Question

Let $L$ be the projection of the line $\dfrac{x-1}{2}=\dfrac{y+1}{-1}=\dfrac{z-3}{4}$ on the plane $x+2y+z=9$ then which of the following is not correct?
A. $\left( \dfrac{1}{2},\dfrac{17}{4},0 \right)$ is a point on $L$.
B. $\left( \dfrac{3}{2},\dfrac{15}{4},0 \right)$ is a point on $L$.
C. $\left( \dfrac{13}{6},\dfrac{4}{3},\dfrac{25}{6} \right)$ is a point on $L$.
D. Direction ratios on $L$ are $\left( 4,-7,10 \right)$

Answer 1

For this problem we need to first calculate the equation of the projection which is denoted by $L$ in this problem. For this we will first consider the given equation of the line and write the coordinates of a point let’s say $B$ which lies in both given line and plane by assuming $\dfrac{x-1}{2}=\dfrac{y+1}{-1}=\dfrac{z-3}{4}=\lambda$ where $\lambda$ is a constant. Here we will get the coordinate of the point $B$ in terms of $\lambda$. So, we will substitute the point $B$ in the given plane equation and calculate the value of $\lambda$ . From this we can also find the coordinates of the point $B$. Now we will write the equation of the normal to the given plane from a point on the given line. Here also we will follow the above procedure to find the coordinates of the point which lies on the plane and normal to the given line. After getting the coordinates of the two points we will use the formula $\dfrac{x-{{x}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\dfrac{y-{{y}_{1}}}{{{y}_{2}}-{{y}_{1}}}=\dfrac{z-{{z}_{1}}}{{{z}_{2}}-{{z}_{1}}}$ to find the equation of the projection. Now we have the equation of the projection $L$. We will consider each option individually and check which one is not correct.

Complete step by step solution:
The equation of the line is $\dfrac{x-1}{2}=\dfrac{y+1}{-1}=\dfrac{z-3}{4}$.
Let us assume $\dfrac{x-1}{2}=\dfrac{y+1}{-1}=\dfrac{z-3}{4}=\lambda$
Let $B$ be the any point lies on the above line, then the coordinates of the point $B$ from the above equation will be
$x=2\lambda +1$, $y=-\lambda -1$, $z=4\lambda +3$.
The point $B$ also lies on the given plane $x+2y+z=9$. So, substituting the point $B$ in the given plane equation, then we will get
$\left( 2\lambda +1 \right)+2\left( -\lambda -1 \right)+\left( 4\lambda +3 \right)=9$
Simplifying the above equation by using mathematical operations, then we will have
$\begin{aligned} & 2\lambda +1-2\lambda -2+4\lambda +3=9 \\\ & \Rightarrow 4\lambda +2=9 \\\ & \Rightarrow 4\lambda =7 \\\ & \Rightarrow \lambda =\dfrac{7}{4} \\\ \end{aligned}$
By substituting the value of $\lambda$ in the coordinates of the point $B$ then the coordinates of the point $B$ becomes as
$x=2\left( \dfrac{7}{4} \right)+1$, $y=-\dfrac{7}{4}-1$, $z=4\left( \dfrac{7}{4} \right)+3$
Simplifying the above equation, then we will have
$x=\dfrac{9}{2}$, $\dfrac{-11}{4}$, $z=10$
So, we have the point $B$ as $\left( \dfrac{9}{2},-\dfrac{11}{4},10 \right)$.
Let point $A$ be a point on the given line $\dfrac{x-1}{2}=\dfrac{y+1}{-1}=\dfrac{z-3}{4}$. We can write the coordinates of the point $A$ from the equation of the line as $A=\left( 1,-1,3 \right)$.
The vector equation of the normal to the given plane $x+2y+z=9$ will be $\vec{n}=\hat{i}+2\hat{j}+\hat{k}$
Now the equation of the normal line to the plane $x+2y+z=9$ passing through the point $A=\left( 1,-1,3 \right)$ is given by
$\dfrac{x-1}{1}=\dfrac{y+1}{2}=\dfrac{z-3}{1}$
Let us assume $\dfrac{x-1}{1}=\dfrac{y+1}{2}=\dfrac{z-3}{1}=t$, where $t$ is a constant.
Let the point ${{A}^{'}}$ be the point which lies on the above normal line and the given plane. Hence the coordinates of the point ${{A}^{'}}$ from the above equation will be
$x=t+1$, $y=2t-1$, $z=t+3$.
The point ${{A}^{'}}$ lies on the$z=\dfrac{7}{6}+3$ given plane. So, substituting the point ${{A}^{'}}$ in the given plane equation which is $x+2y+z=9$, then we will get
$\left( t+1 \right)+2\left( 2t-1 \right)+\left( t+3 \right)=9$
Simplifying the above equation by using mathematical operations, then we will have
$\begin{aligned} & t+1+4t-2+t+3=9 \\\ & \Rightarrow 6t+2=9 \\\ & \Rightarrow 6t=7 \\\ & \Rightarrow t=\dfrac{7}{6} \\\ \end{aligned}$
From the value of $t$, the coordinates of the point ${{A}^{'}}$ are modified as
$x=\dfrac{7}{6}+1$, $y=2\left( \dfrac{7}{6} \right)-1$, $z=\dfrac{7}{6}+3$
Simplifying the above equations, then we will get
$x=\dfrac{13}{6}$, $y=\dfrac{4}{3}$, $z=\dfrac{25}{6}$
Hence the point ${{A}^{'}}$ is given by ${{A}^{'}}=\left( \dfrac{13}{6},\dfrac{4}{3},\dfrac{25}{6} \right)$.
Now the equation of the projection of the line $\dfrac{x-1}{2}=\dfrac{y+1}{-1}=\dfrac{z-3}{4}$ on the plane $x+2y+z=9$ is nothing but the line which is passing through the points ${{A}^{'}}$, $B$.
Hence the equation of the line passing through the points ${{A}^{'}}\left( \dfrac{13}{6},\dfrac{4}{3},\dfrac{25}{6} \right)$, $B\left( \dfrac{9}{2},-\dfrac{11}{4},10 \right)$ from the formula $\dfrac{x-{{x}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\dfrac{y-{{y}_{1}}}{{{y}_{2}}-{{y}_{1}}}=\dfrac{z-{{z}_{1}}}{{{z}_{2}}-{{z}_{1}}}$ is given by
$\dfrac{x-\dfrac{13}{6}}{\dfrac{9}{2}-\dfrac{13}{6}}=\dfrac{y-\dfrac{4}{3}}{-\dfrac{11}{4}-\dfrac{4}{3}}=\dfrac{z-\dfrac{25}{6}}{10-\dfrac{25}{6}}$
Simplifying the above equation, then we will get
$\begin{aligned} & \dfrac{\dfrac{6x-13}{6}}{\dfrac{7}{3}}=\dfrac{\dfrac{3y-4}{3}}{-\dfrac{49}{12}}=\dfrac{\dfrac{6z-25}{6}}{\dfrac{35}{6}} \\\ & \Rightarrow \dfrac{6x-13}{14}=\dfrac{12y-16}{-49}=\dfrac{6z-25}{35} \\\ \end{aligned}$
Hence the equation of the projection which is denoted by $L$ is
$\dfrac{6x-13}{14}=\dfrac{12y-16}{-49}=\dfrac{6z-25}{35}$
Considering the first option which says that the point $\left( \dfrac{1}{2},\dfrac{17}{4},0 \right)$ lies on $L$.
Substituting the point $\left( \dfrac{1}{2},\dfrac{17}{4},0 \right)$ in the equation of $L$, then we will get
$\dfrac{6\left( \dfrac{1}{2} \right)-13}{14}=\dfrac{12\left( \dfrac{17}{4} \right)-16}{-49}=\dfrac{6\left( 0 \right)-25}{35}$
Simplifying the above equation, then we will have
$-\dfrac{5}{7}=-\dfrac{5}{7}=-\dfrac{5}{7}$
Hence the point $\left( \dfrac{1}{2},\dfrac{17}{4},0 \right)$ lies on $L$. That means option – A is not our required answer.

Considering the second option which says that the point $\left( \dfrac{3}{2},\dfrac{15}{4},0 \right)$ lies on $L$.
Substituting the point $\left( \dfrac{3}{2},\dfrac{15}{4},0 \right)$ in the equation of $L$, then we will get
$\dfrac{6\left( \dfrac{3}{2} \right)-13}{14}=\dfrac{12\left( \dfrac{15}{4} \right)-16}{-49}=\dfrac{6\left( 0 \right)-25}{35}$
Simplifying the above equation, then we will have
$-\dfrac{2}{7}\ne \dfrac{29}{49}\ne -\dfrac{5}{7}$
Hence the point $\left( \dfrac{3}{2},\dfrac{15}{4},0 \right)$ does not lies on $L$. So, the option – B is the one of consideration for the required option.
Considering the third option which says that the point $\left( \dfrac{13}{6},\dfrac{4}{3},\dfrac{25}{6} \right)$ lies on $L$. As we can see that the given point $\left( \dfrac{13}{6},\dfrac{4}{3},\dfrac{25}{6} \right)$ is same as the point ${{A}^{'}}$ which lies on the line $L$. Hence the option – C is not a required option.
Considering the fourth option which says that Direction ratios on $L$ are $\left( 4,-7,10 \right)$.
We can also write the equation of the $L$ as
$\dfrac{x-\dfrac{1}{2}}{4}=\dfrac{y-\dfrac{17}{4}}{-7}=\dfrac{z}{10}$
From the above equation the directional rations of the $L$ are $\left( 4,-7,10 \right)$. So, option – D is not our required option.

So, the correct answer is “Option B”.

Note: This type of problem is very lengthy so there are many chances to get diverted from the solution. But these problems have a unique structure such that remembering the process one can easily solve these types of problems.