Question

The image of line \[\dfrac{x-1}{3}=\dfrac{y-3}{1}=\dfrac{z-4}{-5}\] in the plane \[2x-y+z+3=0\] is
(A) \[\dfrac{x-3}{3}=\dfrac{y+5}{1}=\dfrac{z-2}{-5}\]
(B) \[\dfrac{x-3}{-3}=\dfrac{y+5}{-1}=\dfrac{z-2}{5}\]
(C) \[\dfrac{x+3}{3}=\dfrac{y-5}{1}=\dfrac{z-2}{-5}\]
(D) \[\dfrac{x+3}{-3}=\dfrac{y-5}{-1}=\dfrac{z+2}{5}\]

Answer 1

Hint: The direction ratio vector of the line OP \[\dfrac{x-1}{3}=\dfrac{y-3}{1}=\dfrac{z-4}{-5}\] and the plane QRST \[2x-y+z+3=0\] are \[3\hat{i}+1\hat{j}-5\hat{k}\] and \[2\hat{i}-1\hat{j}+1\hat{k}\] respectively. Now, take the dot product of these two vectors. We know the property that the dot product of two perpendicular vectors is zero. The equation of the line perpendicular to the plane QRST is \[\dfrac{x-1}{2}=\dfrac{y-3}{-1}=\dfrac{z-4}{1}\] . Assume \[\dfrac{x-1}{2}=\dfrac{y-3}{-1}=\dfrac{z-4}{1}=k\] . Now, get the coordinate of the general point of the perpendicular line XY and put the value of x, y, and z in the equation of the plane QRST and get the coordinate of the point A. Since point Y is the image of the point X in the plane QRST so, A is the midpoint of the line XY. Now, use the midpoint formula, \[x=\dfrac{{{x}_{1}}+{{x}_{2}}}{2}\] , \[y=\dfrac{{{y}_{1}}+{{y}_{2}}}{2}\] , \[z=\dfrac{{{z}_{1}}+{{z}_{2}}}{2}\] , and get the coordinate of point Y. The line MN is the image of the line OP. The line OP is parallel to the plane QRST. So, the line MN is also parallel to the plane QRST. Therefore, the direction ration of the line MN is the same as of OP. Now, using the direction ratios of the line MN is (3,1,-5) and the coordinate of the passing point Y, gets the equation of the line MN.

Complete step-by-step solution:
According to the question, it is given that we have,
The equation of the line = \[\dfrac{x-1}{3}=\dfrac{y-3}{1}=\dfrac{z-4}{-5}\] ……………………………………….(1)
The direction ratio vector of the line = \[3\hat{i}+1\hat{j}-5\hat{k}\] ………………………………………(2)
The equation of the plane = \[2x-y+z+3=0\] …………………………………………….(3)
The direction ratio vector of the plane = \[2\hat{i}-1\hat{j}+1\hat{k}\] …………………………………………..(4)
Now, first of all, let us check if the line and plane are parallel or perpendicular.
From equation (3) and equation (4), we have the direction ratio vector of the plane and the line.
Now, on applying the dot product to both the direction ratio vector of the plane and the line, we get
\[\begin{align}
  & =\left( 3\hat{i}+1\hat{j}-5\hat{k} \right).\left( 2\hat{i}-1\hat{j}+1\hat{k} \right) \\
 & =3\left( 2 \right)+1\left( -1 \right)+\left( -5 \right)1 \\
 & =6-1-5 \\
 & =6-6 \\
 & =0 \\
\end{align}\]
So, \[\left( 3\hat{i}+1\hat{j}-5\hat{k} \right).\left( 2\hat{i}-1\hat{j}+1\hat{k} \right)=0\] ………………………………….…(5)
We know the property that the dot product of two vectors is zero if both the vectors are perpendicular ………………………………………(6)
From equation (5), we have the dot product of both vectors equal to zero.
Now, using the property in equation (6), we can say that the direction ratio vector of the plane and the line is perpendicular to each other.
We also know that the direction ratio vector of a plane is measured perpendicular to the plane and the direction ratio vector of the line is measured along the line.
It means that in this case the line and the plane are parallel to each other.

The direction ratio of the line XY is the same as the direction-ratios of the plane QRST.
From equation (4), we have the direction ratio vector of the plane QRST.
The equation of the line passing through the point (1,3,4) and the direction ratio (2-,-1,1) is
\[\dfrac{x-1}{2}=\dfrac{y-3}{-1}=\dfrac{z-4}{1}\] …………………………………………….(7)
Let us assume, \[\dfrac{x-1}{2}=\dfrac{y-3}{-1}=\dfrac{z-4}{1}=k\] ………………………….(8)
On solving equation (8), we get
\[x=2k+1\] …………………………………..….(9)
\[y=-k+3\] ……………………………………(10)
\[z=4+k\] ……………………………………….(11)
From equation (9), equation (10), and equation (11), we have the general coordinate of the point on the line XY.
Point A is the intersection of the plane QRST and the line XY.
Now, from equation (3), equation (9), equation (10), and equation (11), we get
\[\begin{align}
  & \Rightarrow 2\left( 2k+1 \right)-1\left( -k+3 \right)+\left( 4+k \right)+3=0 \\
 & \Rightarrow 4k+2+k-3+4+k+3=0 \\
 & \Rightarrow 6k+6=0 \\
 & \Rightarrow k=-1 \\
\end{align}\]
Now, putting \[k=-1\] in equation (9), equation (10), and equation (11), we get
\[x=2\left( -1 \right)+1=-2+1=-1\] …………………………….(12)
\[y=-\left( -1 \right)+3=1+3=4\] ……………………………………(13)
\[z=4+\left( -1 \right)=4-1=3\] ……………………………………….(14)
The coordinate of the point A is (-1,4,3) ……………………………………(15)
From the diagram, we have the coordinate of point X (1,3,4) …………………………..(16)
Point Y is the image of point X and A is the midpoint of the line XY.
We know the midpoint formula, \[x=\dfrac{{{x}_{1}}+{{x}_{2}}}{2}\] , \[y=\dfrac{{{y}_{1}}+{{y}_{2}}}{2}\], and \[z=\dfrac{{{z}_{1}}+{{z}_{2}}}{2}\] ………………………………….(17)
Now, from equation (15), equation (16), and equation (17), we get
\[-1=\dfrac{1+{{x}_{2}}}{2}\] …………………… ….(18)
\[4=\dfrac{3+{{y}_{2}}}{2}\] …………………………(19)
\[3=\dfrac{4+{{z}_{2}}}{2}\] …………………………(20)
Now, solving equation (18), we get
\[\begin{align}
  & -1=\dfrac{1+{{x}_{2}}}{2} \\
 & \Rightarrow -2=1+{{x}_{2}} \\
 & \Rightarrow -3={{x}_{2}} \\
\end{align}\]
Similarly, solving equation (19), we get
\[\begin{align}
  & 4=\dfrac{3+{{y}_{2}}}{2} \\
 & \Rightarrow 8=3+{{y}_{2}} \\
 & \Rightarrow 5={{y}_{2}} \\
\end{align}\]
Similarly, solving equation (20), we get
\[\begin{align}
  & 3=\dfrac{4+{{z}_{2}}}{2} \\
 & \Rightarrow 6=4+{{z}_{2}} \\
 & \Rightarrow 2={{z}_{2}} \\
\end{align}\]
The coordinate of the point Y is (-3, 5, 2).
The line OP is parallel to the plane QRST. So, the line MN is also parallel to the plane QRST. Therefore, the line OP is also parallel to MN.
Since the line MN is parallel to OP so, the direction ratio of both lines will be the same.
From equation (2), we have the direction ratios of the line OP.
So, the direction ratio of the line MN is (3, 1, -5).
Now, the equation of line MN having direction ratio (3, 1, -5) and the coordinate of the passing point Y (-3, 5, 2) is
\[\begin{align}
  & \Rightarrow \dfrac{x-\left( -3 \right)}{3}=\dfrac{y-5}{1}=\dfrac{z-2}{-5} \\
 & \Rightarrow \dfrac{x+3}{3}=\dfrac{y-5}{1}=\dfrac{z-2}{-5} \\
\end{align}\]
Therefore, the equation of the line MN is \[\dfrac{x+3}{3}=\dfrac{y-5}{1}=\dfrac{z-2}{-5}\] .
Hence, the correct option is (C).

Note: In this question, one might get confused because we don’t have any direct formula for the image of a line in a plane. So, first of all, try to find out if the line is parallel or perpendicular by taking the dot product of the direction-ratios of the line and the plane. Then, solve it further.