}_{2}}+2,2{{r}_{2}}-1,-{{r}_{2}}+1 \right)$, we get:
$\begin{align}
  & \left( 3\left( 1 \right)+2,2\left( 1 \right)-1,-\left( 1 \right)+1 \right) \\
 & \Rightarrow \left( 3+2,2-1,0 \right) \\
 & \Rightarrow \left( 5,1,0 \right) \\
\end{align}$
So, we get the coordinates of point Q as (5, 1, 0).
Then, by using the distance formula for the points P(-1, -3, 2) and Q(5, 1, 0), we get the distance between PQ as:
$\begin{align}
  & PQ=\sqrt{{{\left( 5-\left( -1 \right) \right)}^{2}}+{{\left( 1-\left( -3 \right) \right)}^{2}}+{{\left( 0-2 \right)}^{2}}} \\
 & \Rightarrow PQ=\sqrt{{{\left( 5+1 \right)}^{2}}+{{\left( 1+3 \right)}^{2}}+{{\left( -2 \right)}^{2}}} \\
 & \Rightarrow PQ=\sqrt{{{\left( 6 \right)}^{2}}+{{\left( 4 \right)}^{2}}+4} \\
 & \Rightarrow PQ=\sqrt{36+16+4} \\
 & \Rightarrow PQ=\sqrt{56} \\
 & \Rightarrow PQ=2\sqrt{14} \\
\end{align}$
So, we get the distance PQ as $2\sqrt{14}$.

So, the correct answer is “Option A”.

Note: Now, to solve these type of the questions we need to know some of the basic formula for calculating the distance between the two points let A with coordinates $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$and B with coordinates $\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$. So, the distance AB is given by:
$AB=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}+{{\left( {{z}_{2}}-{{z}_{1}} \right)}^{2}}}$